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Commit 456daa9c by JonahStanley
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Deleted Files and annotated tests

parent 4b59a85d
Showing with 21 additions and 3847 deletions
cms/djangoapps/contentstore/tests/test_contentstore.py
......@@ -164,6 +164,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
def test_edit_unit_toy(self):
self.check_edit_unit('toy')
#FIX
def test_edit_unit_full(self):
self.check_edit_unit('full')
......@@ -299,6 +300,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
num_drafts = self._get_draft_counts(course)
self.assertEqual(num_drafts, 1)
#FIX
def test_import_textbook_as_content_element(self):
module_store = modulestore('direct')
import_from_xml(module_store, 'common/test/data/', ['full'])
......@@ -307,6 +309,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
self.assertGreater(len(course.textbooks), 0)
#FIX
def test_static_tab_reordering(self):
module_store = modulestore('direct')
import_from_xml(module_store, 'common/test/data/', ['full'])
......@@ -331,6 +334,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
self.assertEqual(reverse_tabs, course_tabs)
#FIX
def test_import_polls(self):
module_store = modulestore('direct')
import_from_xml(module_store, 'common/test/data/', ['full'])
......@@ -342,6 +346,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
# check that there's actually content in the 'question' field
self.assertGreater(len(items[0].question), 0)
#FIX
def test_xlint_fails(self):
err_cnt = perform_xlint('common/test/data', ['full'])
self.assertGreater(err_cnt, 0)
......@@ -368,6 +373,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
resp = self.client.get(url)
self.assertEqual(resp.status_code, 200)
#FIX
def test_delete(self):
direct_store = modulestore('direct')
import_from_xml(direct_store, 'common/test/data/', ['full'])
......@@ -399,6 +405,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
# make sure the parent no longer points to the child object which was deleted
self.assertFalse(sequential.location.url() in chapter.children)
#FIX
def test_about_overrides(self):
'''
This test case verifies that a course can use specialized override for about data, e.g. /about/Fall_2012/effort.html
......@@ -413,6 +420,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
effort = module_store.get_item(Location(['i4x', 'edX', 'full', 'about', 'end_date', None]))
self.assertEqual(effort.data, 'TBD')
#FIX
def test_remove_hide_progress_tab(self):
module_store = modulestore('direct')
import_from_xml(module_store, 'common/test/data/', ['full'])
......@@ -421,6 +429,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
course = module_store.get_item(source_location)
self.assertFalse(course.hide_progress_tab)
#FIX
def test_asset_import(self):
'''
This test validates that an image asset is imported and a thumbnail was generated for a .gif
......@@ -472,6 +481,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
#
# self.assertIsNotNone(thumbnail)
#FIX
def test_asset_delete_and_restore(self):
'''
This test will exercise the soft delete/restore functionality of the assets
......@@ -528,6 +538,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
thumbnail = content_store.find(thumbnail_location, throw_on_not_found=False)
self.assertIsNotNone(thumbnail)
#FIX
def test_empty_trashcan(self):
'''
This test will exercise the empting of the asset trashcan
......@@ -572,6 +583,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
all_thumbnails = trash_store.get_all_content_thumbnails_for_course(course_location)
self.assertEqual(len(all_thumbnails), 0)
#FIX
def test_clone_course(self):
course_data = {
......@@ -635,6 +647,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
resp = self.client.get('http://localhost:8001/c4x/CDX/123123/asset/&images_circuits_Lab7Solution2.png')
self.assertEqual(resp.status_code, 400)
#FIX
def test_delete_course(self):
"""
This test will import a course, make a draft item, and delete it. This will also assert that the
......@@ -683,6 +696,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
filesystem = OSFS(root_dir / ('test_export/' + dirname))
self.assertTrue(filesystem.exists(item.location.name + filename_suffix))
#FIX
def test_export_course(self):
module_store = modulestore('direct')
draft_store = modulestore('draft')
......@@ -802,6 +816,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
shutil.rmtree(root_dir)
#FIX
def test_course_handouts_rewrites(self):
module_store = modulestore('direct')
......@@ -820,6 +835,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
# note, we know the link it should be because that's what in the 'full' course in the test data
self.assertContains(resp, '/c4x/edX/full/asset/handouts_schematic_tutorial.pdf')
#FIX
def test_prefetch_children(self):
module_store = modulestore('direct')
import_from_xml(module_store, 'common/test/data/', ['full'])
......@@ -842,6 +858,7 @@ class ContentStoreToyCourseTest(ModuleStoreTestCase):
self.assertFalse(Location(['i4x', 'edX', 'full', 'vertical', 'vertical_58', None])
in course.system.module_data)
#FIX
def test_export_course_with_unknown_metadata(self):
module_store = modulestore('direct')
content_store = contentstore()
......@@ -1200,6 +1217,7 @@ class ContentStoreTest(ModuleStoreTestCase):
# make sure we found the item (e.g. it didn't error while loading)
self.assertTrue(did_load_item)
#FIX
def test_forum_id_generation(self):
module_store = modulestore('direct')
import_from_xml(module_store, 'common/test/data/', ['full'])
......@@ -1214,6 +1232,7 @@ class ContentStoreTest(ModuleStoreTestCase):
self.assertNotEquals(new_discussion_item.discussion_id, '$$GUID$$')
#FIX
def test_update_modulestore_signal_did_fire(self):
module_store = modulestore('direct')
import_from_xml(module_store, 'common/test/data/', ['full'])
......@@ -1239,6 +1258,7 @@ class ContentStoreTest(ModuleStoreTestCase):
self.assertTrue(self.got_signal)
#FIX
def test_metadata_inheritance(self):
module_store = modulestore('direct')
import_from_xml(module_store, 'common/test/data/', ['full'])
......
cms/djangoapps/contentstore/tests/test_course_settings.py
......@@ -295,6 +295,7 @@ class CourseMetadataEditingTest(CourseTestCase):
"""
Tests for CourseMetadata.
"""
#FIX
def setUp(self):
CourseTestCase.setUp(self)
# add in the full class too
......
common/test/data/full/README.md deleted 100644 → 0
This is a realistic course, with many different module types and a lot of structure. It is based on 6.002x.
common/test/data/full/about/description.html deleted 100644 → 0
6.002x (Circuits and Electronics) is designed to serve as a first course in an undergraduate electrical engineering (EE), or electrical engineering and computer science (EECS) curriculum. At MIT, 6.002 is in the core of department subjects required for all undergraduates in EECS.
The course introduces engineering in the context of the lumped circuit abstraction. Topics covered include: resistive elements and networks; independent and dependent sources; switches and MOS transistors; digital abstraction; amplifiers; energy storage elements; dynamics of first- and second-order networks; design in the time and frequency domains; and analog and digital circuits and applications. Design and lab exercises are also significant components of the course. You should expect to spend approximately 10 hours per week on the course.
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common/test/data/full/about/effort.html deleted 100644 → 0
12 hours
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common/test/data/full/about/faq.html deleted 100644 → 0
<ul>
<li>What is the format of the class?
<p>The course will consist of 24 lectures, each lasting 50 minutes. There will be regular assignments consisting of map tests and short essays.</p>
</li>
<li>Are there any prerequisites?
<p>No - anyone and everyone is welcome to take this course.</p>
</li>
<li>What textbook should I buy?
<p>Although the lectures are designed to be self-contained, we recommend (but do not require) that students refer to the book Worlds Together, Worlds Apart: A History of the World: From 1000 CE to the Present (W W Norton, 3rd edition) &mdash; Volume II, which was written specifically for this course.</p>
</li>
<li>Does Harvard award credentials or reports regarding my work in this course?
<p>Princeton does not award credentials or issue reports for student work in this course. However, Coursera could maintain a record of your score on the assessments and, with your permission, verify that score for authorized parties.</p>
</li>
</ul>
common/test/data/full/about/more_info.html deleted 100644 → 0
<section class="who-should-take">
<h3>Who should take this?</h3>
<p>If you're one of the many who have a unquenched interest in the worlds history, you'll love this course.</p>
</section>
<section class="who-shouldnt-take">
<h3>Who shouldn't take this?</h3>
<p>No one. Anyone and everyone is welcome to take this course.</p>
</section>
\ No newline at end of file
common/test/data/full/about/requirements.html deleted 100644 → 0
<p>In order to succeed in this course, you must have taken an AP level physics course in electricity and magnetism. You must know basic calculus and linear algebra and have some background in differential equations. Since more advanced mathematics will not show up until the second half of the course, the first half of the course will include an optional remedial differential equations component for those who need it.</p>
<p>The course web site was developed and tested primarily with Google Chrome. We support current versions of Mozilla Firefox as well. The video player is designed to work with Flash. While we provide a partial non-Flash fallback for the video, as well as partial support for Internet Explorer, other browsers, and tablets, portions of the functionality will be unavailable.</p>
\ No newline at end of file
common/test/data/full/about/syllabus.html deleted 100644 → 0
<ul>
<li><strong>Week 1:</strong> What is World History?</li>
<li><strong>Week 2:</strong> Peoples, Plagues and Plunders</li>
<li><strong>Week 3:</strong> Warfare and Motion</li>
<li><strong>Week 4:</strong> Conquests</li>
<li><strong>Week 5:</strong> The Beginnings of Globalization in the Atlantic Worlds</li>
<li><strong>Week 6:</strong> The Beginnings of Globalization in the Indian Ocean Worlds</li>
<li><strong>Week 7:</strong> The Worlds that Merchants Made</li>
<li><strong>Week 8:</strong> The Seventeenth-Century Crisis</li>
<li><strong>Week 9:</strong> Empire and Enlightenment</li>
<li><strong>Week 10:</strong> The Wealth of Nations</li>
<li><strong>Week 11:</strong> The World in Revolution</li>
<li><strong>Week 12:</strong> States and Nations</li>
<li><strong>Week 13:</strong> Global Frontiers</li>
<li><strong>Week 14:</strong> Empires and Nations</li>
<li><strong>Week 15:</strong> Back to the Future</li>
</ul>
common/test/data/full/about/textbook.html deleted 100644 → 0
<p>The course uses the textbook Foundations of Analog and Digital Electronic Circuits, by Anant Agarwal and Jeffrey H. Lang. Morgan Kaufmann Publishers, Elsevier, July 2005. While recommended, the book is not required: relevant sections will be provided electronically as part of the online course for personal use in connection with this course only. The copyright for the book is owned by Elsevier. The book can be purchased on <a href="http://www.amazon.com/exec/obidos/ASIN/1558607358/ref=nosim/mitopencourse-20">Amazon</a>.</p>
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common/test/data/full/about/video.html deleted 100644 → 0
<iframe width="560" height="315" src="http://www.youtube.com/embed/p2Q6BrNhdh8" frameborder="0" allowfullscreen></iframe>
\ No newline at end of file
common/test/data/full/chapter/Overview.xml deleted 100644 → 0
<sequential>
<video url_name="welcome"/>
<sequential filename="System_Usage_Sequence" slug="System_Usage_Sequence" format="Lecture Sequence" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" name="System Usage Sequence"/>
<vertical slug="Lab0_Using_the_tools" format="Lab" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" name="Lab0: Using the tools">
<html slug="html_19" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never"> See the <a href="/section/labintro"> Lab Introduction </a> or <a href="/static/handouts/schematic_tutorial.pdf">Interactive Lab Usage Handout </a> for information on how to do the lab… </html>
<html slug="html_5555" filename="html_5555"/>
<problem filename="Lab_0_Using_the_Tools" slug="Lab_0_Using_the_Tools" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="false" name="Lab 0: Using the Tools"/>
</vertical>
<problem filename="Circuit_Sandbox" slug="Circuit_Sandbox" format="Lab" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="false" name="Circuit Sandbox"/>
</sequential>
common/test/data/full/chapter/Week_1.xml deleted 100644 → 0
<sequential>
<sequential filename="Administrivia_and_Circuit_Elements" slug="Administrivia_and_Circuit_Elements" format="Lecture Sequence" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" name="Administrivia and Circuit Elements"/>
<vertical slug="Basic_Circuit_Analysis" format="Homework" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="closed" rerandomize="per_student" due="March 18" graded="true" name="Basic Circuit Analysis">
<problem filename="H1P1_Energy" slug="H1P1_Energy" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="closed" rerandomize="per_student" due="March 18" graded="true" name="H1P1: Energy"/>
<problem filename="H1P2_Duality" slug="H1P2_Duality" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="closed" rerandomize="per_student" due="March 18" graded="true" name="H1P2: Duality"/>
<problem filename="H1P3_Poor_Workmanship" slug="H1P3_Poor_Workmanship" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="closed" rerandomize="per_student" due="March 18" graded="true" name="H1P3: Poor Workmanship"/>
</vertical>
<problem filename="Resistor_Divider" slug="Resistor_Divider" format="Lab" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" due="March 18" graded="true" name="Resistor Divider"/>
<html filename="Week_1_Tutorials" slug="Week_1_Tutorials" format="Tutorial Index" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" name="Week 1 Tutorials"/>
</sequential>
common/test/data/full/course.xml deleted 100644 → 0
<course filename="6.002_Spring_2012" slug="6.002_Spring_2012" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" name="6.002 Spring 2012" start="2015-07-17T12:00" course="full" org="edX" show_timezone="true" advanced_modules="[&quot;videoalpha&quot;]"/>
common/test/data/full/course/6.002_Spring_2012.xml deleted 100644 → 0
<course>
<textbook title="Textbook" book_url="https://s3.amazonaws.com/edx-textbooks/guttag_computation_v3/"/>
<chapter filename="Overview" slug="Overview" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" name="Overview"/>
<chapter filename="Week_1" slug="Week_1" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" name="Week 1"/>
<chapter slug="Midterm_Exam" graceperiod="1 day 12 hours 59 minutes 59 seconds" showanswer="attempted" rerandomize="never" name="Midterm Exam">
<sequential slug="Midterm_Exam_1121" format="Midterm" graceperiod="0 day 0 hours 5 minutes 0 seconds" showanswer="attempted" rerandomize="per_student" due="April 30, 12:00" graded="true" name="Midterm Exam">
<vertical slug="vertical_1122" graceperiod="0 day 0 hours 5 minutes 0 seconds" showanswer="attempted" rerandomize="per_student" due="April 30, 12:00" graded="true">
<html filename="Midterm_Exam_1123" slug="Midterm_Exam_1123" graceperiod="0 day 0 hours 5 minutes 0 seconds" rerandomize="per_student" due="April 30, 12:00" graded="true" name="Midterm Exam"/>
</vertical>
<vertical filename="vertical_98" slug="vertical_1124" graceperiod="0 day 0 hours 5 minutes 0 seconds" showanswer="attempted" rerandomize="per_student" due="April 30, 12:00" graded="true"/>
</sequential>
</chapter>
</course>
common/test/data/full/course_settings.json deleted 100644 → 0
{
"GRADER" : [
{
"type" : "Homework",
"min_count" : 12,
"drop_count" : 2,
"short_label" : "HW",
"weight" : 0.15
},
{
"type" : "Lab",
"min_count" : 12,
"drop_count" : 2,
"category" : "Labs",
"weight" : 0.15
},
{
"type" : "Midterm",
"name" : "Midterm Exam",
"short_label" : "Midterm",
"weight" : 0.3
},
{
"type" : "Final",
"name" : "Final Exam",
"short_label" : "Final",
"weight" : 0.4
}
],
"GRADE_CUTOFFS" : {
"A" : 0.87,
"B" : 0.7,
"C" : 0.6
}
}
common/test/data/full/custom_tags/book deleted 100644 → 0
More information given in… <a href="/book/${page}">the text</a>.
common/test/data/full/custom_tags/discuss deleted 100644 → 0
<a href='https://6002x.mitx.mit.edu/discussion/questions/scope:all/sort:activity-desc/tags:${tag}/page:1/'> Discussion: ${tag}… </a>
\ No newline at end of file
common/test/data/full/custom_tags/image deleted 100644 → 0
<img src="http://pmitros.csail.mit.edu/tutorial/images/${file}" />
common/test/data/full/custom_tags/slides deleted 100644 → 0
Lecture Slides Handout [<a href="">Clean… </a>][<a href="">Annotated…</a>]
common/test/data/full/custom_tags/videodev deleted 100644 → 0
<video width="560" height="340" controls>
<source src="http://pmitros.csail.mit.edu/tutorial/raw_links/${name}.webm">
</video>
common/test/data/full/grading_policy.js deleted 100644 → 0
{
"GRADER" : [
{
"type" : "Homework",
"min_count" : 3,
"drop_count" : 1,
"short_label" : "HW",
"weight" : 0.5
},
{
"type" : "Final",
"name" : "Final Question",
"short_label" : "Final",
"weight" : 0.5
}
],
"GRADE_CUTOFFS" : {
"A" : 0.8,
"B" : 0.7,
"C" : 0.44
}
}
common/test/data/full/html/Final_Exam_2010.html deleted 100644 → 0
<html>
<body>
<center>
Final Exam
</center>
<p>This exam covers weeks 1-13. You may use your calculator, your notes, your book, the internet, or any other auxiliary materials that you think can help you. However, you are not allowed to communicate with any person on any topic associated with this course while you are taking this exam.</p>
<p>
Once you click on the next tab, you will have 24 hours to complete the examination. For each problem you will be allowed exactly three submissions. Your
answers to that problem will be checked after each submission, so you
can fix mistakes you may have made, within the three-submission limit.
Your most recently checked answer is the answer that will contribute to your grade on the exam.
</p>
<p>
Each answer box on the exam contributes equally to your grade on the exam, regardless of how they are grouped as problems.
</p>
</body>
</html>
common/test/data/full/html/FirstOrderSystemHint.xml deleted 100644 → 0
Hint…
<br/><br/>
Remember that the time evolution of any variable \(x(t)\) governed by
a first-order system with a time-constant \(\tau\) for a time \(t) between an initial
value \(x(t_0)\) and a final value \(x(\infty)\) is the following:
<br/><br/>
\(x(t) = x(\infty) + (x(t_0) - x(\infty)) e^(-(t-t_0)/\tau)\)
<br/><br/>
common/test/data/full/html/Lab2A.html deleted 100644 → 0
<script type="text/javascript">
$(document).ready(function() {
$("#r1_slider").slider({
value: 1, min: 1, max: 10, step: 1, slide: schematic.component_slider,
schematic: "ctrls", component: "R1", property: "r", analysis: "dc",
})
$("#r2_slider").slider({
value: 1, min: 1, max: 10, step: 1, slide: schematic.component_slider,
schematic: "ctrls", component: "R2", property: "r", analysis: "dc",
})
$("#r3_slider").slider({
value: 1, min: 1, max: 10, step: 1, slide: schematic.component_slider,
schematic: "ctrls", component: "R3", property: "r", analysis: "dc",
})
$("#r4_slider").slider({
value: 1, min: 1, max: 10, step: 1, slide: schematic.component_slider,
schematic: "ctrls", component: "R4", property: "r", analysis: "dc",
})
$("#slider").slider(); });
</script>
<b>Lab 2A: Superposition Experiment</b>
<br><br><i>Note: This part of the lab is just to develop your intuition about
superposition. There are no responses that need to be checked.</i>
<br/><br/>Circuits with multiple sources can be hard to analyze as-is. For example, what is the voltage
between the two terminals on the right of Figure 1?
<center>
<input width="425" type="hidden" height="150" id="schematic1" parts="" analyses="" class="schematic ctrls" name="test2" value="[[&quot;w&quot;,[160,64,184,64]],[&quot;w&quot;,[160,16,184,16]],[&quot;w&quot;,[64,16,112,16]],[&quot;w&quot;,[112,64,88,64]],[&quot;w&quot;,[64,64,88,64]],[&quot;g&quot;,[88,64,0],{},[&quot;0&quot;]],[&quot;w&quot;,[112,64,160,64]],[&quot;w&quot;,[16,64,64,64]],[&quot;r&quot;,[160,16,0],{&quot;name&quot;:&quot;R4&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;1&quot;,&quot;0&quot;]],[&quot;r&quot;,[160,16,1],{&quot;name&quot;:&quot;R3&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;1&quot;,&quot;2&quot;]],[&quot;i&quot;,[112,64,6],{&quot;name&quot;:&quot;&quot;,&quot;value&quot;:&quot;6A&quot;},[&quot;0&quot;,&quot;2&quot;]],[&quot;r&quot;,[64,16,0],{&quot;name&quot;:&quot;R2&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;2&quot;,&quot;0&quot;]],[&quot;r&quot;,[64,16,1],{&quot;name&quot;:&quot;R1&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;2&quot;,&quot;3&quot;]],[&quot;v&quot;,[16,16,0],{&quot;name&quot;:&quot;&quot;,&quot;value&quot;:&quot;8V&quot;},[&quot;3&quot;,&quot;0&quot;]],[&quot;view&quot;,-24,0,2]]"/>
Figure 1. Example multi-source circuit
</center>
<br/><br/>We can use superposition to make the analysis much easier.
The circuit in Figure 1 can be decomposed into two separate
subcircuits: one involving only the voltage source and one involving only the
current source. We'll analyze each circuit separately and combine the
results using superposition. Recall that to decompose a circuit for
analysis, we'll pick each source in turn and set all the other sources
to zero (i.e., voltage sources become short circuits and current
sources become open circuits). The circuit above has two sources, so
the decomposition produces two subcircuits, as shown in Figure 2.
<center>
<table><tr><td>
<input style="display:inline;" width="425" type="hidden" height="150" id="schematic2" parts="" analyses="" class="schematic ctrls" name="test2" value="[[&quot;w&quot;,[160,64,184,64]],[&quot;w&quot;,[160,16,184,16]],[&quot;w&quot;,[64,16,112,16]],[&quot;w&quot;,[112,64,88,64]],[&quot;w&quot;,[64,64,88,64]],[&quot;g&quot;,[88,64,0],{},[&quot;0&quot;]],[&quot;w&quot;,[112,64,160,64]],[&quot;w&quot;,[16,64,64,64]],[&quot;r&quot;,[160,16,0],{&quot;name&quot;:&quot;R4&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;1&quot;,&quot;0&quot;]],[&quot;r&quot;,[160,16,1],{&quot;name&quot;:&quot;R3&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;1&quot;,&quot;2&quot;]],[&quot;r&quot;,[64,16,0],{&quot;name&quot;:&quot;R2&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;2&quot;,&quot;0&quot;]],[&quot;r&quot;,[64,16,1],{&quot;name&quot;:&quot;R1&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;2&quot;,&quot;3&quot;]],[&quot;v&quot;,[16,16,0],{&quot;name&quot;:&quot;&quot;,&quot;value&quot;:&quot;8V&quot;},[&quot;3&quot;,&quot;0&quot;]],[&quot;view&quot;,-24,0,2]]"/>
(a) Subcircuit for analyzing contribution of voltage source
</td><td>
<input width="425" type="hidden" height="150" id="schematic3" parts="" analyses="" class="schematic ctrls" name="test2" value="[[&quot;w&quot;,[16,16,16,64]],[&quot;w&quot;,[160,64,184,64]],[&quot;w&quot;,[160,16,184,16]],[&quot;w&quot;,[64,16,112,16]],[&quot;w&quot;,[112,64,88,64]],[&quot;w&quot;,[64,64,88,64]],[&quot;g&quot;,[88,64,0],{},[&quot;0&quot;]],[&quot;w&quot;,[112,64,160,64]],[&quot;w&quot;,[16,64,64,64]],[&quot;r&quot;,[160,16,0],{&quot;name&quot;:&quot;R4&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;1&quot;,&quot;0&quot;]],[&quot;r&quot;,[160,16,1],{&quot;name&quot;:&quot;R3&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;1&quot;,&quot;2&quot;]],[&quot;i&quot;,[112,64,6],{&quot;name&quot;:&quot;&quot;,&quot;value&quot;:&quot;6A&quot;},[&quot;0&quot;,&quot;2&quot;]],[&quot;r&quot;,[64,16,0],{&quot;name&quot;:&quot;R2&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;2&quot;,&quot;0&quot;]],[&quot;r&quot;,[64,16,1],{&quot;name&quot;:&quot;R1&quot;,&quot;r&quot;:&quot;1&quot;},[&quot;2&quot;,&quot;3&quot;]],[&quot;view&quot;,-24,0,2]]"/>
(b) Subcircuit for analyzing contribution of current source
</td></tr></table>
<br>Figure 2. Decomposition of Figure 1 into subcircuits
</center>
<br/>Let's use the DC analysis capability of the schematic tool to see superposition
in action. The sliders below control the resistances of R1, R2, R3 and R4 in all
the diagrams. As you move the sliders, the schematic tool will adjust the appropriate
resistance, perform a DC analysis and display the node voltages on the diagrams. Here's
what you want to observe as you play with the sliders:
<ul style="margin-left:2em;margin-top:1em;margin-right:2em;margin-bottom:1em;">
<i>The voltage for a node in Figure 1 is the sum of the voltages for
that node in Figures 2(a) and 2(b), just as predicted by
superposition. (Note that due to round-off in the display of the
voltages, the sum of the displayed voltages in Figure 2 may only be within
.01 of the voltages displayed in Figure 1.)</i>
</ul>
<br>
<center>
<table><tr valign="top">
<td>
<table>
<tr valign="top">
<td>R1</td>
<td>
<div id="r1_slider" style="width:200px; height:10px; margin-left:15px"></div>
</td>
</tr>
<tr valign="top">
<td>R2</td>
<td>
<div id="r2_slider" style="width:200px; height:10px; margin-left:15px; margin-top:10px;"></div>
</td>
</tr>
<tr valign="top">
<td>R3</td>
<td>
<div id="r3_slider" style="width:200px; height:10px; margin-left:15px; margin-top:10px;"></div>
</td>
</tr>
<tr valign="top">
<td>R4</td>
<td>
<div id="r4_slider" style="width:200px; height:10px; margin-left:15px; margin-top:10px;"></div>
</td>
</tr>
</table>
</td></tr></table>
</center>
common/test/data/full/html/Midterm.html deleted 100644 → 0
<center>
Midterm Exam
</center>
<p>
This is a twenty four hour examination. You can start the exam when it is
convenient for you, but you must complete this examination by 12:00pm (noon)
GMT on April 30th. Please look up what time this is in your local time zone.
</p>
<p>
When you open the next page, you will have started the examination.
You do not need to start now: you will not be timed until you open the
next page. Once you have opened the next page page you must complete
the exam and make your final submission within twenty four hours of starting
the exam.
</p>
<p>
You may use any notes, computational, or auxiliary materials that you
think can help you. However, you may not communicate with any person
about this examination while working on it. Furthermore, you may not
communicate about the exam until the exam has been closed for everyone.
</p>
<p>
For each problem you will be allowed exactly three submissions. Your
answers to that problem will be checked after each submission, so you
can fix mistakes you may have made, within the three-submission limit.
</p>
<p>
If you want to go back and study some more before starting this
exam you can do so. Good Luck!
</p>
common/test/data/full/html/Midterm_Exam_1080.html deleted 100644 → 0
<html>
<body>
<center>
Midterm Exam
</center>
<p>
This is a twenty four hour examination. You can start the exam when it is
convenient for you, but you must complete this examination by 12:00pm (noon)
GMT on April 30th. Please look up what time this is in your local time zone.
</p>
<p>
When you open the next page, you will have started the examination.
You do not need to start now: you will not be timed until you open the
next page. Once you have opened the next page page you must complete
the exam and make your final submission within twenty four hours of starting
the exam.
</p>
<p>
You may use any notes, computational, or auxiliary materials that you
think can help you. However, you may not communicate with any person
about this examination while working on it. Furthermore, you may not
communicate about the exam until the exam has been closed for everyone.
</p>
<p>
For each problem you will be allowed exactly three submissions. Your
answers to that problem will be checked after each submission, so you
can fix mistakes you may have made, within the three-submission limit.
</p>
<p>
If you want to go back and study some more before starting this
exam you can do so. Good Luck!
</p>
</body>
</html>
common/test/data/full/html/Midterm_Exam_1123.html deleted 100644 → 0
<html>
<body>
<center>
Midterm Exam
</center>
<p>
This is a twenty four hour examination. You can start the exam when it is
convenient for you, but you must complete this examination by 12:00pm (noon)
GMT on April 30th. Please look up what time this is in your local time zone.
</p>
<p>
When you open the next page, you will have started the examination.
You do not need to start now: you will not be timed until you open the
next page. Once you have opened the next page page you must complete
the exam and make your final submission within twenty four hours of starting
the exam.
</p>
<p>
You may use any notes, computational, or auxiliary materials that you
think can help you. However, you may not communicate with any person
about this examination while working on it. Furthermore, you may not
communicate about the exam until the exam has been closed for everyone.
</p>
<p>
For each problem you will be allowed exactly three submissions. Your
answers to that problem will be checked after each submission, so you
can fix mistakes you may have made, within the three-submission limit.
</p>
<p>
If you want to go back and study some more before starting this
exam you can do so. Good Luck!
</p>
</body>
</html>
common/test/data/full/html/Week_13_Tutorials.html deleted 100644 → 0
<html>
<body>
<h1>Week 13 Tutorials</h1>
<section class="tutorials">
<h2> Basic Tutorials </h2>
<ul>
<li><a href="/section/wk13_solder">Soldering</a> &mdash; Steve
Finberg, one of the pioneers in from Draper Lab, talks about
soldering. </li>
</ul>
<h2> Bonus Tutorials </h2>
<ul>
<li><a href="/section/wk13_FreqResp">Frequency Response
Curves</a> &mdash; We explain several techniques for understanding
and approximating Bode plots. </li>
</ul>
</section>
</body>
</html>
common/test/data/full/html/Week_1_Tutorials.html deleted 100644 → 0
<html>
<body>
<h1>Week 1 Tutorials</h1>
<section class="tutorials">
<h2> Introduction </h2>
<ul>
<li><a href="/section/intro"> Welcome! </a> - We introduce
ourselves and explain the tutorial format. </li>
</ul>
<h2>Basic Tutorials </h2>
<ul>
<li><a href="/section/circuit_abstraction">The Circuit
Abstraction</a> - We look at a lightbulb, and see that abstracting
away the circuit from the geometry of the wiring has no visible
effect. </li>
<li><a href="/section/a_real_circuit/">Lightbulb Circuit</a> - We
look at the voltage across a lightbulb with a multimeter, and
confirm the device is symmetric. In the process, we see reference
directions for currents, voltages, and polarity for powers. </li>
<li><a href="/section/parallel_resistors">Parallel Resistors</a> -
An explanation of equivalent circuits in the context of parallel
resistors.</li>
<li><a href="/section/series_and_parallel">Combination of Series
and Parallel</a> - Slightly more complex networks of resistors
simplified with equivalent networks.</li>
<li><a href="/section/3r_nodal">Nodal Analysis</a> - A simple
example of nodal analysis.</li>
<li><a href="/section/floating_voltage_3r">Floating Voltage</a> -
Nodal analysis with a floating source.</li>
<li><a href="/section/combination_rules">Combination Rules</a> -
We show how to apply combination rules to solving simple
circuits.</li>
</ul>
<h2>Worked Problems </h2>
<p> For the first week, we work through <a href="http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/assignments/hw1.pdf"> the first problem set from 2007</a></p>
<ul>
<li><a href="/section/exercise_1_1">OCW Exercise 1-1</a>: Find voltages in a simple series and parallel circuit, and show power is conserved</li>
<li><a href="/section/ex_1_2">OCW Exercise 1-2</a> - Synthesize a network of 3/5k and 5/3k from at most four 1k resistors</li>
<li><a href="/section/problem_1-1">OCW Problem 1-1/Textbook 2.7</a> - Find the current through a resistor in a network of resistors</li>
<li><a href="/section/problem_1-2_part1">OCW Problem 1-2</a> - Analyze long chains of resistors (similar to transmission lines or R1R2 ladders) </li>
<li><a href="/section/problem_1_3">OCW Problem 1-3 </a> - Reverse engineer a black-box resistor network</li>
</ul>
<hr/>
<p> Since the course has students from a diverse set of backgrounds, the first week's tutorials includes several extra segments, worked out with greater detail, to help bring everyone up to speed. Gratuitous &ge; entity.</p>
</section>
</body>
</html>
common/test/data/full/html/html_5555.html deleted 100644 → 0
<html slug="html_5555" filename="html_5555> See the <a href="/section/labintro"> Lab Introduction </a> or <a href="/static/handouts/schematic_tutorial.pdf">Interactive Lab Usage Handout </a> for information on how to do the lab. </html>
common/test/data/full/html/linearity_clarify.html deleted 100644 → 0
Clarification of the term "Linear"
<p>
The term "linear" is very clear when applied to a
mathematical function. A function F is linear if and only
if it obeys homogeneity and superposition:
</p><p>
Homogeneity: F(cx) = cF(x)
<br/>
Superposition: F(x+y) = F(x) + F(y)
</p><p>
In the context of what we have seen so far, the only
elements that are linear as mathematical functions are
resistors. An independent voltage source or an independent
current source is not a linear element. (There are also
linear dependent sources, linear capacitors and linear
inductors, but we have not yet introduced them in our class.
You will see them later.)
</p><p>
Formally, a circuit composed of only linear elements is a
linear circuit. When we add independent sources to a linear
circuit as inputs, we get a circuit that is not linear
because it has an offset: its v-i characteristic at a pair
of exposed terminals may not pass through the origin.
However, we can make a Thevenin or Norton equivalent model
of such a circuit: the Thevenin resistance summarizes the
effect of the linear elements and the Thevenin voltage
summarizes the effect of the independent sources.
</p><p>
However the term "linear," when applied to an electrical
circuit often takes on an informal meaning. We often say
that a circuit containing only linear elements and
independent sources is a "linear circuit." So, in the
informal sense, a linear circuit is one where we can apply
the Thevenin or Norton theorems to summarize the behavior at
a pair of exposed terminals.
</p><p>
Sorry for the confusion of words &mdash; natural language is like
that!
</p>
common/test/data/full/html/linearity_clarify.xml deleted 100644 → 0
Clarification of the term "Linear"
<p>
The term "linear" is very clear when applied to a
mathematical function. A function F is linear if and only
if it obeys homogeneity and superposition:
</p><p>
Homogeneity: F(cx) = cF(x)
<br/>
Superposition: F(x+y) = F(x) + F(y)
</p><p>
In the context of what we have seen so far, the only
elements that are linear as mathematical functions are
resistors. An independent voltage source or an independent
current source is not a linear element. (There are also
linear dependent sources, linear
inductors, and other linear elements, but we have not yet introduced them in our class.
You will see them later.)
</p><p>
Formally, a circuit composed of only linear elements is a
linear circuit. When we add independent sources to a linear
circuit as inputs, we get a circuit that is not linear
because it has an offset: its v-i characteristic at a pair
of exposed terminals may not pass through the origin.
However, we can make a Thevenin or Norton equivalent model
of such a circuit: the Thevenin resistance summarizes the
effect of the linear elements and the Thevenin voltage
summarizes the effect of the independent sources.
</p><p>
However the term "linear," when applied to an electrical
circuit often takes on an informal meaning. We often say
that a circuit containing only linear elements and
independent sources is a "linear circuit." So, in the
informal sense, a linear circuit is one where we can apply
the Thevenin or Norton theorems to summarize the behavior at
a pair of exposed terminals.
</p><p>
Sorry for the confusion of words &mdash; natural language is like
that!
</p>
common/test/data/full/html/schematic_tutorial.html deleted 100644 → 0
<p/><b>Usage Summary:</b>
<table style="border-spacing:2em; border-collapse: separate;">
<tr>
<td style="width:10em;">Add a component</td>
<td>Click on the component in the parts bin (the columns of part icons to
the right of the diagram area) and drag it onto the diagram. Release
the mouse when the component is in the correct position.
</td>
</tr>
<!-- note that entities like &mdash; may be used. -->
<tr>
<td>Move a component</td>
<td>Click to select a component in the diagram (it will turn green)
and then drag it to its new location. You can use shift-click to add
a component to the current selection. Or you can click somewhere in
the diagram that is not on top of a component and drag out a selection
rectangle &mdash; components intersecting the rectangle will be added to
the current selection.
</td>
</tr>
<tr>
<td>Rotate a component</td>
<td>Click to select a component in the diagram (it will turn green)
and then type the letter "r" on the keyboard. The component will
be rotated 90 degrees. Additional rotations will move the component
through its eight possible orientations (4 compass points and their
reflections).
</td>
</tr>
<tr>
<td>Delete a component</td>
<td>Click to select the component in the diagram (shift-click to
select multiple components) then type DEL or BACKSPACE on your
keyboard.
</td>
</tr>
<tr>
<td>Change a component's properties</td>
<td>Double click on the component. This will bring up an Edit Properties
window that has input fields for each of the component's properties.
Click OK to change the values. Click CANCEL or the window's close
button to abort the changes. Numeric values can be entered using
engineeering notation:
<center><table style="border-collapse:separate; border-spacing:1em;">
<tr><th>suffix</th><th>multiplier</th> <th style="width:3em"></th> <th>suffix</th><th>multiplier</th></tr>
<tr><td>T</td><td>\(10^{12}\)</td> <td></td> <td>u</td><td>\(10^{-6}\)</td></tr>
<tr><td>G</td><td>\(10^{9}\)</td> <td></td> <td>n</td><td>\(10^{-9}\)</td></tr>
<tr><td>M</td><td>\(10^{6}\)</td> <td></td> <td>p</td><td>\(10^{-12}\)</td></tr>
<tr><td>k</td><td>\(10^{3}\)</td> <td></td> <td>f</td><td>\(10^{-15}\)</td></tr>
<tr><td>m</td><td>\(10^{-3}\)</td></tr>
</table></center>
</td>
</tr>
<tr>
<td>Add a wire</td>
<td>Wires start at connection points, the open circles that
appear at the terminals of components or the ends of wires.
Click on a connection point to start a wire &mdash; a green wire
will appear with one end anchored at the starting point.
Drag the mouse and release the mouse button when the other
end of the wire is positioned as you wish. Once a wire has
been added to the diagram it can be manipulated like any other
component.
</td>
</tr>
</table>
common/test/data/full/html/units_hint.html deleted 100644 → 0
Hint…
<br/><br/>
Be careful of units here. Make sure you notice multipliers such
as u (or &mu;), k, m, M.
common/test/data/full/info/guest_updates.html deleted 100644 → 0
<h1> Updates </h1>
<ol>
<li>
<section class="update-description">
Welcome to 6002x. Please look around.
</section>
</li>
</ol>
common/test/data/full/info/handouts.html deleted 100644 → 0
<a href='/static/handouts/schematic_tutorial.pdf'>Schematic Tutorial</a>
\ No newline at end of file
common/test/data/full/policies/6.002_Spring_2012.json deleted 100644 → 0
{
"course/6.002_Spring_2012": {
"graceperiod": "1 day 12 hours 59 minutes 59 seconds",
"start": "2012-09-21T12:00",
"display_name": "Testing",
"xqa_key": "5HapHs6tHhu1iN1ZX5JGNYKRkXrXh7NC",
"tabs": [
{"type": "courseware"},
{"type": "course_info", "name": "Course Info"},
{"type": "static_tab", "url_slug": "syllabus", "name": "Syllabus"},
{"type": "static_tab", "url_slug": "resources", "name": "Resources"},
{"type": "discussion", "name": "Discussion"},
{"type": "wiki", "name": "Wiki"},
{"type": "progress", "name": "Progress"}
]
}
}
\ No newline at end of file
common/test/data/full/problem/Circuit_Sandbox.xml deleted 100644 → 0
<problem><!-- include ellipses to test non-ascii characters --><startouttext/><p/>Here's a sandbox where you can experiment with all the components
we'll discuss in 6.002x. If you click on CHECK below, your diagram
will be saved on the server and you can return at some later time…
<endouttext/><schematicresponse><p/><center><schematic name="work" value="" width="800" height="600"/></center><answer type="loncapa/python">
correct = ['correct']
</answer></schematicresponse></problem>
common/test/data/full/problem/H1P1_Energy.xml deleted 100644 → 0
<problem><script type="loncapa/python">
i = random.randrange(3)
M = ["Big Mac", "Angus Bacon and Cheese", "Quarter Pounder with Cheese"][i]
C = [540, 790, 510][i]
BM = C*1000*4.2
CB = 50*12*3600
</script><startouttext/>
According to McDonald's
<a href="http://nutrition.mcdonalds.com/getnutrition/nutritionfacts.pdf">nutrition
facts</a> a \($M\) provides \($C\) kiloCalories of energy. (1 dietitian's
calorie is 1000 physicist's calories.) A typical automobile lead-acid
battery can supply 50 Ampere-hours at 12 Volts. We can compare these by
converting the energy specifications to common units. The SI unit of
energy is the Joule=1 Watt-second. One physicist's calorie is 4.2 Joules.
<endouttext/>
<startouttext/>
<br/>
How many Joules are there in a \($M\)?
<endouttext/>
<numericalresponse answer="$BM"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
How many Joules are there in a car battery?
<endouttext/>
<numericalresponse answer="$CB"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
Which tastes better? :-)
<endouttext/>
</problem>
common/test/data/full/problem/H1P2_Duality.xml deleted 100644 → 0
<problem><script type="loncapa/python">
scale = float(random.randrange(1000,2000,10))/1000
V = 10*scale
R1 = 3*scale
R2 = 2*scale
vV = V
R1R2S = R1+R2
i1 = V/R1R2S
i2 = i1
iV = -i1
v1 = i1*R1
v2 = i2*R2
Ps= vV*iV + v1*i1 + v2*i2
I = V
R3 = 1/R1
R4 = 1/R2
iI = -I
R3R4P = (R3*R4)/(R3+R4)
vI = I*R3R4P
v3 = vI
v4 = vI
i3 = v3/R3
i4 = v4/R4
Pp = vI*iI + v3*i3 + v4*i4
</script><startouttext/>
In this problem we will investigate a fun idea called "duality."
<br/>
Consider the series circuit in the diagram shown.
<center><img src="/static/circuits/duality.gif"/></center>
We are given device parameters \(V=$V\)V, \(R_1=$R1\Omega\), and \(R_2=$R2\Omega\).
All of the unknown voltages and currents are labeled in associated reference
directions. Solve this circuit for the unknowns and enter them into
the boxes given.
<br/>
The value (in Volts) of \(v_1\) is:
<endouttext/>
<numericalresponse answer="$v1"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Amperes) of \(i_1\) is:
<endouttext/>
<numericalresponse answer="$i1"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Volts) of \(v_2\) is:
<endouttext/>
<numericalresponse answer="$v2"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Amperes) of \(i_2\) is:
<endouttext/>
<numericalresponse answer="$i2"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Volts) of \(v_V\) is:
<endouttext/>
<numericalresponse answer="$vV"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Amperes) of \(i_V\) is:
<endouttext/>
<numericalresponse answer="$iV"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The sum of the powers (in Watts) entering all of the elements is:
<endouttext/>
<numericalresponse answer="$Ps"><responseparam type="tolerance" default="0.00001" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/> <br/> Now, let's turn our attentions to the parallel circuit.
The numerical value of the strength of the current source is set to be
the same as the numerical value of the strength of the voltage source:
\(I=$I\)A. The resistances are set to be reciprocals of the
resistances in the series circuit: \(R_3=$R3\Omega\) and
\(R_4=$R4\Omega\).
<br/>
Solve this circuit for the unknowns and enter them into
the boxes given.
<br/>
The value (in Volts) of \(v_3\) is:
<endouttext/>
<numericalresponse answer="$v3"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Amperes) of \(i_3\) is:
<endouttext/>
<numericalresponse answer="$i3"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Volts) of \(v_4\) is:
<endouttext/>
<numericalresponse answer="$v4"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Amperes) of \(i_4\) is:
<endouttext/>
<numericalresponse answer="$i4"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value (in Volts) of \(v_I\) is:
<endouttext/>
<numericalresponse answer="$vI"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The value of (in Amperes) \(i_I\) is:
<endouttext/>
<numericalresponse answer="$iI"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
The sum of the powers (in Watts) entering all of the elements is:
<endouttext/>
<numericalresponse answer="$Pp"><responseparam type="tolerance" default="0.00001" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/> <br/> Look carefully at the numbers you have derived.
Compare the series and parallel circuit. Do you see the pattern?
<endouttext/>
</problem>
common/test/data/full/problem/H1P3_Poor_Workmanship.xml deleted 100644 → 0
<problem><script type="loncapa/python">
V = 240
P = float(random.randrange(800,1600,10))
Pmax = 3*P
Imax = Pmax/V
Iq = Pmax/120
R = (V**2)/P
Rbad = 3*R/2
Pbad = (V**2)/Rbad
Ibad = Pbad/V
PH1 = (Ibad**2)*R
PH2 = ((Ibad/2)**2)*R
PH3 = PH2
</script><startouttext/>
Joe wants to heat his 12'X20' workshop with electric heat. He has
hired the HACME electrician company to build the system. They propose
to use three \($P\)W 240V baseboard heaters to provide a total heating
capacity of \($Pmax\)W. (A heater is basically a
resistor. This is not quite true, because there is a thermostatic
switch incorporated into the heater and because the resistance of a
heater varies a bit with its temperature. But we will use a linear
resistor as a model of a heater.)
<br/>
In the proposed system the heaters are connected in parallel with the
240V 60Hz AC power line (modeled by a voltage source) as shown in the
diagram:
<center><img src="/static/circuits/heaters-parallel.gif"/></center>
Remember (from exercise S1E3: AC POWER) that AC power-line voltages
and currents are specified as RMS values. So 120V AC heats a given
resistance exactly as much as 120V DC would heat that same
resistance.
<endouttext/>
<startouttext/>
<br/>
How much current is expected to be drawn from the power line by this
heating system when all three heaters are on?
<endouttext/>
<numericalresponse answer="$Imax"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
If instead, HACME chose to implement the system with 120V heaters, how
much current would have been needed?
<endouttext/>
<numericalresponse answer="$Iq"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
Notice that this would require much heavier and more expensive wire to
distribute the power.
<br/> <br/>
Back to the original plan with 240V power.
<br/>
Unfortunately, Sparky, who works for HACME, was a little sleepy that
day. He accidentally connected the heaters as shown below:
<center><img src="/static/circuits/heaters-bad.gif"/></center>
<br/>
As a consequence, Joe found his workshop too cold. H1 was weak; H2
and H3 barely warmed up.
<br/>
What power was being dissipated in H1?
<endouttext/>
<numericalresponse answer="$PH1"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
What power was being dissipated in H2 (or in H3)?
<endouttext/>
<numericalresponse answer="$PH2"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
So the total heating power in Joe's shop was:
<endouttext/>
<numericalresponse answer="$Pbad"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/>
<!-- add non-ascii utf-8 character here -->
No wonder Joe was cold…
<endouttext/>
</problem>
common/test/data/full/problem/Lab_0_Using_the_Tools.xml deleted 100644 → 0
<problem><startouttext/><br/>In this first lab, the goal is to familiarize yourself with the
interactive tool that will be used to enter and analyze circuits. Here's
the example schematic used below:
<br/><center><input type="hidden" class="schematic" analyses="" parts="" origin_x="50" origin_y="20" width="400" height="350" name="fig1" value="[[&quot;w&quot;,[120,160,144,160]],[&quot;v&quot;,[48,64,0],{&quot;name&quot;:&quot;Vsupply&quot;,&quot;value&quot;:&quot;dc(3)&quot;,&quot;_json_&quot;:1},[&quot;A&quot;,&quot;D&quot;]],[&quot;r&quot;,[120,64,0],{&quot;name&quot;:&quot;R2&quot;,&quot;r&quot;:&quot;3k&quot;,&quot;_json_&quot;:2},[&quot;B&quot;,&quot;C&quot;]],[&quot;r&quot;,[120,16,0],{&quot;name&quot;:&quot;R1&quot;,&quot;r&quot;:&quot;1k&quot;,&quot;_json_&quot;:3},[&quot;A&quot;,&quot;B&quot;]],[&quot;r&quot;,[120,112,0],{&quot;name&quot;:&quot;R3&quot;,&quot;r&quot;:&quot;2k&quot;,&quot;_json_&quot;:4},[&quot;C&quot;,&quot;D&quot;]],[&quot;w&quot;,[48,64,48,16]],[&quot;w&quot;,[48,16,120,16]],[&quot;w&quot;,[48,112,48,160]],[&quot;w&quot;,[48,160,120,160]],[&quot;w&quot;,[120,112,144,112]],[&quot;w&quot;,[120,64,144,64]],[&quot;w&quot;,[120,16,144,16]],[&quot;L&quot;,[144,16,3],{&quot;label&quot;:&quot;A&quot;,&quot;_json_&quot;:12},[&quot;A&quot;]],[&quot;L&quot;,[144,64,3],{&quot;label&quot;:&quot;B&quot;,&quot;_json_&quot;:13},[&quot;B&quot;]],[&quot;L&quot;,[144,112,3],{&quot;label&quot;:&quot;C&quot;,&quot;_json_&quot;:14},[&quot;C&quot;]],[&quot;g&quot;,[48,160,0],{&quot;_json_&quot;:15},[&quot;D&quot;]],[&quot;L&quot;,[144,160,3],{&quot;label&quot;:&quot;D&quot;,&quot;_json_&quot;:16},[&quot;D&quot;]],[&quot;view&quot;,0,0,2,&quot;5&quot;,&quot;10&quot;,&quot;10MEG&quot;,null,&quot;100&quot;,&quot;1&quot;]]"/>
Figure 1. Test Circuit for Lab 0
</center><br/>And here's the interactive tool to be used in the tasks that follow:
<endouttext/><schematicresponse><startouttext/><br/><center><schematic analyses="dc,tran" parts="g,L,v,r,s" name="work" value="" width="600" height="400"/></center><endouttext/><answer type="loncapa/python">
correct = ['correct']
</answer></schematicresponse><startouttext/><br/>Please do the following tasks. When you're done, remember to click CHECK to preserve
your work and to verify your answers.
<ol style="list-style-type:decimal; margin-left: 2em;"><br/><li><b>Enter schematic.</b> Add the voltage source, resistors,
wires and ground reference to your diagram, connected as shown
in <nobr>Figure 1</nobr>.</li><br/><li><b>Measure voltages.</b> Use the tool to run a DC analysis on
the circuit, i.e.., click on the DC button in the toolbar. This will
annotate each node/voltage source in the circuit with a
voltage/current. If the components are wired correctly, the
voltage for node D should be reported as 0V since it's
connected to the ground reference and the voltage for node A
should be reported as 3V since it is connected directly to the
positive terminal of a 3V voltage source whose negative terminal
is connected to ground. Please enter the voltages reported for nodes
B and C.
<br/><div style="margin-left: 4em;">
Voltage for node B (volts):
<numericalresponse answer="2.5"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
<div style="margin-left: 4em;">
Voltage for node C (volts):
<numericalresponse answer="1.0"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
</li><br/><li><b>Measure current.</b>
The DC analysis annotates each voltage source with the current it is
providing to the rest of the circuit. The simulator denotes the sign
of the source current as positive if the current is flowing out of the
positive terminal of the source, through the rest of the circuit, and
then back in to the negative terminal. And the source current sign is
negative if the current is flowing out of the negative terminal of the
source, through the circuit, and into the positive terminal of the
source. Please report the total resistance of the path connecting the
positive and negative terminals of the source, the signed current
flowing from the source into the circuit, and (using the voltage
value for the source) verify that Ohm's law is obeyed.
<br/><div style="margin-left: 4em;">
Total resistance between pos and neg terminals (ohms):
<numericalresponse answer="6000"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
<div style="margin-left: 4em;">
Current flowing from source into the circuit (amps, with correct sign):
<numericalresponse answer="500e-6"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
</li><br/><li><b>Verify KVL.</b> The four components in the schematic form a
large loop -- let's verify that Kirchoff's Voltage Law holds by
summing the voltage changes across the devices. Starting with node A,
travel clockwise around the loop of components, entering the voltage
change across each component. Choose the sign of the change for a
component using the first terminal you come to in the clockwise
traversal as the reference node. Verify that these four values sum
to 0.
<br/>
<div style="margin-left: 4em;">
Voltage change across R1 (volts, with correct sign):
<numericalresponse answer="-0.5"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
<div style="margin-left: 4em;">
Voltage change across R2 (volts, with correct sign):
<numericalresponse answer="-1.5"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
<div style="margin-left: 4em;">
Voltage change across R3 (volts, with correct sign):
<numericalresponse answer="-1.0"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
<div style="margin-left: 4em;">
Voltage change across Vsupply (volts, with correct sign):
<numericalresponse answer="3.0"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
</li><br/><li><b>Transient analysis.</b> Change the waveform produced by
the voltage source from a DC value of 3V to a sinusoid with a 1V
amplitude, an offset voltage of 1V, and a frequency of 1kHz. Add
scope probes to nodes A, B and C and edit their properties so that the
plots will be different colors. Now run a transient analysis for 5ms.
Move the mouse over the plot until the marker (a vertical dashed line
that follows the mouse when it's over the plot) is at approximately
1.25ms. Please report the measured voltages for nodes A, B and C…
<br/>
<div style="margin-left: 4em;">
Voltage for node A (volts):
<numericalresponse answer="1.995"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
<div style="margin-left: 4em;">
Voltage for node B (volts):
<numericalresponse answer="1.662"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
<div style="margin-left: 4em;">
Voltage for node C (volts):
<numericalresponse answer=".665"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</div>
</li></ol><endouttext/></problem>
common/test/data/full/problem/Resistor_Divider.xml deleted 100644 → 0
<problem><startouttext/><br/><br/>You have a 6-volt battery (assumed ideal) and a 1.5-volt flashlight
bulb, which is known to draw \(0.5 A\) when the bulb voltage is \(1.5 V\) (see
figure below). Design a network of resistors to go between the battery
and the bulb to give \(v_s = 1.5 V\) when the bulb is connected, yet
ensures that \(v_s\) does not rise above \(2 V\) when the bulb is disconnected.
<br/><br/><center><img src="/static/circuits/Lab1_1.png"/></center><br/><br/><i>Hint</i>: use a two-resistor voltage divider to create the voltage for node A. You'll
have two unknowns (R1 and R2) which can be determined by solving the two equations for
\(v_s\) derived from the constraints above: one involving R1, R2 and Rbulb where \(v_s = 1.5\), and
one involving R1 and R2 where \(v_s = 2\).
<br/><br/>There are two schematic diagrams below. Please enter the network
of resistors you've designed into both diagrams. The top diagram is
the model when the bulb is connected; the bottom diagram is the model
when the bulb is disconnected.
<br/><br/>Run a DC analysis on both diagrams to
show that the node labeled "A" has a voltage of approximately
\(1.5 V\) in the top diagram and less than \(2 V\) in the bottom
adiagram. Submit your results <i>after</i> the DC analyses have
been run (so the results of the analyses will be submitted too).
<endouttext/>
<schematicresponse><startouttext/>
Schematic model when bulb is connected:
<center><schematic height="270" width="400" parts="r" analyses="dc" initial_value="[[&quot;w&quot;,[48,88,48,112]],[&quot;L&quot;,[160,16,3],{&quot;label&quot;:&quot;A&quot;},[&quot;A&quot;]],[&quot;g&quot;,[48,112,0],{},[&quot;0&quot;]],[&quot;w&quot;,[160,112,136,112]],[&quot;w&quot;,[160,88,160,112]],[&quot;w&quot;,[160,16,136,16]],[&quot;w&quot;,[160,40,160,16]],[&quot;r&quot;,[160,40,0],{&quot;name&quot;:&quot;Bulb&quot;,&quot;r&quot;:&quot;3&quot;},[&quot;A&quot;,&quot;1&quot;]],[&quot;w&quot;,[48,112,72,112]],[&quot;w&quot;,[48,16,72,16]],[&quot;w&quot;,[48,40,48,16]],[&quot;v&quot;,[48,40,0],{&quot;name&quot;:&quot;Battery&quot;,&quot;value&quot;:&quot;6V&quot;},[&quot;2&quot;,&quot;0&quot;]],[&quot;view&quot;,0,0,2]]"/></center>
Schematic model when bulb is disconnected:
<center><schematic height="270" width="400" parts="r" analyses="dc" initial_value="[[&quot;w&quot;,[48,88,48,112]],[&quot;L&quot;,[160,16,3],{&quot;label&quot;:&quot;A&quot;},[&quot;A&quot;]],[&quot;g&quot;,[48,112,0],{},[&quot;0&quot;]],[&quot;w&quot;,[160,112,136,112]],[&quot;w&quot;,[160,88,160,112]],[&quot;w&quot;,[160,16,136,16]],[&quot;w&quot;,[160,40,160,16]],[&quot;w&quot;,[48,112,72,112]],[&quot;w&quot;,[48,16,72,16]],[&quot;w&quot;,[48,40,48,16]],[&quot;v&quot;,[48,40,0],{&quot;name&quot;:&quot;Battery&quot;,&quot;value&quot;:&quot;6V&quot;},[&quot;2&quot;,&quot;0&quot;]],[&quot;view&quot;,0,0,2]]"/></center>
<endouttext/>
<answer type="loncapa/python">
# for a schematic response, submission[i] is the json representation
# of the diagram and analysis results for the i-th schematic tag
correct = ['incorrect', 'incorrect'] # optimistic default :)
def get_dc(json):
for element in json:
if element[0] == 'dc': return element[1]
return None
dc_with_bulb = get_dc(submission[0])
if dc_with_bulb:
v = dc_with_bulb['A']
if v &gt;= 1.4 and v &lt;= 1.6: # want 1.5
correct[0] = 'correct'
dc_without_bulb = get_dc(submission[1])
if dc_without_bulb:
v = dc_without_bulb['A']
if v &lt;= 2.1: # want 2
correct[1] = 'correct'
</answer>
</schematicresponse>
</problem>
common/test/data/full/problem/S1E3_AC_power.xml deleted 100644 → 0
<problem><script type="loncapa/python">
R = float(random.randrange(100,200,10))
Vrms = 120
Vp = math.sqrt(2)*Vrms
Pp = (Vp**2)/R
Prms = (Vrms**2)/R
</script><startouttext/>
The plot shows 1/10 second of the voltage waveform of a 120V 60Hz AC
(Alternating Current)
power circuit, like that delivered to residences in the United States.
<center><img src="/static/circuits/120V60Hz.gif"/></center>
The actual voltage is \(120*sqrt(2)*cos(2*\pi*60*t)\)Volts. If we apply
this voltage across a resistor of resistance \($R\Omega\) the resistor
will dissipate a time-varying power.
What is the peak power (in Watts) dissipated by the resistor?
<endouttext/>
<numericalresponse answer="$Pp"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/> What is the average power (in Watts) dissipated by the resistor? (Hint: you
compute the average power by integrating the power over one cycle of
the waveform.)
<endouttext/>
<numericalresponse answer="$Prms"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/> What would be the power (in Watts) dissipated by the resistor if the voltage
was a constant value of 120V?
<endouttext/>
<numericalresponse answer="$Prms"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
<startouttext/>
<br/> If a time-varying voltage dissipates the same power in a resistor
as a constant voltage would dissipate, we say that the time-varying voltage has
the RMS value of the constant. RMS is an abbreviation for Root-Mean-Square.
<endouttext/>
</problem>
common/test/data/full/problem/Sample_Algebraic_Problem.xml deleted 100644 → 0
<problem><script type="loncapa/python">
z = "A*x^2 + sqrt(y)"
</script><startouttext/>
Enter the algebraic expression \(A x^2 + \sqrt{y}\) in the box below. The
entry is case sensitive. The product must be indicated with an
asterisk, and the exponentation with a caret, so you must write
"A*x^2 + sqrt(y)"…
<endouttext/>
<formularesponse type="cs" samples="A,x,y@1,1,1:3,3,3#10" answer="$z"><responseparam description="Numerical Tolerance" type="tolerance" default="0.00001" name="tol"/><textline size="40"/></formularesponse>
</problem>
common/test/data/full/problem/Sample_Numeric_Problem.xml deleted 100644 → 0
<problem><startouttext/>
Enter the numerical value of the expression \(x + y\) where
\(x = 3\) and \(y = 5\)…
<endouttext/>
<numericalresponse answer="8"><responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance"/><textline/></numericalresponse>
</problem>
common/test/data/full/problem/choiceresponse_demo.xml deleted 100644 → 0
<problem display_name="S3E2: Lorentz Force">
<startouttext/>
<p>Consider a hypothetical magnetic field pointing out of your computer screen. Now imagine an electron traveling from right to left in the plane of your screen. A diagram of this situation is show below…</p>
<center><img width="400" src="/static/images/LSQimages/LSQ_W01_8.png"/></center>
<p>a. The magnitude of the force experienced by the electron is proportional the product of which of the following? (Select all that apply.)</p>
<endouttext/>
<choiceresponse>
<checkboxgroup>
<!-- include ellipses to test non-ascii characters -->
<choice correct="true"><text>Magnetic field strength…</text></choice>
<choice correct="false"><text>Electric field strength…</text></choice>
<choice correct="true"><text>Electric charge of the electron…</text></choice>
<choice correct="false"><text>Radius of the electron…</text></choice>
<choice correct="false"><text>Mass of the electron…</text></choice>
<choice correct="true"><text>Velocity of the electron…</text></choice>
</checkboxgroup>
</choiceresponse>
<startouttext/>
<p>b. The direction of the force experienced by the electron is _______.</p>
<endouttext/>
<choiceresponse>
<checkboxgroup>
<choice correct="false"><text>up, in the plane of the screen.</text></choice>
<choice correct="true"><text>down, in the plane of the screen.</text></choice>
<choice correct="false"><text>into the plane of the screen.</text></choice>
<choice correct="false"><text>out of the plane of the screen.</text></choice>
<choice correct="false"><text>towards the right, in the plane of the screen.</text></choice>
<choice correct="false"><text>towards the left, in the plane of the screen.</text></choice>
</checkboxgroup>
</choiceresponse>
</problem>
common/test/data/full/problem/codeinput_demo.xml deleted 100644 → 0
<?xml version="1.0"?>
<problem display_name="L4 Problem 1">
<text>
<p>
<!-- include ellipses to test non-ascii characters -->
<b class="bfseries">Part 1: Function Types…</b>
</p>
<p>
For each of the following functions, specify the type of its <b class="bfseries">output</b>. You can assume each function is called with an appropriate argument, as specified by its docstring. </p>
<p>
If the output can be either an int or a float, select num, which isn't a real Python type, but which we'll use to indicate that either basic numeric type is legal. </p>
<p>
In fact, in Python, booleans True and False can be operated on as if they were the integers 1 and 0; but it is ugly and confusing to take advantage of this fact, and we will resolutely pretend that it isn't true. </p> <!-- so why even mention it? -->
<p>
<section class="hints">
<div class="collapsible">
<header>
<a href="#" id="id41">What are those lines under the function definitions?</a>
</header>
<section id="id41">
<p>
In this and future problems, you'll see function definitions that look like this: </p>
<pre>
def a(x):
'''
x: int or float.
'''
return x + 1
</pre>
<p>
What are those three lines between <code>def a(x):</code> and <code>return x + 1</code>? These lines are called the <i class="it">docstring</i> of the function. A docstring is a special type of comment that is used to document what your function is doing. Typically, docstrings will explain what the function expects the type(s) of the argument(s) to be, and what the function is returning. </p>
<p>
In Python, docstrings appear immediately after the <code>def</code> line of a function, before the body. Docstrings start and end with triple quotes - this can be triple single quotes or triple double quotes, it doesn't matter as long as they match. To sum up this general form: </p>
<pre>
def myFunction(argument):
"""
Docstring goes here. Explain what type argument(s) should have, and what your function
is going to return.
"""
&lt; Code for your function (the body of the function) goes here &gt;
</pre>
<p>
As you begin coding your own functions, we strongly encourage you to document all your functions by using properly-formatted docstrings! </p>
</section>
</div>
</section>
</p>
<ol class="enumerate">
<li>
<pre>
def a(x):
'''
x: int or float.
'''
return x + 1
</pre>
<p>
Indicate the type of the output that the function <code>a</code> will yield. <optionresponse><optioninput options="('NoneType','num','int','float','boolean')" correct="num"/></optionresponse> </p>
</li>
<li>
<pre>
def b(x):
'''
x: int or float.
'''
return x + 1.0
</pre>
<p>
Indicate the type of the output that the function <code>b</code> will yield. <optionresponse><optioninput options="('NoneType','num','int','float','boolean')" correct="float"/></optionresponse> </p>
</li>
<li>
<pre>
def c(x, y):
'''
x: int or float.
y: int or float.
'''
return x + y
</pre>
<p>
Indicate the type of the output that the function <code>c</code> will yield. <optionresponse><optioninput options="('NoneType','num','int','float','boolean')" correct="num"/></optionresponse> </p>
</li>
<li>
<pre>
def d(x, y):
'''
x: Can be of any type.
y: Can be of any type.
'''
return x &gt; y
</pre>
<p>
Indicate the type of the output that the function <code>d</code> will yield. <optionresponse><optioninput options="('NoneType','num','int','float','boolean')" correct="boolean"/></optionresponse> </p>
</li>
<li>
<pre>
def e(x, y, z):
'''
x: Can be of any type.
y: Can be of any type.
z: Can be of any type.
'''
return x &gt;= y and x &lt;= z
</pre>
<p>
Indicate the type of the output that the function <code>e</code> will yield. <optionresponse><optioninput options="('NoneType','num','int','float','boolean')" correct="boolean"/></optionresponse> </p>
</li>
<li>
<pre>
def f(x, y):
'''
x: int or float.
y: int or float
'''
x + y - 2
</pre>
<p>
Indicate the type of the output that the function <code>f</code> will yield. <optionresponse><optioninput options="('NoneType','num','int','float','boolean')" correct="NoneType"/></optionresponse> </p>
</li>
</ol>
<p>
<b class="bfseries">Part 2: Transcript</b>
</p>
<p>
Below is a transcript of a session with the Python shell. Assume the functions from Part 1 (above) have been defined. Provide the type and value of the expressions being evaluated. If evaluating an expression would cause an error, select NoneType and write 'error' in the box. If the value of an expression is a function, select function as the type and write 'function' in the box. </p>
<ol class="enumerate">
<li>
<p>
<code>a(6)</code>
<optionresponse>
<optioninput options="('NoneType','num','int','float','boolean','function')" correct="int"/>
</optionresponse>
<stringresponse answer="7">
<textline size="50"/>
</stringresponse>
</p>
</li>
<li>
<p>
<code>a(-5.3)</code>
<optionresponse>
<optioninput options="('NoneType','num','int','float','boolean','function')" correct="float"/>
</optionresponse>
<stringresponse answer="-4.3">
<textline size="50"/>
</stringresponse>
</p>
</li>
<li>
<p>
<code>a(a(a(6)))</code>
<optionresponse>
<optioninput options="('NoneType','num','int','float','boolean','function')" correct="int"/>
</optionresponse>
<stringresponse answer="9">
<textline size="50"/>
</stringresponse>
</p>
</li>
<li>
<p>
<code>c(a(1), b(1))</code>
<optionresponse>
<optioninput options="('NoneType','num','int','float','boolean','function')" correct="float"/>
</optionresponse>
<stringresponse answer="4.0">
<textline size="50"/>
</stringresponse>
</p>
</li>
<li>
<p>
<code>d('apple', 11.1)</code>
<optionresponse>
<optioninput options="('NoneType','num','int','float','boolean','function')" correct="boolean"/>
</optionresponse>
<stringresponse answer="True">
<textline size="50"/>
</stringresponse>
</p>
</li>
<li>
<p>
<code>e(a(3), b(4), c(3, 4))</code>
<optionresponse>
<optioninput options="('NoneType','num','int','float','boolean','function')" correct="boolean"/>
</optionresponse>
<stringresponse answer="False">
<textline size="50"/>
</stringresponse>
</p>
</li>
<li>
<p>
<code>f</code>
<optionresponse>
<optioninput options="('NoneType','num','int','float','boolean','function')" correct="function"/>
</optionresponse>
<stringresponse answer="function">
<textline size="50"/>
</stringresponse>
</p>
</li>
</ol>
</text>
</problem>
common/test/data/full/problem/multiplechoicelresponse_demo.xml deleted 100644 → 0
<problem display_name="S3E3: Deflection of a Proton">
<startouttext/>
<p>Suppose you were to pass a hydroge ion (H+) through an electric field and measure the deflection of the ion. How would its deflection compare to the deflection of an electron passed through the same electric field? </p>
<endouttext/>
<multiplechoiceresponse direction="vertical" randomize="yes">
<choicegroup type="MultipleChoice">
<choice location="random" correct="false" name="1"><text>The ion would deflect identically as both species have the same charge.</text></choice>
<choice location="random" correct="false" name="2"><text>The ion would deflect by an amount identical in magnitude but opposite in direction because the charges are of opposite sign but equal in magnitude.</text></choice>
<choice location="random" correct="false" name="3"><text>The deflection will be in the same direction but smaller in magnitude due to the larger mass of the ion.</text></choice>
<choice location="random" correct="true" name="4"><text>The deflection will be smaller in magnitude due to the increased mass of the ion, as well as in the opposite direction due to the opposite sign of the charge.</text></choice>
</choicegroup>
</multiplechoiceresponse>
</problem>
\ No newline at end of file
common/test/data/full/problem/numericalresponse_demo.xml deleted 100644 → 0
<problem display_name="S2E2: Hydrogen Combustion">
<text>
<p>Gaseous hydrogen can be reacted with gaseous oxygen in the presence of a flame to produce water vapor according to the following reaction:</p>
[mathjax] \text{ a } \text{H}_{2 } + \text{ b }\text{O}_{2 } = \text{ c }\text{H}_{2 }\text{O }[/mathjax]
<p>Balance the equation i.e., specify the values of <i> a, b, </i> and <i>c</i>. Use the lowest whole number coefficients.</p>
<p>a =</p>
<numericalresponse answer="2">
<responseparam type="tolerance" default="1%" name="tol" description="Numerical Tolerance" />
<textline correct_answer="2" math="1" />
</numericalresponse>
<p>b =
<numericalresponse answer="1">
<responseparam type="tolerance" default="1%" name="tol" description="Numerical Tolerance" />
<textline math="1" />
</numericalresponse> </p>
<p>c =
<numericalresponse answer="2">
<responseparam type="tolerance" default="1%" name="tol" description="Numerical Tolerance" />
<textline math="1" />
</numericalresponse> </p>
</text>
</problem>
common/test/data/full/problem/test_files/chemicalformularesponse.xml deleted 100644 → 0
<problem>
<text>
<p>Metallothermic production of zirconium (Zr) would involve the reaction of sodium (Na) with zirconium tetrachloride (ZrCl<sub>4</sub>).</p>
<p><b>(a)</b> Write a balanced chemical equation.</p>
<!-- chemformularesponse
[mathjax] \text{a ZrCl}_{4 } + \text{ b }\text{Na} -> \text{ c }\text{NaCl } + \text{ d } \text{Zr} [/mathjax]
ratio of (a, b, c, d) = (1, 4, 4, 1)
-->
<customresponse>
<chemicalequationinput size="50"/>
<answer type="loncapa/python">
if chemcalc.chemical_equations_equal(submission[0], 'ZrCl4 + 4 Na -&gt; 4 NaCl + Zr'):
correct = ['correct']
else:
correct = ['incorrect']
</answer>
</customresponse>
<p><b>(b)</b> Write the balanced chemical equation using lowest whole numbers.</p>
<!-- chemformularesponse
as above, with (a, b, c, d) = (1, 4, 4, 1)
-->
<customresponse>
<chemicalequationinput size="50"/>
<answer type="loncapa/python">
if chemcalc.chemical_equations_equal(submission[0], 'ZrCl4 + 4 Na -&gt; 4 NaCl + Zr', exact=True):
correct = ['correct']
else:
correct = ['incorrect']
</answer>
</customresponse>
<p><b>(c)</b> Calculate the amount of zirconium produced (in kg) if a reactor were charged with 111 kg [mathjaxinline]\text{ZrCl}_4[/mathjaxinline] and 11.1 kg [mathjaxinline]\text{Na}[/mathjaxinline].
<numericalresponse answer="11.0">
<responseparam type="tolerance" default="5%" name="tol" description="Numerical Tolerance" />
<textline math="1" />
</numericalresponse> </p>
<br/>
<solution>
<p><b>SOLUTION:</b></p>
<p><b>(b)</b>[mathjax] \text{ZrCl}_{4 } + \mathbf{4 }\text{Na} = \mathbf{4 }\text{NaCl } + \mathbf{1 } \text{Zr} [/mathjax] </p>
<p><b>(c)</b></p>
<p>[mathjaxinline]111\text{kg ZrCl}_4 = 111000/[91.22 + (4 \times 35.45)] = 476 \text{ moles ZrCl}_4[/mathjaxinline] </p>
<p>[mathjaxinline]11.1\text{ kg Na} = 11100/22.99 = 483\text{ moles C}[/mathjaxinline]</p>
<p>stoichiometry of reaction further dictates that on a molar basis there needs to be 4 x amount of [mathjaxinline]\text{Na}[/mathjaxinline] as there is [mathjaxinline]\text{ZrCl}_4[/mathjaxinline], but calculations show there to be a shortfall of [mathjaxinline]\text{Na}[/mathjaxinline] </p>
<p>[mathjaxinline]\therefore[/mathjaxinline] reaction yield is contrained by [mathjaxinline]\text{Na}[/mathjaxinline] present </p>
<p>stoichiometry of reaction further dictates that if [mathjaxinline]\text{Na}[/mathjaxinline] controls the yield, then the amount of [mathjaxinline]\text{Zr}[/mathjaxinline] produced = 1/4 x molar quantity of [mathjaxinline]\text{Na}[/mathjaxinline] = 1/4 x 483 moles = 121 </p>
[mathjaxinline]121 \text{ moles Zr} \times 91.22\text{ g / mol } \mathbf{= 11.0 \text{ kg Zr} }[/mathjaxinline]<br/>
</solution>
</text>
</problem>
common/test/data/full/problem/test_files/choiceresponse_checkbox.xml deleted 100644 → 0
<problem>
<choiceresponse>
<checkboxgroup>
<choice correct="false">
<startouttext />This is foil One.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Two.<endouttext />
</choice>
<choice correct="true">
<startouttext />This is foil Three.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Four.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Five.<endouttext />
</choice>
</checkboxgroup>
</choiceresponse>
<choiceresponse>
<checkboxgroup>
<choice correct="false">
<startouttext />This is foil One.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Two.<endouttext />
</choice>
<choice correct="true">
<startouttext />This is foil Three.<endouttext />
</choice>
<choice correct="true">
<startouttext />This is foil Four.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Five.<endouttext />
</choice>
</checkboxgroup>
</choiceresponse>
<choiceresponse>
<checkboxgroup>
<choice correct="false">
<startouttext />This is foil One.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Two.<endouttext />
</choice>
<choice correct="true">
<startouttext />This is foil Three.<endouttext />
</choice>
<choice correct="true">
<startouttext />This is foil Four.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Five.<endouttext />
</choice>
</checkboxgroup>
</choiceresponse>
</problem>
common/test/data/full/problem/test_files/choiceresponse_radio.xml deleted 100644 → 0
<problem>
<choiceresponse>
<radiogroup>
<choice correct="false">
<startouttext />This is foil One.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Two.<endouttext />
</choice>
<choice correct="true">
<startouttext />This is foil Three.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Four.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Five.<endouttext />
</choice>
</radiogroup>
</choiceresponse>
<choiceresponse>
<radiogroup>
<choice correct="false">
<startouttext />This is foil One.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Two.<endouttext />
</choice>
<choice correct="true">
<startouttext />This is foil Three.<endouttext />
</choice>
<choice correct="true">
<startouttext />This is foil Four.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Five.<endouttext />
</choice>
</radiogroup>
</choiceresponse>
</problem>
common/test/data/full/problem/test_files/coderesponse.xml deleted 100644 → 0
<problem>
<text>
<h2>Code response</h2>
<p>
</p>
<text>
Write a program to compute the square of a number
<coderesponse tests="repeat:2,generate">
<textbox rows="10" cols="70" mode="python"/>
<codeparam>
<initial_display>def square(x):</initial_display>
<answer_display>answer</answer_display>
<grader_payload>grader stuff</grader_payload>
</codeparam>
</coderesponse>
</text>
<text>
Write a program to compute the square of a number
<coderesponse tests="repeat:2,generate">
<textbox rows="10" cols="70" mode="python"/>
<codeparam>
<initial_display>def square(x):</initial_display>
<answer_display>answer</answer_display>
<grader_payload>grader stuff</grader_payload>
</codeparam>
</coderesponse>
</text>
</text>
</problem>
common/test/data/full/problem/test_files/coderesponse_externalresponseformat.xml deleted 100644 → 0
<problem>
<text>
<h2>Code response</h2>
<p>
</p>
<text>
Write a program to compute the square of a number
<coderesponse tests="repeat:2,generate">
<textbox rows="10" cols="70" mode="python"/>
<answer><![CDATA[
initial_display = """
def square(n):
"""
answer = """
def square(n):
return n**2
"""
preamble = """
import sys, time
"""
test_program = """
import random
import operator
def testSquare(n = None):
if n is None:
n = random.randint(2, 20)
print 'Test is: square(%d)'%n
return str(square(n))
def main():
f = os.fdopen(3,'w')
test = int(sys.argv[1])
rndlist = map(int,os.getenv('rndlist').split(','))
random.seed(rndlist[0])
if test == 1: f.write(testSquare(0))
elif test == 2: f.write(testSquare(1))
else: f.write(testSquare())
f.close()
main()
sys.exit(0)
"""
]]>
</answer>
</coderesponse>
</text>
<text>
Write a program to compute the cube of a number
<coderesponse tests="repeat:2,generate">
<textbox rows="10" cols="70" mode="python"/>
<answer><![CDATA[
initial_display = """
def cube(n):
"""
answer = """
def cube(n):
return n**3
"""
preamble = """
import sys, time
"""
test_program = """
import random
import operator
def testCube(n = None):
if n is None:
n = random.randint(2, 20)
print 'Test is: cube(%d)'%n
return str(cube(n))
def main():
f = os.fdopen(3,'w')
test = int(sys.argv[1])
rndlist = map(int,os.getenv('rndlist').split(','))
random.seed(rndlist[0])
if test == 1: f.write(testCube(0))
elif test == 2: f.write(testCube(1))
else: f.write(testCube())
f.close()
main()
sys.exit(0)
"""
]]>
</answer>
</coderesponse>
</text>
</text>
</problem>
common/test/data/full/problem/test_files/formularesponse_with_hint.xml deleted 100644 → 0
<problem>
<script type="loncapa/python">
# from loncapa import *
x1 = 4 # lc_random(2,4,1)
y1 = 5 # lc_random(3,7,1)
x2 = 10 # lc_random(x1+1,9,1)
y2 = 20 # lc_random(y1+1,15,1)
m = (y2-y1)/(x2-x1)
b = y1 - m*x1
answer = "%s*x+%s" % (m,b)
answer = answer.replace('+-','-')
inverted_m = (x2-x1)/(y2-y1)
inverted_b = b
wrongans = "%s*x+%s" % (inverted_m,inverted_b)
wrongans = wrongans.replace('+-','-')
</script>
<text>
<p>Hints can be provided to students, based on the last response given, as well as the history of responses given. Here is an example of a hint produced by a Formula Response problem.</p>
<p>
What is the equation of the line which passess through ($x1,$y1) and
($x2,$y2)?</p>
<p>The correct answer is <tt>$answer</tt>. A common error is to invert the equation for the slope. Enter <tt>
$wrongans</tt> to see a hint.</p>
</text>
<formularesponse samples="x@-5:5#11" id="11" answer="$answer">
<responseparam description="Numerical Tolerance" type="tolerance" default="0.001" name="tol" />
<text>y = <textline size="25" /></text>
<hintgroup>
<formulahint samples="x@-5:5#11" answer="$wrongans" name="inversegrad">
</formulahint>
<hintpart on="inversegrad">
<text>You have inverted the slope in the question.</text>
</hintpart>
</hintgroup>
</formularesponse>
</problem>
common/test/data/full/problem/test_files/imageresponse.xml deleted 100644 → 0
<problem>
<text><p>
Two skiers are on frictionless black diamond ski slopes.
Hello</p></text>
<imageresponse max="1" loncapaid="11">
<imageinput src="/static/Physics801/Figures/Skier-conservation of energy.jpg" width="560" height="388" rectangle="(490,11)-(556,98)"/>
<text>Click on the image where the top skier will stop momentarily if the top skier starts from rest.</text>
<imageinput src="/static/Physics801/Figures/Skier-conservation of energy.jpg" width="560" height="388" rectangle="(242,202)-(296,276)"/>
<text>Click on the image where the lower skier will stop momentarily if the lower skier starts from rest.</text>
<imageinput src="/static/Physics801/Figures/Skier-conservation of energy.jpg" width="560" height="388" rectangle="(490,11)-(556,98);(242,202)-(296,276)"/>
<text>Click on either of the two positions as discussed previously.</text>
<imageinput src="/static/Physics801/Figures/Skier-conservation of energy.jpg" width="560" height="388" rectangle="(490,11)-(556,98);(242,202)-(296,276)"/>
<text>Click on either of the two positions as discussed previously.</text>
<imageinput src="/static/Physics801/Figures/Skier-conservation of energy.jpg" width="560" height="388" rectangle="(490,11)-(556,98);(242,202)-(296,276)"/>
<text>Click on either of the two positions as discussed previously.</text>
<hintgroup showoncorrect="no">
<text><p>Use conservation of energy.</p></text>
</hintgroup>
</imageresponse>
</problem>
common/test/data/full/problem/test_files/javascriptresponse.xml deleted 100644 → 0
<problem>
<javascriptresponse>
<generator src="test_problem_generator.js"/>
<grader src="test_problem_grader.js"/>
<display class="TestProblemDisplay" src="test_problem_display.js"/>
<responseparam name="value" value="4"/>
<javascriptinput>
</javascriptinput>
</javascriptresponse>
</problem>
common/test/data/full/problem/test_files/multi_bare.xml deleted 100644 → 0
<problem>
<multiplechoiceresponse>
<choicegroup>
<choice correct="false" >
<startouttext />This is foil One.<endouttext />
</choice>
<choice correct="false" >
<startouttext />This is foil Two.<endouttext />
</choice>
<choice correct="true" >
<startouttext />This is foil Three.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Four.<endouttext />
</choice>
<choice correct="false">
<startouttext />This is foil Five.<endouttext />
</choice>
</choicegroup>
</multiplechoiceresponse>
</problem>
common/test/data/full/problem/test_files/multichoice.xml deleted 100644 → 0
<problem>
<multiplechoiceresponse>
<choicegroup>
<choice correct="false" name="foil1">
<startouttext />This is foil One.<endouttext />
</choice>
<choice correct="false" name="foil2">
<startouttext />This is foil Two.<endouttext />
</choice>
<choice correct="true" name="foil3">
<startouttext />This is foil Three.<endouttext />
</choice>
<choice correct="false" name="foil4">
<startouttext />This is foil Four.<endouttext />
</choice>
<choice correct="false" name="foil5">
<startouttext />This is foil Five.<endouttext />
</choice>
</choicegroup>
</multiplechoiceresponse>
</problem>
common/test/data/full/problem/test_files/multiplechoicelresponse_demo.xml deleted 100644 → 0
<problem display_name="S3E3: Deflection of a Proton">
<startouttext/>
<p>Suppose you were to pass a hydroge ion (H+) through an electric field and measure the deflection of the ion. How would its deflection compare to the deflection of an electron passed through the same electric field? </p>
<endouttext/>
<multiplechoiceresponse direction="vertical" randomize="yes">
<choicegroup type="MultipleChoice">
<choice location="random" correct="false" name="1"><text>The ion would deflect identically as both species have the same charge.</text></choice>
<choice location="random" correct="false" name="2"><text>The ion would deflect by an amount identical in magnitude but opposite in direction because the charges are of opposite sign but equal in magnitude.</text></choice>
<choice location="random" correct="false" name="3"><text>The deflection will be in the same direction but smaller in magnitude due to the larger mass of the ion.</text></choice>
<choice location="random" correct="true" name="4"><text>The deflection will be smaller in magnitude due to the increased mass of the ion, as well as in the opposite direction due to the opposite sign of the charge.</text></choice>
</choicegroup>
</multiplechoiceresponse>
</problem>
\ No newline at end of file
common/test/data/full/problem/test_files/numericalresponse_demo.xml deleted 100644 → 0
<problem display_name="S2E2: Hydrogen Combustion">
<text>
<p>Gaseous hydrogen can be reacted with gaseous oxygen in the presence of a flame to produce water vapor according to the following reaction:</p>
[mathjax] \text{ a } \text{H}_{2 } + \text{ b }\text{O}_{2 } = \text{ c }\text{H}_{2 }\text{O }[/mathjax]
<p>Balance the equation i.e., specify the values of <i> a, b, </i> and <i>c</i>. Use the lowest whole number coefficients.</p>
<p>a =</p>
<numericalresponse answer="2">
<responseparam type="tolerance" default="1%" name="tol" description="Numerical Tolerance" />
<textline correct_answer="2" math="1" />
</numericalresponse>
<p>b =
<numericalresponse answer="1">
<responseparam type="tolerance" default="1%" name="tol" description="Numerical Tolerance" />
<textline math="1" />
</numericalresponse> </p>
<p>c =
<numericalresponse answer="2">
<responseparam type="tolerance" default="1%" name="tol" description="Numerical Tolerance" />
<textline math="1" />
</numericalresponse> </p>
</text>
</problem>
common/test/data/full/problem/test_files/optionresponse.xml deleted 100644 → 0
<problem>
<text>
<p>
Why do bicycles benefit from having larger wheels when going up a bump as shown in the picture? <br/>
Assume that for both bicycles:<br/>
1.) The tires have equal air pressure.<br/>
2.) The bicycles never leave the contact with the bump.<br/>
3.) The bicycles have the same mass. The bicycle tires (regardless of size) have the same mass.<br/>
</p>
</text>
<optionresponse texlayout="horizontal" max="10" randomize="yes">
<ul>
<li>
<text>
<p>The bicycles with larger wheels have more time to go over the bump. This decreases the magnitude of the force needed to lift the bicycle.</p>
</text>
<optioninput name="Foil1" location="random" options="('True','False')" correct="True">
</optioninput>
</li>
<li>
<text>
<p>The bicycles with larger wheels always have a smaller vertical displacement regardless of speed.</p>
</text>
<optioninput name="Foil2" location="random" options="('True','False')" correct="False">
</optioninput>
</li>
<li>
<text>
<p>The bicycles with larger wheels experience a force backward with less magnitude for the same amount of time.</p>
</text>
<optioninput name="Foil3" location="random" options="('True','False')" correct="False">
</optioninput>
</li>
<li>
<text>
<p>The bicycles with larger wheels experience a force backward with less magnitude for a greater amount of time.</p>
</text>
<optioninput name="Foil4" location="random" options="('True','False')" correct="True">
</optioninput>
</li>
<li>
<text>
<p>The bicycles with larger wheels have more kinetic energy turned into gravitational potential energy.</p>
</text>
<optioninput name="Foil5" location="random" options="('True','False')" correct="False">
</optioninput>
</li>
<li>
<text>
<p>The bicycles with larger wheels have more rotational kinetic energy, so the horizontal velocity of the biker changes less.</p>
</text>
<optioninput name="Foil6" location="random" options="('True','False')" correct="False">
</optioninput>
</li>
</ul>
<hintgroup showoncorrect="no">
<text>
<br/>
<br/>
</text>
</hintgroup>
</optionresponse>
</problem>
common/test/data/full/problem/test_files/stringresponse_with_hint.xml deleted 100644 → 0
<problem >
<text><h2>Example: String Response Problem</h2>
<br/>
</text>
<text>Which US state has Lansing as its capital?</text>
<stringresponse answer="Michigan" type="ci">
<textline size="20" />
<hintgroup>
<stringhint answer="wisconsin" type="cs" name="wisc">
</stringhint>
<stringhint answer="minnesota" type="cs" name="minn">
</stringhint>
<hintpart on="wisc">
<text>The state capital of Wisconsin is Madison.</text>
</hintpart>
<hintpart on="minn">
<text>The state capital of Minnesota is St. Paul.</text>
</hintpart>
<hintpart on="default">
<text>The state you are looking for is also known as the 'Great Lakes State'</text>
</hintpart>
</hintgroup>
</stringresponse>
</problem>
common/test/data/full/problem/test_files/symbolicresponse.xml deleted 100644 → 0
<problem>
<text>
<h2>Example: Symbolic Math Response Problem</h2>
<p>
A symbolic math response problem presents one or more symbolic math
input fields for input. Correctness of input is evaluated based on
the symbolic properties of the expression entered. The student enters
text, but sees a proper symbolic rendition of the entered formula, in
real time, next to the input box.
</p>
<p>This is a correct answer which may be entered below: </p>
<p><tt>cos(theta)*[[1,0],[0,1]] + i*sin(theta)*[[0,1],[1,0]]</tt></p>
<script>
from symmath import *
</script>
<text>Compute [mathjax] U = \exp\left( i \theta \left[ \begin{matrix} 0 &amp; 1 \\ 1 &amp; 0 \end{matrix} \right] \right) [/mathjax]
and give the resulting \(2 \times 2\) matrix. <br/>
Your input should be typed in as a list of lists, eg <tt>[[1,2],[3,4]]</tt>. <br/>
[mathjax]U=[/mathjax] <symbolicresponse cfn="symmath_check" answer="[[cos(theta),i*sin(theta)],[i*sin(theta),cos(theta)]]" options="matrix,imaginary" id="filenamedogi0VpEBOWedxsymmathresponse_1" state="unsubmitted">
<textline size="80" math="1" response_id="2" answer_id="1" id="filenamedogi0VpEBOWedxsymmathresponse_2_1"/>
</symbolicresponse>
<br/>
</text>
</text>
</problem>
common/test/data/full/problem/test_files/truefalse.xml deleted 100644 → 0
<problem>
<truefalseresponse max="10" randomize="yes">
<choicegroup>
<choice location="random" correct="true" name="foil1">
<startouttext />This is foil One.<endouttext />
</choice>
<choice location="random" correct="true" name="foil2">
<startouttext />This is foil Two.<endouttext />
</choice>
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<choice location="random" correct="false" name="foil4">
<startouttext />This is foil Four.<endouttext />
</choice>
<choice location="random" correct="false" name="foil5">
<startouttext />This is foil Five.<endouttext />
</choice>
</choicegroup>
</truefalseresponse>
</problem>
common/test/data/full/sequential/Administrivia_and_Circuit_Elements.xml deleted 100644 → 0
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<html slug="html_68"> S1E4 has been removed…</html>
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<html slug="html_95">Minor correction: Six elements (five resistors)…
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common/test/data/full/sequential/System_Usage_Sequence.xml deleted 100644 → 0
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common/test/data/full/static/js/schematic.js deleted 100644 → 0
This source diff could not be displayed because it is too large. You can view the blob instead.
common/test/data/full/subs/3NIegrCmA5k.srt.sjson deleted 100644 → 0
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common/test/data/full/subs/BGU1poJDgOY.srt.sjson deleted 100644 → 0
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null,
"Let us summarize what we&#39;ve\nlearned so far.",
"So we started out with building\nthis playground, this",
"EECS playground.",
"And within this playground,\nwe&#39;ve all decided that we will",
"abide by the lumped matter\ndiscipline, the LMD.",
"And these are simply constraints\nthat we impose",
"upon ourselves to simplify\nour analysis.",
"Notice that, you really\ndon&#39;t have to do that.",
"You are welcome to hang out here\nand do whatever you want.",
"But you might have a\nlot of fun as well.",
"But, you will not be able to\nbuild things as simply as you",
"would if you were inside\nthe EECS playground.",
"But, Within this playground\nyou have promised that you",
"will follow the lumped\nmatter discipline.",
"Lumped matter discipline has\nthree components to it.",
"One is that the elements will\nhave del phi b by del t equals",
"zero, outside the elements.",
"In other words, when you build\ncircuits, you want to make",
"sure that the del phi b by del\nt is going to be zero.",
"And inside the elements, we are\ngoing to make sure that",
"when you define something--\ndescribe something as a lumped",
"element, del q by del t,\nfor that element--",
"the net del q by del t for\nthat element inside the",
"element is going to be zero.",
"And that&#39;s going to be true\nfor all your elements,",
"registers, sources,\nwires, and so on.",
"And the third element is that\nyou will only focus on signal",
"speeds of interest that\nwill be way lower than",
"the speed of light.",
"Now one of the interesting facts\nof life is that as our",
"microprocessors get faster and\nfaster, some of the chips",
"inside your iPhones, and your\ndesktops, and so on, are",
"clocking at speeds in excess\nof three to four gigahertz.",
"When signal speeds are that\nfast, we are now getting to a",
"point where the signal speeds\ninside our chips are",
"approaching the speed\nof light.",
"So they&#39;re not all\nthat far off.",
"And so therefore, over the next\ndecade, as people try to",
"push speeds to even higher\nvalues, various things are",
"going to start breaking unless\nwe change things",
"dramatically somehow.",
"What this does is these\ndiscipline elements, the",
"lumped matter discipline,\nenables us to create the",
"lumped circuit abstraction.",
"Look at our lumped elements.",
"Lumped elements will\nbe described",
"by their v-i relation.",
"So each of these lumped elements\nof that sum i, that",
"would be sum function of\nv, the v-i relation.",
"The relation could be linear,\ncould be non-linear.",
"Could be, in general, could be\nanything, just that they have",
"some v-i relation.",
"For the register you saw that\ni was v divided by R.",
"For some other element, it could\nbe that i is 6.7 times v",
"squared plus v divided by\n2 plus some constant 1.",
"I mean, who knows.",
"That could be some function of\nv. Could be linear, could be",
"non-linear.",
"And for this element, the power\nconsumed by the element",
"is given by vi.",
"Finally, we showed that when\nyou make the lumped matter",
"discipline assumptions,\nMaxwell&#39;s equations simplify",
"to extremely simple\nKVL and KCL.",
"So, for example, KVL simply\nsays that the sum of the",
"voltages around a\nloop is zero.",
"While KCL says that the sum of\nthe current that go into a",
"node is zero.",
"You can also say the currents\ncoming out of a node are zero",
"by the same principle.",
"What is amazing about this is\nthat by making the lumped",
"matter discipline, by abiding\nby the lumped matter",
"discipline, we have been able\nto convert the use of",
"Maxwell&#39;s equation into the use\nof very simple algebra.",
"So we replaced our del phi&#39;s and\ndel t&#39;s, and all of that",
"stuff, with extremely\nsimple algebra.",
"And in the next set of segments\nI&#39;m going to show you",
"how we can simplify and analyze\ncircuits extremely",
"easily using very, very\nsimple algebra.",
"And, in fact, for the first half\nof this course, you are",
"not going to see any\nsquigglies at all.",
"It&#39;s going to be very,\nvery simple.",
"And even later, we&#39;re not going\nto be seeing any partial",
"differential equations.",
"We will see some fairly simple\ndifferential equations, but by",
"making the LMD assumptions, we\nhave really gotten rid of a",
"lot of the complexity.",
"And things are going to be\nreally, really, really simple.",
null
]
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"text": [
"SPEAKER 1: So I want to show\nyou a simple circuit that",
"looks like this.",
"And let&#39;s go ahead and measure\nsome voltages and currents.",
"In terms of terminology,\nremember",
"this is called a loop.",
"So if I start from the point\nC, and I travel through the",
"voltage source, come to the node\nA, down through R1 and",
"all the way down through R2\nback to C, that&#39;s a loop.",
"Similarly, this point A is a\nnode where the register R1,",
"the voltage source V0 and\nR4 are connected.",
"Just make sure your terminology\nis correct.",
"So what I&#39;ll do is let me make\nsome quick measurements for",
"you, and show you that these KVL\nand KCL are indeed true.",
"So the circuit&#39;s up there.",
"So let me take some\nmeasurements.",
"And why don&#39;t you write down\nwhat I measure on the board.",
null,
"Let me borrow another\npiece of chalk here.",
"What I&#39;ll do is focus on this\nloop here and focus on this",
"node and make some\nmeasurements.",
"So you see the circuit\nup there.",
"[INAUDIBLE]",
null,
"So I get three volts for the\nvoltage from C to A. So why",
"don&#39;t you write down\nthree volts.",
null,
"OK, so the next one\nis minus 1.6.",
null,
"And so that will be\nI&#39;m doing AB, VAB.",
null,
"And then let me do\nthe last one.",
"And it is minus 1.37.",
"Within the bonds of experimental\nerror, notice",
"that if I add up these\nthree voltages, they",
"nicely sum up to zero.",
"OK, next let me focus\non this node here.",
"And at this node, let\nme go ahead and",
"measure some currents.",
null,
"What I&#39;ll do now is change to an\nAC voltage so that I can go",
"ahead and measure the current\nwithout breaking my circuit.",
"This time around, you&#39;ll get to\nsee the measurements that",
"I&#39;m taking as well.",
"What I have here is three wires\nthat I have pulled out",
"from D. And this is the node D.\nSo three wires coming into",
"the node D just to make it a\nlittle bit easier for me to",
"measure stuff.",
"So everybody keep your fingers\ncrossed so I don&#39;t",
"look like a fool here.",
"Hope this works out.",
"So it&#39;s about 10 millivolts\npeak to peak out there.",
"Let&#39;s say that if the wave form\nrises on the left hand",
"side, it&#39;s positive.",
"So it&#39;s positive\n10 millivolts.",
"And another positive\n10 millivolts.",
"So that&#39;s 20 millivolts.",
"And this time it&#39;s a negative,\nroughly 20, I",
"guess, negative 20.",
"So I&#39;m getting, in terms of\ncurrents, I have a positive",
"10, a positive 10 and a negative\n20 that adds up to 0.",
"But more interestingly, I can\nshow you the same thing by",
"holding this current\nmeasuring probe",
"directly across that node.",
"And notice that the net current\nthat is entering into",
"this node here is zero.",
"So that should just show you\nthat KCL does indeed hold in",
"practice, and it is not just a\nfigment of our imaginations.",
null
]
}
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"SPEAKER 1: We just learned the fundamental rules of",
"electrical circuits.",
"For example, if we have a circuit like this, which is",
"called a bridge and which has a voltage source and five",
"resistors--",
"SPEAKER 2: And I'll show the circuit in lecture.",
"SPEAKER 1: That's right.",
"We had a bunch of voltage current constraints, one for",
"each of the branches, each of the elements in the network.",
"For example, this one, the voltage across it is the",
"voltage intrinsic to it.",
"And we have Kirchoff's Current Laws, one for each of the",
"possible nodes in the network.",
"So that's four of them here.",
"It turns out one of them is redundant.",
"It can be derived from the others.",
"And we have Kirchoff Voltage Laws, which are some of the",
"voltages around the loop is zero.",
"And I have four of them written down here, and only",
"three of them are independent.",
"But we can have many more because there's a loop I",
"didn't show, which is the one that looks like this, for",
"example, OK?",
"And the other thing we have is that some of the powers on the",
"network entering each the elements is zero.",
"So there's no power gained or lost. This is a general method",
"of solving all circuit problems. Unfortunately, even",
"for a simple circuit like this, we have a mess of",
"equations here.",
"This is not a good idea.",
"SPEAKER 2: So in the next lecture, what Anant will show",
"you is a much more coherent, a much better organized",
"technique for solving these circuits.",
"So again, if we circle this circuit--",
"this is the same brood circuit Anant showed in the lecture--",
"we have six resistors, one, two, three, four nodes.",
"And what we do is we take each of these nodes and for those",
"nodes we'll write Kirchoff's Current Law.",
"And that's called nodal analysis.",
"So step zero is we pick one of these nodes and we label that",
"node as ground.",
"And labeling it ground just says the voltage on",
"that node is zero.",
"SPEAKER 1: The potential on that node is zero.",
"Voltages are differences of potential.",
"SPEAKER 2: Then we take a node, such as b.",
"And for that node, b, we write down an equation that says the",
"currents going into that node are o0.",
"So IR1 is plus IR2 plus IR3 is equal to zero.",
"We take this node, d, where we also write down Kirchoff's",
"Current Law, IR3 plus IR4 plus IR5 is equal to zero.",
"And at that point, we substitute",
"in for those voltages.",
"So when we have IR3, that's just b minus d over R3.",
"That's the current going into it.",
"For R5 over here, we have just d--",
"IS is 0 minus d.",
"So that's just minus d over R5.",
"And for IR4, we have a minus d over R4.",
"So we have an equation like this.",
"SPEAKER 1: So these labels, a, b and d, are within node",
"potentials, OK?",
"SPEAKER 2: We do the same thing over here for--",
"SPEAKER 1: Right, yes.",
"SPEAKER 2: We do the same thing over here for IR2 is--",
"SPEAKER 1: It's IR1.",
"SPEAKER 2: IR1 is a minus b over R1.",
"IR3 is d minus b over R3.",
"And finally, IR2 is minus b over R2.",
"So now we have two equations.",
"And the last thing that we do is we label this node a to",
"have the voltage v. We say a is equal to v.",
"Since this is a zero, instead of a, it's v, we make that",
"substitution everywhere here.",
"We have two equations, two unknowns.",
"We solve it.",
"And that gives us our answer.",
"SPEAKER 1: So the equations are b and d because all the",
"A's are substituted for.",
"And all of the KCLs that were written with the KCLs for some",
"of the currents entering the node is zero.",
"You could have equally written them as some of the currents",
"leaving the node as zero and get the same answer because",
"that would just multiply everything by minus 1.",
"SPEAKER 2: So you'll see this next week.",
"That's just a preview of coming attractions."
]
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"SPEAKER 1: In this sequence, are\ngoing to look at a whole",
"bunch of techniques for\nanalyzing circuits.",
"Think of these as your tool\nchest. So you are embarking on",
"6.002 and going on to build\nlots of useful electronic",
"systems.",
"But before we go and build these\nsystems, much like a",
"carpenter who needs to go build\na house, much like a",
"carpenter has to go and buy a\nset of tools from the local",
"store, we need to go and make\nsure that you have the tools",
"in your tool chest to analyze\nthese circuits.",
"This tool chest will include\nthings like the KVL KCL",
"method, the Node method\nand so on.",
"But before I begin talking about\nthese techniques for",
"your tool chest, I&#39;d like to do\na extremely quick review--",
"as is my usual practice--",
"of what you&#39;ve covered so far.",
"So remember our EECS\nplayground.",
"We&#39;ve gotten into this place by\npromising that we are going",
"to observe the lumped\nmatter discipline.",
"By observing the discipline\nrecall that the complexities",
"of Maxwell&#39;s equations and the\ndifferential equations that",
"you have to solve-- the\npartial differential",
"equations-- goes away.",
"And you&#39;re left with very\nsimple linear equations.",
"And you will see a lot\nof that today.",
"As you make the lumped matter\ndiscipline, you end up with",
"these lumped elements, like\nresistors and voltage",
"sources and so on.",
"And what you do with these\nlumped elements is we can",
"label what are called the\nbranch variables or the",
"terminal variables for these\nlumped elements, OK?",
"So for example, v is the voltage\nacross the element.",
"And i is the current through\nthe element.",
"The power consumed by the\nelement is given by vi.",
"Continuing with our review, the\nlumped matter discipline",
"enables us to create the lumped\ncircuit abstraction.",
"So you take these lumped\nelements and you connect them",
"with ideal wires.",
"So in this case, I have a\nresistor which is a lumped",
"element, a voltage source.",
"And I connect them with\nthese ideal wires.",
"And what you end up with is\ncalled a lumped circuit.",
"But this is your lumped\ncircuit abstraction.",
"Well as we made the transition\nfrom physics to EECS by",
"adhering to the lumped\nmatter discipline.",
"I had discussed in the last\nsequence that Maxwell&#39;s",
"equations turn into very simple\nalgebraic equations.",
"And these are captured by\nKirchoff&#39;s voltage and current",
"laws, called KVL and KCL.",
"So KVL says that for all loops\nin your circuit, the sum of",
"the voltages around the\nloops add up to zero.",
"Similarly for all the nodes in\nthe circuit, the currents that",
"enter into a node\nadd up to zero.",
"And in the same matter, if I\njust summed the currents",
"leaving a node, they also\nadd up to zero.",
"So whether you sum the currents\nentering the node or",
"currents leaving a node, they\nwould add up to zero by KCL.",
"And the beauty of this was by\nadhering to lump matter",
"discipline, recall the\ncomplexities Maxwell&#39;s",
"equations whether an\nintegral form or",
"differential form went away.",
"And we were now left with\nextremely simple algebraic",
"expressions of the sort.",
"And you will see shortly that\nyou will end up solving",
"extremely simple linear\nequations",
"as we analyze circuits.",
"So as one example here\nis a linear circuit.",
"It has one, two, three, four,\nfive, six elements.",
"It&#39;s got a voltage source\nof a voltage v zero.",
"It&#39;s got a resistor, r one\nand so on and so forth.",
"So for this circuit according\nto our abstraction and the",
"laws of KVL and KCL,\nthe following",
"are going to be true.",
"So for instance if I look at\nnode a, then according to KCL,",
"the currents entering the\nnode must be zero.",
"So in other words, the\ncurrent here is ica.",
"And I can sum to that\nthe current",
"entering from the d direction.",
"So that would be plus ida.",
"And then there is a current from\nthe ba direction and iba.",
"Those are going to\nsum to zero.",
"Now I can multiply the whole\nthing by minus 1.",
"And that would give me the sum\nof the currents leaving the",
"node are also going\nto be zero.",
"So that is KCL.",
"Now I can also write KVL in this\ncase for this loop, for",
"instance, here.",
"For this loop I can add up the\nvoltages around the loop.",
"And by KVL they must\nadd up to zero.",
"So for example if I look at\nthe voltage vca, I can add",
"that to the next voltage here.",
"That is to vab and then\nadd that to the",
"final voltage, vbc.",
"And by KCL--",
null,
"so all of those must\nadd up to be zero.",
"Now if you go back to our\ncircuit here that you see on",
"the left hand side, I am going\nto show you a little demo that",
"will look at these currents,\nica, ida, and iba, and also",
"look at the voltages, vcabab\nand bbc and actually go and",
"look at the circuit, make the\nmeasurements, and show that",
"they indeed sum up to zero.",
"Not surprising.",
"KVL and KCL really work.",
"And so the demo should\nalso work.",
"OK.",
"So let&#39;s go and do our demo.",
null
]
}
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null,
"PROFESSOR ANANT AGARWAL:\nWelcome to MITx, MIT&#39;s",
"worldwide online learning\ninitiative.",
"We are delighted to have you\njoin us for 6.002x: Circuits",
"and Electronics, the inaugural\ncourse of MITx.",
"OK, welcome to 6.002.",
"6.002x is the same as Electronic\nCircuits course",
"taken by MIT students as the\nfirst course in an EE or EECS",
"major at MIT.",
"This course will help you make\nthe transition from physics to",
"electrical engineering.",
"In this course, you will learn\nabout a lot of fun things,",
"including the foundations behind\nmany cool devices, such",
"as the smartphone.",
"The course will be hard and we\nwant you to stick with us.",
"It is our hope that you\nwill have a lot",
"of fun in this course.",
"Let me start by introducing\nyou to our team.",
"I am Anant Agarwal.",
"I am the director of MIT&#39;s\nComputer Science and",
"Artificial Intelligence\nLaboratory or CSAIL.",
"Like most MIT faculty, I have\na wide range of interests,",
"including a fascination\naround chainsaws.",
null,
"GERRY SUSSMAN: Hello, I&#39;m Gerry\nSussman, professor of",
"electrical engineering at MIT.",
"Been a faculty member\nhere since 1973.",
"I build electronics, optics,\ncomputer software, and I",
"repair watches as a hobby.",
null,
"PIOTR MITROS: I&#39;m\nPiotr Mitros.",
"I&#39;m a research scientist at MIT,\nand I was in charge of",
"developing the software\ninfrastructure for MITx.",
"As a graduate student, I spent a\nlot of time traveling around",
"India, Nigeria, and China,\nmostly as part of educational",
"technology projects.",
"So I&#39;m very excited to be\nhelping out with this course.",
"PROFESSOR ANANT AGARWAL: Whether\nwe like chainsaws or",
"watches, we are all passionate\neducators.",
"This course is free.",
"You will have the opportunity\nto earn an MITx pilot",
"certificate with your grade\nat the end of this course.",
"In this experimental first\ncourse of MITx, the",
"certificates are also free.",
"Now, this is a prototype course\nand we welcome your",
"participation and any feedback\nyou might have to improving",
"the learning experience.",
"Let&#39;s get started.",
null
]
}
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"SPEAKER 1: In this next segment,\nI&#39;d like to show you",
"a little demo of a set of lumped\nelements that you will",
"find in 6002x and in building\nelectronic circuits and",
"systems in general.",
"And these lumped elements have\nbehavior that is completely",
"captured by their V-I\nrelationship.",
"I also want to point out that\nthe behavior of these lumped",
"elements is captured by the\nV-I relationship only for",
"those sorts of questions that\nwe, as electrical engineers,",
"would like to ask.",
"Understand that the V-I relation\ndoesn&#39;t capture all",
"the possible behaviors\nof the element.",
"It only captures those possible\nbehaviors that we, as",
"EEs, generally like to ask.",
"Once I&#39;ve shown you these basic\nV-I relationships, we",
"will do some other fun stuff.",
"I&#39;m going to show you some\nexamples of behavior that the",
"V-I relationship will\nnot capture.",
"We&#39;ll show you what\nwe can predict.",
"For example, we&#39;ll\nshow you how the",
"resistors don&#39;t work sometimes.",
"And sometimes we will show you\nanother example of a cucumber",
"pickle that may behave in ways\nthat you don&#39;t quite expect",
"and whose behavior need not\nnecessarily be captured by",
"your V-I relationship.",
"So I don&#39;t want you to go around\nsaying that you don&#39;t",
"have any fun in this course.",
"So hopefully, you will\nenjoy what you are",
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null
]
}
\ No newline at end of file
common/test/data/full/tabs/resources.html deleted 100644 → 0
<p>This is another sample tab</p>
\ No newline at end of file
common/test/data/full/tabs/syllabus.html deleted 100644 → 0
<h1>This is a sample tab</h1>
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common/test/data/full/to_delete deleted 100644 → 0
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common/test/data/full/vertical/test_problems.xml deleted 100644 → 0
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<problem url_name="test_files:coderesponse_externalresponseformat"/>
<problem url_name="test_files:formularesponse_with_hint"/>
<problem url_name="test_files:imageresponse"/>
<problem url_name="test_files:javascriptresponse"/>
<problem url_name="test_files:multi_bare"/>
<problem url_name="test_files:multichoice"/>
<problem url_name="test_files:multiplechoicelresponse_demo"/>
<problem url_name="test_files:numericalresponse_demo"/>
<problem url_name="test_files:optionresponse"/>
<problem url_name="test_files:stringresponse_with_hint"/>
<problem url_name="test_files:symbolicresponse"/>
<problem url_name="test_files:truefalse"/>
<problem url_name="test_files:chemicalequationresponse"/>
<problem url_name="test_files:filesubmission"/>
</vertical>
\ No newline at end of file
common/test/data/full/vertical/vertical_58.xml deleted 100644 → 0
<sequential>
<video youtube_id_1_5="&quot;8kARlsUt9lM&quot;" youtube_id_1_25="&quot;4cLA-IME32w&quot;" youtube_id_1_0="&quot;pFOrD8k9_p4&quot;" youtube_id_0_75="&quot;CcgAYu0n0bg&quot;" slug="S1V9_Demo_Setup_-_Lumped_Elements" name="S1V9: Demo Setup - Lumped Elements"/>
<customtag tag="S1" slug="discuss_59" impl="discuss"/>
<customtag page="29" slug="book_60" impl="book"/>
<customtag lecnum="1" slug="slides_61" impl="slides"/>
</sequential>
common/test/data/full/vertical/vertical_89.xml deleted 100644 → 0
<sequential>
<html slug="html_90">
<!-- UTF-8 characters are acceptable… HTML entities are not -->
<h1>Inline content…</h1>
</html>
<video youtube_id_1_5="&quot;vl9xrfxcr38&quot;" youtube_id_1_25="&quot;qxNX4REGqx4&quot;" youtube_id_1_0="&quot;BGU1poJDgOY&quot;" youtube_id_0_75="&quot;8rK9vnpystQ&quot;" slug="S1V14_Summary" name="S1V14: Summary"/>
<customtag tag="S1" slug="discuss_91" impl="discuss"/>
<customtag page="70" slug="book_92" impl="book"/>
<customtag lecnum="1" slug="slides_93" impl="slides"/>
<poll_question name="T1_changemind_poll_foo" display_name="Change your answer" reset="false">
<p>Have you changed your mind?</p>
<answer id="yes">Yes</answer>
<answer id="no">No</answer>
</poll_question>
</sequential>
common/test/data/full/vertical/vertical_94.xml deleted 100644 → 0
<sequential>
<video youtube_id_0_75="&quot;3NIegrCmA5k&quot;" youtube_id_1_0="&quot;eLAyO33baQ8&quot;" youtube_id_1_25="&quot;m1zWi_sh4Aw&quot;" youtube_id_1_5="&quot;EG-fRTJln_E&quot;" slug="S2V1_Review_KVL_KCL" name="S2V1: Review KVL, KCL"/>
<customtag tag="S2" slug="discuss_95" impl="discuss"/>
<customtag page="54" slug="book_96" impl="book"/>
<customtag lecnum="2" slug="slides_97" impl="slides"/>
</sequential>
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