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Course-v1:Arbisoft+CS200+Eng/en/block-v1:Arbisoft+CS200+Eng+type@problem+block@3370b6c207a1473e8a55c982ccd5fb4d
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{"@metadata": {"sourceLanguage": "en", "priorityLanguages": ["eo"], "allowOnlyPriorityLanguages": true}, "problem.p.1": "\n Circuit schematic problems allow students to create virtual circuits by\n arranging elements such as voltage sources, capacitors, resistors, and\n MOSFETs on an interactive grid. The system evaluates a DC, AC, or\n transient analysis of the circuit.\n ", "problem.p.2": "\n For more information, see\n ", "problem.p.2.a": "\n Circuit Schematic Builder Problem", "problem.p.2.i": "Building and Running an edX Course", "problem.p.3": "\n When you add the problem, be sure to select ", "problem.p.3.strong.1": "Settings", "problem.p.3.strong.2": "Display Name", "problem.p.4": "You can use the following example problems as models.", "problem.schematicresponse.1.p": "Make a voltage divider that splits the provided voltage evenly.", "problem.schematicresponse.1.answer": "\ndc_value = \"dc analysis not found\"\nfor response in submission[0]:\n if response[0] == 'dc':\n for node in response[1:]:\n dc_value = node['output']\n\nif dc_value == .5:\n correct = ['correct']\nelse:\n correct = ['incorrect']\n", "problem.schematicresponse.1.solution.div.p.1": "Explanation", "problem.schematicresponse.1.solution.div.p.2": "\n You can form a voltage divider that evenly divides the input\n voltage with two identically valued resistors, with the sampled\n voltage taken in between the two.\n ", "problem.schematicresponse.2.p": "Make a high-pass filter.", "problem.schematicresponse.2.answer": "\nac_values = None\nfor response in submission[0]:\n if response[0] == 'ac':\n for node in response[1:]:\n ac_values = node['NodeA']\nprint \"the ac analysis value:\", ac_values\nif ac_values == None:\n correct = ['incorrect']\nelif ac_values[0][1] < ac_values[1][1]:\n correct = ['correct']\nelse:\n correct = ['incorrect']\n", "problem.schematicresponse.2.solution.div.p.1": "Explanation", "problem.schematicresponse.2.solution.div.p.2": "\n You can form a simple high-pass filter without any further\n constraints by simply putting a resistor in series with a\n capacitor. The actual values of the components do not really\n matter in this problem.\n ", "display_name": "Circuit Schematic Builder"}
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