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MIT 16 01 - Light Bulb Concept Test

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Light Bulb II Concept Test SCAB24 V12 VThe light bulbs in the circuit are identical. When the switch is closed, 1. The intensities of light bulbs B and C increase. 2. The intensities of light bulbs B and C decrease. 3. The intensity of light bulb A increases. 4. The intensity of light bulb A decreases. 5. A combination of 1 and 4 6. A combination of 2 and 3 7. Nothing happens 8. None of the aboveLight Bulb II Solution 0 5 10 15 20 25 30 35 4012345678Number of StudentsAnswerLecture S5: Light Bulb II0 10 20 30 40 5012345678Number of StudentsAnswerLecture S5: Light Bulb IIBefore the switch is closed, the current through the three bulbs must be the same. Thus, the voltage drop across each bulb is the same. Because the total voltage drop across the three bulbs is 24 V, the drop across each bulb must be 8 V. Therefore, if we designate the lower node as ground, thepotential of the node to the left of the switch is 16 V. After the switch is closed, the potential of the node to the left of the switch must be 12 V, since the upper terminal of the battery is 12 V above ground potential. Therefore, the voltage across bulb A increases, and therefore its intensity increases. Likewise, the total voltage across bulbs B and C decreases, and hence there intensities decrease. This is a combination of #2 and #3, so answer #6 is correct. By the second try, most students had either answer #2 (which is partially correct) or #6 (the correct answer). The answer is a little surprsing — adding a battery in parallel to the lower bulbs makes them dimmer, not brighter.Light Bulb III Concept Test SCAB24 V12 VThe light bulbs in the circuit are identical. After the switch is closed, the 12 V battery 1. Supplies power to the circuit. 2. Absorbs power. 3. Neither of the above.Light Bulb III Solution 0 10 20 30 40123Number of StudentsAnswerLecture S5: Light Bulb IIIBefore the switch is closed, the current through the three bulbs must be the same. Thus, the voltage drop across each bulb is the same. Because the total voltage drop across the three bulbs is 24 V, the drop across each bulb must be 8 V. Therfore, if we designate the lower node as ground, the potential of the node to the left of the switch is 16 V. After the switch is closed, the potential of the node to the left of the switch must be 12 V, since the upper terminal of the battery is 12 V above ground potential. Therefore, the voltage across bulb A increases, and therefore its the current through it increases. Likewise, the total voltage across bulbs B and C decreases, and hence the current through them decreases. So the current through bulb A is greater than the current through bulb B. To satisfy KCL at the node to the left of the switch, there most be current to the right through the switch, and down through the 12 V battery. With current .o wing into the positive terminal, the current through the battery is positive (if we are using the passive sign convention), and therefore the power iv > 0, and hence the battery absorbs power. The correct answer is #2. In effect, this circuit is a sort of battery charger, using the 24 V battery to charge the 12 V battery.Loop Method Concept Test R2R4R5R3IVe1e2_+iaib-IFor the network above, .nd the equation that expresses Kirchhoff’s Voltage Law for the loop ib. My answer was 1. −R3ia + (R3 + R4 + R5)ib = R5I 2. −R3ia + (R3 + R4 + R5)ib = −R5I 3. −R3ia + (R3 + R4)ib = R5I 4. Not among the answers above 5. I was unable to get an answerLoop Method Solution 0 5 10 15 20 25 30 35 4012345Number of StudentsAnswerLecture S5: Loop MethodThe correct answer is Number 2. The voltage drop across R4 (in the ib direction) is R4ib. The voltage drop across R5 (in the ib direction) is R5(ib − (−I)) = R5(ib + I). The voltage drop across R3 (in the ib direction) is R3(ib − ia). The total voltage drop around the loop is zero, so −R3ia + (R3 + R4)ib + R5I = 0 Therefore, −R3ia + (R3 + R4)ib = −R5I Many students answered #1. It is a little tricky. To correctly get the minus sign on the right hand side, you must recognize that there are three minus signs .oating around — one because the current in the right hand loop is −I to match the direction of the current source, one because the loop as drawn .o ws in the opposite direction as ib through R5, and one because the source term has been moved to the right hand side of the


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MIT 16 01 - Light Bulb Concept Test

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