HW05-solutions - Munoz (gm7794) HW05 TSOI (92515) This...

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Munoz (gm7794) – HW05 – TSOI – (92515) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points Abatteryw ithanemFoF15 . 6Vandinterna l resistance oF 0 . 573 Ω is connected across a load resistor R . IF the current in the circuit is 2 . 23 A, what is the value oF R ? Correct answer: 6 . 42252 Ω. Explanation: The electromotive Force E is given by, E = I ( R + R i ) Thus the resistance in question is, R = E I - R i =6 . 42252 Ω The power dissipation due to the internal re- sistance is, P = I 2 R i =2 . 84947 W 002 (part 2 oF 2) 10.0 points What power is dissipated in the internal re- sistance oF the battery? Correct answer: 2 . 84947 W. Explanation: P = I 2 R i . 003 10.0 points Unlike most real bulbs, the resistance oF the bulb in this question does not change as the current through it changes. A capacitor, a bulb, and a switch are in the circuit as shown below. C R S The switch is initially open as shown in the above diagram, and the capacitor is charged. C R S Which oF the Following correctly describes what happens to the bulb when the switch is closed? 1. The bulb is dim and remains dim. 2. The bulb is bright and remains bright. 3. At frst the bulb is bright and it gets dimmer and dimmer until it goes o±. correct 4. At frst the bulb is dim and it gets brighter and brighter until the brightness levels o±. 5. None oF these is correct. Explanation: When the switch is closed, the bulb has the same potential di±erence as the capacitor. As the capacitor discharges, the potential diF- Ference across the bulb decreases and it gets dimmer and dimmer. 004 (part 1 oF 2) 10.0 points The currents are ²owing in the direction in- dicated by the arrows. A negative current denotes ²ow opposite to the direction oF the arrow. Assume the batteries have zero inter- nal resistance. 14 Ω 24 . 5 . 8V 18 . 5V I I I
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Munoz (gm7794) – HW05 – TSOI – (92515) 2 Find the current through the 14 Ω resistor and the 5 . 8Vbatteryatthetopofthecircuit. Correct answer: 1 . 73571 A. Explanation: Let : R 1 =14Ω , R 2 =24 . , E 1 =5 . 8V , and E 2 =18 . 5V . R 1 R 2 E 1 E 2 I 1 I 2 I 3 At nodes, we have I 1 - I 2 - I 3 =0 . (1) Pay attention to the sign of the battery and the direction of the current in the ±gure. Us- ing the lower circuit in the ±gure, we get E 2 + I 2 R 2 (2) so I 2 = -E 2 R 2 = - 18 .
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HW05-solutions - Munoz (gm7794) HW05 TSOI (92515) This...

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