# OHW 15 - terry (ect328) oldhomework 15 Turner (59130) This...

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terry (ect328) – oldhomework 15 – Turner – (59130) 1 This print-out should have 11 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 The internal resistance r oF a battery with emF E is connected to a load resistor with resistance R . I 10 V A B 21 Ω 144 Ω internal resistance ±ind the potential di²erence V BA = V A - V B . Correct answer: 8 . 72727 V. Explanation: Let : E = 10 V , R = 144 Ω , and r = 21 Ω . The current through the circuit is i = E r + R , so the potential di²erence V A - V B is V A - V B = i R = E R r + R = (10 V) (144 Ω) 21 Ω + 144 Ω = 8 . 72727 V . 002 (part 2 oF 2) 10.0 points What is the power P dissipated by the load resistor R ? 1. P = E 2 r 2. P = p E r + R P 2 r 3. P = r E 2 4. P = p E r + R P 2 R correct 5. P = E 2 r + R 6. P = E 2 R r 7. P = E 2 r R 8. P = p E R P 2 r 9. P = ( r + R ) E 2 10. P = R E 2 Explanation: The power dissipated by the load is P = i 2 R = p E r + R P 2 R . 003 (part 1 oF 3) 10.0 points An electric heater is rated at 1220 W, a toaster is rated at 1030 W, and an electric grill is rated at 1530 W. The three appliances are connected in parallel across a 112 V emF source. ±ind the current in the heater. Correct answer: 10 . 8929 A. Explanation: Let : P H = 1220 W , P T = 1030 W , P G = 1530 W and Δ V = 112 V . P H = I H Δ V I H = P H Δ V = 1220 W 112 V = 10 . 8929 A .

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terry (ect328) – oldhomework 15 – Turner – (59130) 2 004 (part 2 of 3) 10.0 points Find the current in the toaster. Correct answer: 9
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## This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.

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OHW 15 - terry (ect328) oldhomework 15 Turner (59130) This...

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