24 - Kapoor(mk9499 – oldhomework 24 – Turner –(60230 1 This print-out should have 9 questions Multiple-choice questions may continue on the

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Unformatted text preview: Kapoor (mk9499) – oldhomework 24 – Turner – (60230) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points The switch in the circuit shown is closed at time t = 0. 4 H 6 Ω 20 V S I At what rate is energy being dissipated as Joule heat in the resistor after an elapsed time equal to the time constant of the circuit? Correct answer: 26 . 6384 W. Explanation: L R E S I Let : R = 6 Ω and E = 20 V . The current as a function of time in an RL circuit that has zero initial current is given by I = E R parenleftBig 1- e − t/τ parenrightBig which for the case where t = τ , gives I = E R ( 1- e − 1 ) = 20 V 6 Ω parenleftbigg 1- 1 e parenrightbigg = 2 . 10707 A . The power dissipated in the resistor is thus P = I 2 R = (2 . 10707 A) 2 (6 Ω) = 26 . 6384 W . 002 (part 2 of 4) 10.0 points Calculate the rate at which energy is being stored in the inductor at this time. Correct answer: 15 . 5029 W. Explanation: The energy stored in the inductor is U L = 1 2 L I 2 , with I ( t ) given in the previous question. To find the time rate of change we take the derivative with respect to time, giving us the rate energy being stored in the inductor P L = d U L dt = L I d I dt = L I E R τ e − t/τ = L I E R τ e , since t = τ = L R , so P L = L I E R R L 1 e = I E 1 e = (2 . 10707 A) (20 V) 1 e = 15 . 5029 W ....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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24 - Kapoor(mk9499 – oldhomework 24 – Turner –(60230 1 This print-out should have 9 questions Multiple-choice questions may continue on the

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