# 24 - Shie Gary – Homework 24 – Due Nov 1 2004 4:00 am...

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Unformatted text preview: Shie, Gary – Homework 24 – Due: Nov 1 2004, 4:00 am – Inst: Turner 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A certain circuit consists of an inductor of 19 mH in series with a resistor of 145 Ω. At a moment when the current in the circuit is 24 A, a switch in the circuit is opened. How long will it take for the current to fall to 8 . 88 A? Correct answer: 0 . 000131034 s. Explanation: The time constant τ of a circuit gives the time required for the current to fall to 0.37 times its initial value. For an RL-circuit, τ = L R . Here you can easily check that I f I = 0 . 37, so that the decay time to I f is given by τ = . 019 H 145 Ω = 0 . 000131034 s . If you did not notice this, then note that in general the current at time t in such a circuit is given by I ( t ) = I e- t R/L = I e- t/τ , which can be solved for t as t = τ ln µ I I f ¶ . 002 (part 1 of 1) 10 points An inductor that has a resistance of 98 kΩ is connected to an ideal battery of 94 V. 2 . 6 ms seconds after the switch is thrown the current in the circuit is 0 . 326122 mA. Calculate the inductance. Correct answer: 613 . 214 H. Explanation: Basic Concepts: RL circuits The magni- tude of the current in the circuit as a function of time is I ( t ) = E R 1- e- R L t . Knowing that at t 1 = 2 . 6 ms the current is I 1 = 0 . 326122 mA, we can solve for L . First, e- R L t = 1- I 1 R E and finally, L =- t R ln µ 1- I 1 R E ¶ = 613 . 214 H ....
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24 - Shie Gary – Homework 24 – Due Nov 1 2004 4:00 am...

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