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Unformatted text preview: Wong, Lemuel Homework 12 Due: Apr 23 2007, 11:00 pm Inst: Matt Mackie 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. NOTE the due time is 11pm central, which is MIDNIGHT our time. 001 (part 1 of 2) 10 points At t = 0, a 15 . 8 V battery is connected to a series circuit containing a 10 resistor and a 2 . 6 H inductor. What will be the current in the circuit, a long time after the circuit is established? Correct answer: 1 . 58 A. Explanation: Let : E = 15 . 8 V and R = 10 . After a long time, we have a simple circuit of a battery with E and a resistor of R . Then, E = I R and thus the current is I = E R = 15 . 8 V 10 = 1 . 58 A . 002 (part 2 of 2) 10 points How long will it take the current to reach 50 percent of its final value? Correct answer: 0 . 180218 s. Explanation: Let : L = 2 . 6 H and I = 0 . 5 . The time constant is = L R . The current in an RL circuit is given by I = I s 1 e t/ , where I s is the steady state current E R and is the time constant of the circuit. Solving the above equation for t , we obtain t = ln 1 I I s = L R ln 1 . 5 I s I s = 2 . 6 H 10 ln(1 . 5) Therefore, the time at which I = 0 . 5 I s is t 1 = (0 . 26 s) ln(1 . 5) = . 180218 s . keywords: 003 (part 1 of 1) 10 points A certain circuit consists of an inductor of 45 mH in series with a resistor of 120 . At a moment when the current in the circuit is 21 A, a switch in the circuit is opened. How long will it take for the current to fall to 7 . 77 A? Correct answer: 0 . 000375 s. Explanation: Let : L = 45 mH , R = 120 , I = 21 A , and I f = 7 . 77 A . The time constant of a circuit gives the time required for the current to fall to 0.37 times its initial value. For an RLcircuit, = L R . Here you can easily check that I f I = 0 . 37, so that the decay time to I f is given by = . 045 H 120 = . 000375 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/ , which can be solved for t as t = ln I I f . Wong, Lemuel Homework 12 Due: Apr 23 2007, 11:00 pm Inst: Matt Mackie 2 keywords: 004 (part 1 of 2) 10 points An inductor and a resistor are connected with a double pole switch to a battery as shown in the figure. The switch has been in position b for a long period of time. 161 mH 6 . 75 6 . 5 V S b a If the switch is thrown from position b to position a (connecting the battery), how much time elapses before the current reaches 154 mA? Correct answer: 4 . 15647 ms....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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