HW-16Solutions-03-03-08

# HW-16Solutions-03-03-08 - Lehigh University HW-16 Solutions...

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1 Lehigh University Physics 21, Spring 2008 March 3, 2008 HW-16 Solutions 16-1. (HRW 30-59 ) In Fig. 30-70 , after switch S is closed at time , the emf of the source is automatically adjusted to maintain a constant current i through S. (a) Find the current through the inductor as a function of time. (b) At what time is the current through the resistor equal to the current through the inductor? SSM Figure 30-70 Solution: 59. (a) We assume i is from left to right through the closed switch. We let i 1 be the current in the resistor and take it to be downward. Let i 2 be the current in the inductor, also assumed downward. The junction rule gives i = i 1 + i 2 and the loop rule gives i 1 R – L ( di 2 / dt ) = 0. According to the junction rule, ( di 1 / dt ) = – ( di 2 / dt ). We substitute into the loop equation to obtain L di dt iR 1 1 0 += . This equation is similar to Eq. 30-46, and its solution is the function given as Eq. 30-47: ii e Rt L 10 = , where i 0 is the current through the resistor at t = 0, just after the switch is closed. Now just after the switch is closed, the inductor prevents the rapid build-up of current in its branch, so at that moment i 2 = 0 and i 1 = i . Thus i 0 = i , so ( ) 12 1 ,1 . Rt L Rt L e ii e −− == = (b) When i 2 = i 1 , 1 1.

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HW-16Solutions-03-03-08 - Lehigh University HW-16 Solutions...

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