ch30-p059

# ch30-p059 - t = 0 just after the switch is closed Now just...

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59 (a) Assume i is from left to right through the closed switch. Let i 1 be the current in the resistor and take it to be downward. Let i 2 be the current in the inductor and also take it to be downward. The junction rule gives i = i 1 + i 2 and the loop rule gives i 1 R L ( di 2 /dt ) = 0. Since di/dt =0 , the junction rule yields ( di 1 /dt )= ( di 2 /dt ). Substitute into the loop equation to obtain L di 1 dt + i 1 R =0 . This equation is similar to Eq. 30 44, and its solution is the function given as Eq. 30 45: i 1 = i 0 e Rt/L , where i 0
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Unformatted text preview: 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 time, i 2 = 0 and i 1 = i . Thus i = i , so i 1 = ie − Rt/L and i 2 = i − i 1 = i ± 1 − e − Rt/L ² . (b) When i 2 = i 1 , e − Rt/L = 1 − e − Rt/L , so e − Rt/L = 1 2 . Take the natural logarithm of both sides and use ln(1 / 2) = − ln 2 to obtain ( Rt/L ) = ln 2 or t = L R ln 2 ....
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## This test prep was uploaded on 04/15/2008 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.

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