In the figure below, suppose that = 63.0 V, R = 125 , and L = 0.160 H. With swit

Question

In the figure below, suppose that = 63.0 V, R = 125 , and L = 0.160 H. With switch S2 open, switch S1 is left closed until a constant current is established. Then S2 is closed and S1 opened, taking the battery out of the circuit. S, S2 w 0000 (a) What is the initial current in the resistor, just after S2 is closed and S1 is opened? (b) What is the current in the resistor at t = 4.00 x 10-4 s? (c) What is the potential difference between points b and c at t 4.00 x 10-4 s? VWhich point is at a higher potential? Point c is at a higher potential. Both points are at the same potential Point b is at a higher potential. (d) How long does it take the current to decrease to half its initial value?

Explanation / Answer

Here,

E = 63 V

R = 125 Ohm

L = 0.160 H

a) just when S2 is closed and S1 is open , the current in the resistance will not change

I = E/R = 63/125

I = 0.504 A

b)time constant , T = L/R

T = 0.160/125 = 1.28 *10^-3 s

let the current in resistor is I

I = 0.504 * e^(-t/T)

I = 0.504 * e^(4 *10^-4/(-1.28 *10^-3))

I= 0.37 A

the current is 0.37 A

c)

as the direction of current in the inductor will be to the right and hence , the point a will be at the higher potential.

d)

let the time taken is t

0.504/4 = 0.504 * e^(t/(-1.28 *10^-3))

solving for t

t = 0.00177 s

the time taken is 0.00177 s

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