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In the figure below, suppose that ? = 56.?V, R = 145 ?, and L = 0.160 H. With sw

ID: 2032797 • Letter: I

Question

In the figure below, suppose that ? = 56.?V, R = 145 ?, 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. W 0000 (a) What is the initial current in the resistor, just after S2 is closed and Si 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-4.00x10-4 s? VWhich point is at a higher potential? Both points are at the same potential Point b is at a higher potential Point c is at a higher potential. (d) How long does it take the current to decrease to half its initial value?

Explanation / Answer

a) when s1 is closed for long time  

inductor will act as a short circuit

=> constant current in the circuit , I = E/R = 56/145 = 0.386 A  

b) NOW s2 is closed and s1 is opened

=> now this is LR discharging circuit without source.

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

at t = 4.0*10^-4 sec , time constant T = L/R = 0.160/145 = 1.103*10^-3 sec

=> I(t) = 0.386*e^(-4*10^-4/(1.103*10^-3)) = 0.268 A  

c) at t = 4.0*10^-4 sec

Vbc = - i*R = 0.268*145 = - 38.86 V  

point c is at heigher potential.

d) I(t) =I/2

=> I(t) = I*e^(-t/T)  

=> 1/2 = e^(-t/T)

=> t = T*0.693 = 1.103*10^-3*0.693 = 0.764*10^-4 sec