In the figure below after switch S is closed at time t 0 th

In the figure below, after switch S is closed at time t = 0, the emf of the source is automatically adjusted to maintain a constant current j = 1. 34 A through S. (a) Find the current through the inductor at time t = 10TL. (b) What is the time when the current through the resistor equal to the current through the inductor (assume that TL = 4. 0 s)?

Solution

After the switch is closed, the initial current will only flow through the resistor, since current through an inductor cannot change instantaneously. The steady state current will only flow through the inductor due to lower impedance. To make things easier, we can change the parallel current source and resistor to their thevenin equivalent series voltage source and resistor.
Vth = I*R
Rth = R

So now the series equations will apply:
v.L(t) = (IR)exp(-tR/L) = L*di/dt
i.L(t) = I*(1-exp(-tR/L)) = 1.34 (1-e^-10) = 1.339 A

Resistor & inductor currents are equal when:
i.L(t) = I*(1-exp(-tR/L)) = 0.5*I
Solving for time t gives:
t = -(L/R)ln(0.5) = 4* .693 = 2.77 s