## Chapter 5

## Q. 5.10

**Inductor Voltage**

Find an expression for v_L (t) in the circuit shown in Figure 5.29. Switch is opened at t = 0 after being closed for a long time.

## Step-by-Step

## Verified Solution

This problem can be solved in a way similar to the previous example using Equation (5.31). The time constant is:

\begin{aligned}i_{L}(t) &=I_{s}\left(1-e^{-\frac{t}{L / R}}\right) \\&=I_{s}\left(1-e^{-\frac{t}{\tau}}\right)\end{aligned} (5.31)

\begin{aligned}\tau &=\frac{L}{R}=\frac{20 \times 10^{-3}}{5 \times 10^3}=4 \mu s \\i_{L}(t) &=0.01\left(1-e^{-\frac{t}{4 \times 10^{-6}}}\right)=0.01\left(1-e^{-2.5 \times 10^5 t}\right)\end{aligned}For the circuit shown in Figure 5.29:

\begin{aligned}v_{L}(t) &=L \frac{di_{L}(t)}{dt} \\&=20 \times 10^{-3} \times 0.01 \frac{d\left(1-e^{-2.5 \times 10^5 t}\right)}{dt} \\&=0.2 \times 10^{-3}\left(0+2.5 \times 10^5 e^{-2.5 \times 10^5 t}\right) \\&=50 e^{-2.5 \times 10^5 t} V\end{aligned}