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## Q. 5.13

Inductor Current and Voltage

Consider the circuit shown in Figure 5.34. The switch closes at t = 0 after being opened for a long time period. Find $i_{L}\left(0^{+}\right), v_{L}\left(0^{+}\right)$, and $i_{L}(\infty)$.

## Verified Solution

For t < 0, the switch is open for a long time period. Therefore, the inductor can be represented by a short circuit and the equivalent circuit is shown in Figure 5.34(b). The inductor current can be calculated as follows:

\begin{aligned}i_{L}\left(0^{+}\right)=i_{L}\left(0^{-}\right) &=\frac{30}{100+50} \\&=0.2 A\end{aligned}

Closing the switch short-circuits the 100-Ω resistor. As a result:

\begin{aligned}v_{L}\left(0^{+}\right) &=30-50 \times i_{L}\left(0^{+}\right) \\&=30-50 \times 0.2 \\&=20 V\end{aligned}

At t = ∞ , or steady state, the inductor becomes a short circuit. Thus:

$i_{L}(\infty)=\frac{30}{50}=0.6 A$