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## Q. 5.9

Inductor Current

Find an expression for $i_L (t)$ in the circuit shown in Figure 5.28. Assume the switch has been closed for a long time and opens at t = 0.

## Verified Solution

KCL for Node A corresponds to:

$-i_{s}+i_{R}(t)+i_{L}(t)=0$

Because $v_{R}(t)=v_{L}(t)$

$i_{R}(t)=\frac{v_{L}(t)}{6.4 k \Omega}$

$v_{L}(t)=200 \mu H \frac{di_{L}(t)}{dt}$

Replacing for $i_{R}(t)$ and $v_{L}(t)$ in the node A KCL equation:

$-25 \times 10^{-3}+\frac{200 \times 10^{-6}}{6400} \frac{di_{L}(t)}{dt}+i_{L}(t)=0$

A simple mathematical manipulation leads to:

$\frac{di_{L}(t)}{i_{L}(t)-25 \times 10^{-3}}=-32 \times 10^6 dt$

Integrating both sides of this equation yields (Note that the switch has been closed for a long time; thus, $i_{L}(\bar{o})=0$.):

$\ln \left(\frac{i_{L}(t)-25 \times 10^{-3}}{0-25 \times 10^{-3}}\right)=-32 \times 10^6(t-0)$

Solving for $i_{L}(t)$

$i_{L}(t)=25 \times 10^{-3}\left(1-e^{-32 \times 10^6 t}\right)$