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

KCL for Node A corresponds to: [latex]-i_{s}+i_{R}(t)+i_{L}(t)=0[/latex] Because [latex]v_{R}(t)=v_{L}(t)[/latex] [latex]i_{R}(t)=\frac{v_{L}(t)}{6.4 k \Omega}[/latex] In addition: [latex]v_{L}(t)=200 \mu H \frac{di_{L}(t)}{dt}[/latex] Replacing for [latex]i_{R}(t)[/latex] and [latex]v_{L}(t)[/latex] in the node A KCL equation: [latex]-25 \times 10^{-3}+\frac{200 \times 10^{-6}}{6400} \frac{di_{L}(t)}{dt}+i_{L}(t)=0[/latex] A simple mathematical manipulation leads to: [latex]\frac{di_{L}(t)}{i_{L}(t)-25 \times 10^{-3}}=-32 \times 10^6 dt[/latex] Integrating both sides of this equation yields (Note that the switch has been closed for a long time; thus, [latex]i_{L}(\bar{o})=0[/latex].): [latex]\ln \left(\frac{i_{L}(t)-25 \times 10^{-3}}{0-25 \times 10^{-3}}\right)=-32 \times 10^6(t-0)[/latex] Solving for [latex]i_{L}(t)[/latex] [latex]i_{L}(t)=25 \times 10^{-3}\left(1-e^{-32 \times 10^6 t}\right)[/latex]

Question 5.9: Inductor Current Find an expression for iL (t) in the circui...

Electrical Engineering: Concepts and Applications

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.

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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}

In addition:

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)

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