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Chapter 5

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.

5.9

Step-by-Step

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}

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)