Question 8.49: Determine i(t) for t > 0 in the circuit of Fig. 8.96.

Determine i(t) for t > 0 in the circuit of Fig. 8.96.

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For \mathrm{t}=0^{-}, \mathrm{i}(0)=3+12 / 4=6 and \mathrm{v}(0)=0
For \mathrm{t}>0, we have a parallel R L C circuit with a step input.

\begin{array}{l}\alpha=1 /(2 \mathrm{RC})=(1) /(2 \times 5 \times 0.05)=2 \\\\\omega_{0}=1 / \sqrt{\mathrm{LC}}=1 / \sqrt{5 \times 0.05}=2\end{array}

since \alpha=\omega_{0}, we have a critically damped response.

\mathrm{s}_{1,2}=-2

Thus,

\mathrm{i}(\mathrm{t})=\mathrm{I}_{\mathrm{s}}+\left[(\mathrm{A}+\mathrm{Bt}) \mathrm{e}^{-2 \mathrm{t}}\right], \quad \mathrm{I}_{\mathrm{s}}=3

\mathrm{i}(0)=6=3+\mathrm{A} \text { or } \mathrm{A}=3 \\
\mathrm{v}=\mathrm{Ldi} / \mathrm{dt} \text { or } \mathrm{v} / \mathrm{L}=\mathrm{di} / \mathrm{dt}=\left[\mathrm{Be}^{-2 \mathrm{t}}\right]+\left[-2(\mathrm{A}+\mathrm{Bt}) \mathrm{e}^{-2 \mathrm{t}}\right] \\
\mathrm{v}(0) / \mathrm{L}=0=\operatorname{di}(0) / \mathrm{dt}=\mathrm{B}-2 \times 3 \text { or } \mathrm{B}=6 \\
\text { Thus, } .\mathrm{i}(\mathrm{t})=3+[(3+6 \mathrm{t}) \mathrm{e}^{-2 \mathrm{t}}] \mathrm{A}

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