Obtain v(t) and i(t) for t 0 in the circuit of Fig. 8.84.

Question

Obtain v(t) and i(t) for t > 0 in the circuit of Fig. 8.84.

Accepted Answer

For t = 0-, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V.

For t > 0, we have the series RLC circuit shown below.

[latex]\begin{aligned}&\alpha=\mathrm{R} /(2 \mathrm{L})=(2+5+1) /(2 imes 5)=0.8\\&\omega_{0}=1 / \sqrt{\mathrm{LC}}=1 / \sqrt{5 imes 0.2}=1\\&\mathrm{s}_{1,2}=-\alpha \pm \sqrt{\alpha^{2}-\omega_{\mathrm{o}}^{2}}=-0.8 \pm \mathrm{j} 0.6\\&\mathrm{v}(\mathrm{t})=\mathrm{V}_{\mathrm{s}}+\left[(\mathrm{Acos} 0.6 \mathrm{t}+\mathrm{B} \sin 0.6 \mathrm{t}) \mathrm{e}^{-0.8 t} ight]\\&\mathrm{V}_{\mathrm{s}}=15+20=35 \mathrm{V} ext { and } \mathrm{v}(0)=20=35+\mathrm{A} ext { or } \mathrm{A}=-15\\&\mathrm{i}(0)=\mathrm{Cdv}(0) / \mathrm{dt}=0\\& ext { But } \mathrm{d} \mathrm{v} / \mathrm{dt}=\left[-0.8(\mathrm{Acos} 0.6 \mathrm{t}+\mathrm{B} \sin 0.6 \mathrm{t}) \mathrm{e}^{-0.8 \mathrm{t}} ight]+\left[0.6(-\mathrm{A} \sin 0.6 \mathrm{t}+\mathrm{B} \cos 0.6 \mathrm{t}) \mathrm{e}^{-0.8 \mathrm{t}} ight]\\&0=\mathrm{dv}(0) / \mathrm{dt}=-0.8 \mathrm{A}+0.6 \mathrm{B} ext { which leads to } \mathrm{B}=0.8 imes(-15) / 0.6=-20\\&\mathrm{v}(\mathrm{t})=\left\{35-\left[(15 \cos 0.6 \mathrm{t}+20 \sin 0.6 \mathrm{t}) \mathrm{e}^{-0.8 t} ight] ight\} \mathrm{V}\\&\mathrm{i}=\mathrm{Cdv} / \mathrm{dt}=0.2\left\{\left[0.8(15 \cos 0.6 \mathrm{t}+20 \sin 0.6 \mathrm{t}) \mathrm{e}^{-0.8 \mathrm{t}} ight]+\left[0.6(15 \sin 0.6 \mathrm{t}-20 \cos 0.6 \mathrm{t}) \mathrm{e}^{-0.8 \mathrm{t}} ight] ight\}\\&\mathrm{i}(\mathrm{t})=\left[(5 \sin 0.6 \mathrm{t}) \mathrm{e}^{-0.8 \mathrm{t}} ight] \mathrm{A}\end{aligned}[/latex]

Question 8.36: Obtain v(t) and i(t) for t > 0 in the circuit of Fig. 8....

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