For the circuit in Fig. 8.97, find i(t) for t 0

Question

For the circuit in Fig. 8.97, find i(t) for t >0

Accepted Answer

For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A.

For t > 0, we have a parallel RLC circuit.

[latex]\mathrm{I}_{\mathrm{s}}=3+6=9 \mathrm{A}[/latex] and [latex]\mathrm{R}=10 \| 40=8 \mathrm{ohms}[/latex]

[latex]\begin{array}{l}\alpha=1 /(2 \mathrm{RC})=(1) /(2 imes 8 imes 0.01)=25 / 4=6.25 \\\omega_{\mathrm{o}}=1 / \sqrt{\mathrm{LC}}=1 / \sqrt{4 imes 0.01}=5\end{array}[/latex]

since [latex]\alpha>\omega_{0},[/latex] we have a overdamped response.

[latex]\mathrm{s}_{1,2}=-\alpha \pm \sqrt{\alpha^{2}-\omega_{\mathrm{o}}^{2}}=-10,-2.5[/latex]

Thus, [latex]\mathrm{i}(\mathrm{t})=\mathrm{I}_{\mathrm{s}}+\left[\mathrm{Ae}^{-10 \mathrm{t}} ight]+\left[\mathrm{Be}^{-2.5 \mathrm{t}} ight], \quad \mathrm{I}_{\mathrm{s}}=9[/latex]

[latex]\begin{array}{l}\mathrm{i}(0)=3=9+\mathrm{A}+\mathrm{B} ext { or } \mathrm{A}+\mathrm{B}=-6 \\\mathrm{di} / \mathrm{dt}=\left[-10 \mathrm{Ae}^{-10 \mathrm{t}} ight]+\left[-2.5 \mathrm{Be}^{-2.5 \mathrm{t}} ight]\\end{array}[/latex]

[latex]\mathrm{v}(0)=0=\operatorname{Ldi}(0) / \mathrm{dt}[/latex] or [latex]\mathrm{di}(0) / \mathrm{dt}=0=-10 \mathrm{A}-2.5 \mathrm{B}[/latex] or [latex]\mathrm{B}=-4 \mathrm{A}[/latex]
Thus, A=2 and B=-8
Clearly,
[latex]\mathrm{i}(\mathrm{t})={9+[2 \mathrm{e}^{-10 t}]+[-8 \mathrm{e}^{-2.5 t}]}[/latex] A

Question 8.50: For the circuit in Fig. 8.97, find i(t) for t >0

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