Question 3.24: In the circuit of Fig. 1, the switch remains in position-1 f......

Case i : Switch in position-1
In position-1, the circuit has attained a steady state. Hence, we can perform steady state analysis. The steady state of the RC circuit with switch in position-1 is shown in Fig. 2.

The standard form of sinusoidal source is, E_m sin (ωt ± \theta).

Here, E_m sin (ωt ± \theta)=100 sin(200t + 45^o ) V

\therefore E_m=100 V , \omega=200 rad / s , \phi=45^{\circ}

\bar{E}_m=100 \angle 45^{\circ} V

∴ Rms value of voltage {{E}}={\frac{{E}_{\mathrm{m}}}{\sqrt{2}}}={\frac{100}{\sqrt{2}}}\angle45^{\circ}V

Let \overline{V}_0 be the voltage across the capacitor in steady state. By voltage division rule,

\overline{{{V}}}_{0}\,=\,\frac{100}{\sqrt{2}}\angle45^{\circ}\times\frac{-j250}{50-j250}

=\,{\frac{100}{\sqrt{2}}}\angle45^{\circ}\times{\frac{250\angle-90^{\circ}}{254.951\angle-78.7^{\circ}}}

\,=\,69.3375\angle 33.7^{\circ}V

Let,{V}_{0}\;=\;\left|{\overline{{{V}}}}_{0}\right|\;=\;69.3375\,V

This steady state voltage V_0 = 69.3375 \ V will be the initial voltage when the switch is moved from position-1 to position-2.

Case ii : Switch in position-2

When the switch is changed from position-1 to position-2, a steady voltage of V_0 (and hence a charge of Q_0 ) exists across the capacitance. Since the capacitance does not allow a sudden change in voltage, this steady voltage V_0 will be the initial voltage when the switch is closed to position-2.

∴ v_{\mathrm{{C}}}(0^{-})=v_{\mathrm{{C}}}(0^{+})=V_{0}=69.3375\,V

The time domain and s-domain RC circuits with the switch in position-2 are shown in Figs 3 and 4, respectively.

Let, i(t) be the current through the RC circuit, when the switch is closed to position-2 at t = 0.
Let, I(\mathbf{s})=\mathcal{L}\{i(t)\}

With reference to Fig. 4, we can write,

{I}(\mathrm{s})\,=\,\frac{\frac{69.3375}{s} }{50+50+\frac{1}{20\times 10^{-6}s}}

=\;{\frac{69.3375}{s}}\times{\frac{1}{100+{\frac{1}{20\times10^{-6}\mathrm{s}}}}}

=\;\frac{69.3375}{100s+\frac{1}{20\times10^{-6}}}=\frac{69.3375}{100\left(s+\frac{1}{100\times20\times10^{-6}}\right)}=\frac{0.693375}{(s+500)}

∴ I(s)\,=\,{\frac{0.693375}{s+500}}

On taking the inverse Laplace transform of I (s), we get,

\mathcal{L}{{{\{\mathrm{e}^{-\mathrm{at}}\}}}}\:=\:\frac{1}{\mathrm{s~+~a}}