Question 3.23: An RC circuit is excited by a sinusoidal source of voltage 5......

Given that, e(t) = 50 sin 314t V

R = 100 Ω and {C}=20\mu F=20\times10^{-6}F

The time domain RC circuit is shown in Fig. 1.

Let, {\mathcal{L}}\{{\mathcal{e}}(t)\}={E}({S})\ {\mathrm{and}} \ \mathcal{L}\{i(t)\}=I({s})

∴ E=\;{\mathcal{L}}\{e(\mathbf{t})\}\;={\mathcal{L}}\{50{\mathrm{~sin~}}314\mathbf{t}\}

=\;50\times{\frac{314}{{\mathrm{s}}^{2}+314^{2}}}={\frac{15700}{{\mathrm{s}}^{2}+314^{2}}}

The s-domain equivalent of the RC circuit is shown in Fig. 2. With reference to Fig. 2, we can write,

I(\mathbf{s})={\frac{\operatorname{E}\left({s}\right)}{\operatorname{R}+{\frac{1}{\mathbf{s}C}}}}

=\,{\frac{\operatorname{E}\left(\mathbf{s}\right)}{{\frac{sRC+1}{sC}}{} }}\,=\,\operatorname{E}(\mathbf{s})\times{\frac{\operatorname{sC}}{\operatorname{RC}\left(\mathbf{s}+{\frac{1}{\mathrm{RC}}}\right)}}

=\;\frac{15700}{s^{2}+314^{2}}\times\frac{s}{\mathrm{R}\Big(\mathrm{s}+\frac{1}{\mathrm{RC}}\Big)}

Here,\mathrm{RC}=100\times20\times10^{-6}=0.002\,\mathrm{sec}o n d.

∴ I(\mathrm{s})\,=\,\frac{15700}{\mathrm{s}^{2}+314^{2}}\times\frac{s}{100\Big(\mathrm{s}+\frac{1}{0.002}\Big)}

=\;{\frac{157s}{(s+500)\left(s^{2}+314^{2}\right)}}

By partial fraction expansion technique, we can write,

I(\mathrm{s})\,=\,{\frac{157s}{(\mathrm{s+500})\,(\mathrm{s^{2}+314^{2}})}}\,=\,{\frac{\mathrm{K}_{1}}{\mathrm{s+500}}}+{\frac{\mathrm{K}_{2}\mathrm{s}+\mathrm{K}_{3}}{\mathrm{s^{2}+3\,14^{2}}}}                   …….(1)

\mathrm{K_{1}\;=\;{\frac{157s}{(\mathrm{s+500})\left(\mathrm{s}^{2}+314^{2}\right)}}\times\left(\mathrm{s}+500\right)}\Bigg|_{\mathrm{s=-500}}=\,\frac{157\times(-500)}{(-500)^{2}+314^{2}}\,=\,-\,0.2252

On cross-multiplying equation (1), we get,

\mathrm{157s~}=\mathrm{K}_{\mathrm{1}}(\mathrm{s}^{2}+\mathrm{314}^{2})~+~(\mathrm{K}_{\mathrm{2}}\mathrm{s}+\mathrm{K}_{\mathrm{3}})~(\mathrm{s}+\mathrm{500})

{{157s}}=\ K_{1}{{s}^{2}}\,+\,314^{2}{ K}_{1}\,+\,K_{2}{{s}^{2}}\,+\,{\,500\,}K_{2}{\bf s}\,+\,K_{3}{ s}\,+\,500{\bf k}_{3}

\mathrm{157s}\;=\;({K}_{1}+\mathrm{K}_{2})\mathrm{s}^{2}\:+\:(\mathrm{500~K}_{2}+\mathrm{K}_{3})\mathrm{s}\:+\:\stackrel{}{314}^{2}\mathrm{K}_{1}\:+\:\mathrm{\;500K}_{3}\;        ………(2)

On equating coefficients of s² of equation (2), we get,

{K}_{{1}}\ +\ {K_2}\ =\ 0

\therefore K _2=- K _1=0.2252

On equating coefficients of s of equation (2), we get

∴ {K}_{3}={157}-{500 \ K_2}

=\ 157\ -\ 500\ \times\ 0.2252\ =\ 44.4

∴ I(\mathrm{s})\,=\,{\frac{-0.2252}{\mathrm{s}+500}}+{\frac{0.2252\mathrm{s}+44.4}{\mathrm{s}^{2}+3.14^{2}}}

Let us take the inverse Laplace transform of I (s).

∴i({ t})\;=\;-0.2252\,{e}^{-500t}\:+\:0.2252\,\cos314{t}\:+\:0.1414\,\mathrm{sin}\,314{ t}\;.

Let us construct a right-angled triangle with 0.1414 and 0.2252 as
two sides as shown in Fig. 3. With reference to Fig. 3, we can write,

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

{\mathcal L}\{\mathrm{cos \omega t }\}\,=\,\frac{s}{s^2+\omega^2}

{\mathcal L}\left\{{\mathrm{sin}}\,{\mathrm{\omega t}}\right\}\;=\;{\frac{\mathrm{\omega}}{\mathrm{s}^{2}\;+\;\mathrm{\omega}^{2}}}

\tan\phi\ =\ {\frac{0.2252}{0.1414}}\ =\ 1.5926

∴\phi\ =\ \tan^{-1}(1.5926)\ =\ 57.9^{\circ}\approx58^{\circ}

Also, \cos{\phi}\,=\,{\frac{0.1414}{0.2659}}\quad\Longrightarrow\quad0.1414\,=\,0.2659\,\mathrm{cos}\,\phi\,=\,0.2659\,\mathrm{cos}\,58^{\circ}

∴i({ t})\;=\;-0.2252\,{\mathrm e}^{-500t}\;+[\sin314{\bf t}\times0.2659\,\cos58^{\circ}+\cos3\,14{\bf t}\times0.2659\,\sin58^{\circ}]

=\;-0.2252\,\mathrm{e}^{-500t}\;+\;0.2659\;[\sin314t\cos58^{\circ}+\cos314t \sin58^{\circ}]

=\;-0.2252\;{\mathrm{e}}^{-500t}\;+\;0.2659\;{\mathrm{sin}}\;(314\pm58^{\circ})A

sin(A + B) = sinA cosB + cosA sinB

If i (t) is the current through the capacitor then the voltage across the capacitor v_C (t) is given by,

v_{\mathrm{{C}}}({t})\;=\;{\frac{1}{\mathrm{{C}}}}\int{i({t}){{d}}t}

∴v_{\mathrm{{c}}}({t})\,=\,{\frac{1}{\mathbf{C}}}\int\left[-0.2252\,{\mathrm{e}}^{-500\mathbf{t}}+0.2659\,\sin(314\mathbf{t}+58^{\circ})\right]\mathrm{d}\mathbf{t}

=\;\frac{1}{20\times10^{-6}}\left[\frac{-0.2252\,\mathrm{e}^{-500t}}{-500}\;-\;\frac{0.2659\cos(314t+58^{\circ})}{314}\right]

\sin( θ – 90^{\circ}) = –\cosθ

=\;\frac{-\,0.2252}{20\times10^{-6}\times(-500)}e^{-500t}\;+\;\frac{0.2659}{20\times10^{-6}\times314}\sin(314t-32^{\circ})V

= 22.52 e^{− 500t} + 42.34 \sin (314 t − 32^o ) V

RESULT

i(t)=-0.2252\mathrm{{e}}^{-500t}+0.2659\,\mathrm{{sin}}\left(314\mathrm{{t}}+58^{\circ}\right)A

V_{\mathrm{{C}}}(t)=22.52\,{\mathrm{e}}^{-500t}+42.34\,\sin{(3144t-32^{\circ})}\,V