Question 3.8: An RL series circuit excited by a sinusoidal source e (t) = ......
An RL series circuit excited by a sinusoidal source e (t) = 10 sin 100t V by closing the switch at t = 0. Take R = 10 Ω and L = 0.1 H. Determine the current i (t) flowing through the RL circuit.
Given that, e (t) = 10 sin 100t V
Let, {E}(\mathbf{s})={\mathcal{L}}\{e(t)\}
∴ {E}(\mathbf{s})\;=\;\mathcal{L}\{{e}({t})\}\;=\;\mathcal{L}\{{10}\sin100{t}\}\;=10\times{\frac{100}{{\mathrm{s}}^{2}+100^{2}}}={\frac{10^{3}}{{\mathrm{s}}^{2}+10^{4}}}
The time domain and s-domain RL circuits excited by a sinusoidal source are shown in Figs 1 and 2.
With reference to Fig. 2, we can write,
I({ s})\,=\,\frac{{E}\left({ s}\right)}{10+0.1s}
=\;\frac{10^{3}{s^2+10^4}}{0.1(\frac{10}{0.1}+s)^{4}}=\frac{10^{3}/0.1}{(s+100)\left({s}^{2}+10^{4}\right)}=\frac{10^{4}}{(\mathrm{s+100}){(s^{2}+10^{4})}}By partial fraction expansion technique, I (s) can be expressed as,
I(\mathrm{s})\,=\,{\frac{10^{4}}{\left(\mathrm{s+100}\right)\left(\mathrm{s^{2}+10^{4}}\right)}}\,=\,{\frac{\mathrm{K}_{1}}{\mathrm{s+100}}}\,+\,{\frac{\mathrm{K}_{2}\mathrm{s+K}_{3}}{\mathrm{s^{2}+10^{4}}}} ……….(1)
K_1=\left.\,{\frac{10^{4}}{\cancel{(\mathrm{s+100})}\left(\mathrm{s^{2}+10^{4}}\right)}}\times\cancel{(\mathrm{s+100})}\right|_{\mathrm{s}=-100}=\left.\frac{10^{4}}{s^{2}+10^{4}}\right|_{\mathrm{s}=-100}={\frac{10^{4}}{(-\,100)^{2}+10^{4}}}\,=\,0.5On cross-multiplying equation (1), we get,
10^{4}=\mathrm{K_{1}}\left(\mathrm{s}^{2}+10^{4}\right)+\left(\mathrm{K_{2}}\mathrm{s}+\mathrm{K_{3}}\right)\left(\mathrm{s}+100\right)
10^{4}=\mathrm{K}_{1}{s}^{2}+10^{4}\;\mathrm{K}_{1}\mathrm{K}_{2}\mathrm{s}^{2}+100\;\mathrm{K}_{2}\;\mathrm{s}+\mathrm{K}_{3}\;\mathrm{s}+100\;\mathrm{K}_{3}
10^{4}=\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)\mathrm{s}^{2}+\left(100\,\mathrm{K}_{2}+\mathrm{K}_{3}\right)\mathrm{s}+\left(10^{4}\,\mathrm{K}_{1}+100\,\mathrm{K}_{3}\right) …..(2)
On equating coefficients of s² of equation (2 ), we get,
K_{1}+K_{2}=0∴ K_{2}=-K _{1}=-0.5
On equating coefficients of s of equation (2 ), we get,
100\ {\mathrm{K}}_{2}+{\mathrm{K}}_{3}=0∴ { K}_{3}=-\,100\,{K}_{2}=-\,100\times\,(-0.5)=50
∴ I(\mathrm{s})\,=\,\frac{0.5}{\mathrm{s}+100}\,+\,\frac{-0.5\mathrm{s}+50}{\mathrm{s}^{2}+10^{4}}
=\;0.5{\frac{1}{s+100}}\;-0.5{\frac{s}{s^{2}+100^{2}}}\;+\;{\frac{50}{100}}\;{\frac{100}{s^{2}+100^{2}}}
=\;0.5{\frac{1}{\mathrm{s}+100}}\;-0.5{\frac{\mathrm{s}}{\mathrm{s}^{2}+100^{2}}}\;+\;0.5{\frac{100}{\mathrm{s}^{2}+100^{2}}}
Let us take the inverse Laplace transform of I (s).
∴ \mathcal{L}^{-1}\left\{I(\mathrm{s})\right\}\,=\,\mathcal{L}^{-1}\left\{0.5\frac{1}{\mathrm{s}+100}-0.5\frac{\mathrm{s}}{\mathrm{s}^{2}+100^{2}}+0.5\frac{100}{\mathrm{s}^{2}+100^{2}}\right\}
\therefore i( t )=0.5 L ^{-1}\left\{\frac{1}{ s +100}\right\}-0.5 L ^{-1}\left\{\frac{ s }{ s ^2+100^2}\right\}+0.5 L ^{-1}\left\{\frac{100}{ s ^2+100^2}\right\}=0.5 e^{-100 t}-0.5 \cos 100 t+0.5 \sin 100 t
=0.5 e ^{-100 t }+[\sin 100 t \times 0.5-\cos 100 t \times 0.5]
Let us construct a right-angled triangle with 0.5 as two sides as shown in Fig. 3.With reference to Fig. 3,
∴ \mathrm{tan}\phi\;=\;{\frac{0.5}{0.5}}\;=\;1
∴ \phi\ =\ \tan^{-1}\;1\ =\ 45^{\circ}
Also, \cos\phi\ =\ \frac{0.5}{0.7071} ⇒
0.5\;=\;0.7071\cos\phi\;=\;0.7071\cos45^{\circ}\sin\phi\ =\ \frac{0.5}{0.7071} ⇒
0.5\;=\;0.7071\cos\phi\;=\;0.7071\sin45^{\circ}∴ i({t})\,=\,0.5\mathrm{e}^{-100{\bf t}}+\left[\sin100{\bf t}\times0.7071\,\cos45^{\circ}-\cos100{\bf t}\times0.7071\,\sin45^{\circ}\right]
=\;0.5\mathrm{e}^{-100t}+0.7071\;[\sin100t\;\mathrm{cos\;}45^{\circ}-\mathrm{cos100t\;sin\;}45^{\circ}]
\underbrace{0 .5e^{ -100t}}_{Transient \ part} +\underbrace{0 7071 sin ( 100t- 45^\omicron )Ae^{ -100t}}_{Steady \ state \ part}
sin(A – B) = sinA cosB – cosA sinB
\nu_{\mathrm{{L}}}(t)\,=L{\frac{\mathrm{d}\mathbb{i}(t)}{\mathrm{d}t}}\,=L{\frac{\mathrm{d}}{\mathrm{d}t}}[0.5\mathrm{e}^{-\,100t}+0.707\sin(100\mathrm{t}-45^{o})]
=\;0.1\big[0.5\mathrm{e}^{-\,100t}\times(-\,100)+0.707\cos(\,100\mathrm{t-}\,45^{\circ})\times100\big]
=\;-5\mathrm{e}^{-\,100t}+7.07\cos{(\,100t-45^{\circ})}
=\;-\,5\,\mathrm{e}^{-\,100t}+7.07\sin\left(100\mathrm{t-}\,45^{\circ}+90^{\circ}\right)
\because cos θ = sin (θ + 90°)\underbrace{0 .5e^{ -100t}}_{Transient \ part} +\underbrace{0 707 \sin ( 100t+ 45^\omicron )V}_{Steady \ state \ part}
