Question 3.Q.1: In the circuit of Fig. 1, a steady current of 5 A is establi......
In the circuit of Fig. 1, a steady current of 5 A is established through the inductance by connecting it to a current source. At time t = 0, the current source is disconnected and switch S_1 is closed to connect a 20 Ω resistance across the inductance. Find the expression for i (t) and sketch the response. Also draw the initial and final condition of the circuit.
Let,{\mathcal{L}}\{i(t)\}\ =I({{s}})
The s-domain equivalent circuit is shown in Fig. 2. With reference to Fig.2, we can write,
I(\mathrm{s})\,=\,\frac{10}{2\mathrm{s}+20}\,=\,\frac{10}{2\Big(\mathrm{s}+\frac{20}{2}\Big)}
∴I(\mathrm{s})\,=\,\frac{5}{\mathrm{s}+10} ………..(1)
Equation (1) is the s-domain response of the circuit. Let us take the inverse Laplace transform of I (s) to get the time domain response.
\mathcal{L}^{-1}\left\{I(\mathbf{s})\right\}\,=\,\mathcal{L}^{-1}\left\{\frac{\mathcal{5}}{{5}+10}\right\}
∴i({t})~=~{5}~{ e}^{-10t}A
{\mathcal{L}}\{e^{-\mathrm{{at}}}\}\ =\ {\frac{1}{{s+a}}} {\mathcal{L}}\{e^{-\mathrm{{at}}}\}\ =\ {\frac{1}{{s+a}}} \nu_{\mathrm{{L}}}(t)\;=\;\mathrm{{L}}\,{\frac{\mathrm{di}\left(t\right)}{\mathrm{dt}}} =\;2\times{\frac{\mathrm{d}}{\mathrm{dt}}}\left(5e^{-10\mathrm{t}}\right) =\,-\,100\,\mathrm{e}^{-\,\mathrm{10t}}\,\,V \nu_{\mathrm{L}}(\infty)\;=-\;100\;\mathrm{e}^{-\infty}\;=\;0since, v_{ L }(\infty)=0, the inductance is represented by short circuit in final condition circuit.
Time constant,\tau\,=\,\frac{L}{{R}}\,=\,\frac{2}{20}\,=\,0.1 second
∴ i(t)\;=\;5\:e^{-\frac{\mathrm{t}}{1/10}\;}=\;5\:e^{-\frac{\mathrm{t}}{0.1}}\:A
At \mathbf{t}\;=\;0^{+},\;i(0^{+})\;=\;5\times\mathbf{e}^{0}=5 \ A
At \mathbf{t}\ =\ \tau\ =\ 0.1sec, i({\bf t})\,=\,5\times\mathrm{e}^{-\frac{0.1}{0.1}}\,=\,5\times\mathrm{e}^{-1}\,=\,5\times0.3679\,=\,1.8395\,A
At \begin{array}{r c l}{{\mathrm{t~}=\infty ,~i(\infty )~=~5\times{\mathbf{e}}^{-\infty }~=~5\times0~=~0}}\end{array}
From the above analysis, we can say that at t = 0^+, the initial current is 5 A and this current of 5 A exponentially decays to zero as t tends to infinity.
