Question 3.31: In the circuit of Fig. 1, the switch is closed to position-1......
In the circuit of Fig. 1, the switch is closed to position-1 for a long time. At t = 0, the switch position is changed from 1 to 2. Determine the current response.
Case i : Switch in position-1
In position-1, the circuit should have attained steady state because the switch is closed for a long time. The steady state condition of the given circuit with switch in position-1 is shown in Fig. 2. Here the capacitor is fully charged and behaves as an open circuit. Hence, the voltage across the capacitor V_0 will be equal to the supply voltage 20 V.
\therefore V _0=20 V
This voltage V_0 will be the initial voltage when the switch position is moved from 1 to 2.
Case ii : Switch in position-2
Let, i (t) be the current through the circuit when the switch is closed to position-2.
Let, I ( s )= \mathcal{L}\{i( t )\}
The time domain and s-domain circuits with the switch in position-2 are shown in Figs 3 and 4, respectively.
With reference to Fig. 4, we can write,
I(s)=\frac{\frac{20}{s}}{0.2 s+10+\frac{1}{100 \times 10^{-6} s}} \\ =\frac{20}{0.2 s^2+10 s+\frac{1}{100 \times 10^{-6}}}=\frac{20}{0.2\left(s^2+\frac{10}{0.2} s+\frac{1}{0.2 \times 100 \times 10^{-6}}\right)} =\frac{100}{s^2+50 s+50000}=\frac{100}{\left(s^2+2 \times 25 s+25^2\right)+50000-25^2} =\frac{100}{(s+25)^2+49375}=\frac{100}{(s+25)^2 (\sqrt{49375})^2}=\frac{100}{(s+25)^2+(222.2)^2} =\frac{100}{222.2} \times \frac{222.2}{(s+25)^2+(222.2)^2} \\=0.45 \times \frac{222.2}{(s+25)^2+222.2^2}Add and subtract 25^2
(a+b)^2=a^2+2 a b+b^2
Let us take the inverse Laplace transform of I (s).
\mathcal{L}^{-1}\{ I ( s )\}= \mathcal{L} ^{-1}\left\{0.45 \times \frac{222.2}{( s +25)^2+222.2^2}\right\} \\i( t )=0.45 e ^{-25 t } \sin (222.2 t ) A\mathcal{L} \left\{ e ^{- at } \sin \omega t \right\}=\frac{\omega}{( s + a )^2+\omega^2}
