Question

Consider the circult shown in figure. Assume that at $t=0,-1$ A flows through the inductor and $+5V$ is across the capacitor. If $V_s=10u\left(t\right)V,$ then the value of the voltage across the capacitor is

• A. $(25e^{-t}-20e^{-2t})u\left(t\right)V$
• B. $(17.5e^{-t}-12.5e^{-2t})u\left(t\right)V$
• C. $(30e^t-25e^{2t})u\left(t\right)V$
• D. $(35e^{-t}-30e^{-2t})u\left(t\right)V$

Answer 1

The correct option is D $(35e^{-t}-30e^{-2t})u\left(t\right)V$

$i\left(0\right)=-1A$
$V\left(0\right)=5V$
Apply node analysis

$\frac{(V_1-\frac{10}{s})}{\frac{10}{3}}+\frac{V_1}{5s}-\frac{1}{s}+\frac{(V_1-\frac{5}{s})}{\left(\frac{10}{s}\right)}=0$
$V_1(\frac{3}{10}+\frac{1}{5s}+\frac{s}{10})-\frac{10\times 3}{s\times 10}-\frac{1}{s}-\frac{5}{s}\times \frac{s}{10}=0$
$V_1\left(\frac{3s+2+s^2}{10s}\right)=(\frac{3}{s}+\frac{1}{s}+\frac{0.5s}{s})$
$V_1=\frac{10s}{(s^2+3s+2)}\times \frac{(0.5s+4)}{s}$
$V_1=\frac{(5s+40)}{s^2+3s+2}=\frac{5(s+8)}{(s+1)(s+2)}$
$V_1=5(\frac{7}{s+1}-\frac{6}{s+2})$
$v_1\left(t\right)=(35e^{-t}-30e^{-2t})u\left(t\right)$