Question

If the Laplace transform of the voltage across a capacitor of value of $\frac{1}{2}F$ is
$V_C\left(s\right)=\frac{s+1}{s^3+s^2+s+1}$,
the value of the current through the capacitor at $t=0^{+}$ is

• A. 0 A
• B. 2 A
• C. $\frac{1}{2}A$
• D. 1 A

Answer 1

The correct option is C $\frac{1}{2}A$

$Z_C\left(s\right)=\frac{1}{Cs}=\frac{2}{s}$

$I_C\left(s\right)=\frac{V_C\left(s\right)}{Z_C\left(s\right)}=\frac{s(s+1)}{2(s^3+s^2+s+1)}$

$=\frac{s(s+1)}{2(s^2+1)(s+1)}$

$I_C\left(s\right)=\frac{s}{2(s^2+1)}$

$i\left(0^{+}\right)={lim}_{s\to \infty }S{I}_C\left(s\right)={lim}_{s\to \infty }\frac{s^2}{2(s^2+1)}$

$=\frac{1}{2+0}=\frac{1}{2}Amp.$