Topic: Network Theory
In the circuit shown the switch closes at t = 0. The voltage across 4 $\mu$F capacitor in ideal condition changes to
- A. 0
- B. 16 V
- C. 15 V
- D. 24 V
Answer: B. 16 V
Explanation: $I(s)=\frac{\frac{V(0^+)}{s}}{\frac{1}{2s}+\frac{1}{4s}}$\n$V_c(s)=\frac{\frac{V(0^+)}{s}}{\frac{1}{2s}+\frac{1}{4s}}\cdot\frac{1}{4s} = \frac{\frac{24}{s}}{\frac{3}{4s}}\times\frac{1}{4s}=\frac{8}{s}$.\nAt steady state, final value theorem is applicable, so\n$V=\lim_{s\rightarrow0}s\cdot v_c(s)=s\cdot\frac{8}{s}=8~V$.\nChange in voltage $=(24-8)V=16~V$.