Topic: Network Theory
For a network shown in figure, a steady state is reached with switch k is open. Switch is closed at time t=0, Calculate $(di_1/dt)$ and $(di_2/dt)$ at $t=0+$
- A. 16.67 A/sec and $-4.16\times10^4 A/sec$
- B. 16.67 A/sec and $-3.33\times10^4 A/sec$
- C. 25 A/sec and $-4.16\times10^4 A/sec$
- D. 25 A/sec and $-3.33\times10^4 A/sec$
Answer: A. 16.67 A/sec and $-4.16\times10^4 A/sec$
Explanation: For $t < 0$ (steady state with switch open), the $100$V source is in series with the $10\Omega$ resistor. The inductor acts as a short circuit and the capacitor as an open circuit. The initial current $i_1(0^-) = 100 / (10 + 20) = 100 / 30 = 10/3$ A. The initial capacitor voltage $V_c(0^-) = 100 \times 20 / 30 = 200/3$ V. When the switch closes at $t=0$, the $10\Omega$ resistor is shorted, applying $100$V across both parallel branches. For the inductor branch: $100 = 20 i_1 + L \frac{di_1}{dt}$. At $t=0^+$, $100 = 20(10/3) + 2 \frac{di_1}{dt} \implies \frac{di_1}{dt} = 50/3 \approx 16.67$ A/sec. For the capacitor branch: $100 = 20 i_2 + V_c$. At $t=0^+$, $i_2(0^+) = (100 - 200/3)/20 = 5/3$ A. Differentiating the voltage equation gives $0 = 20 \frac{di_2}{dt} + \frac{dV_c}{dt}$ where $\frac{dV_c}{dt} = \frac{i_2}{C}$. So, $\frac{di_2}{dt} = -\frac{i_2(0^+)}{20 C} = -\frac{5/3}{20 \times 2\times 10^{-6}} = -41666.67$ A/sec $\approx -4.16 \times 10^4$ A/sec.