Topic: Network Theory
For a circuit given in figure, switch K is closed to position 1 at $t=0$. After $t=1 TC$ (Time Constant), switch is moved to position 2. Find The Current I at $1TC(-)$ (just before one-time constant time) and $1TC(+)$ (just after one-time constant time) respectively (assume $1/e\approx0.37$)
(Position 1 connects 20V, Position 2 connects 40V, R = $500\Omega$, C = $0.5\mu F$)
- A. 0.0148 A and 0.055 A
- B. 0.0148 A and -0.055 A
- C. 0.0296 A and 0.11 A
- D. 0.0296 A and -0.11 A
Answer: A. 0.0148 A and 0.055 A
Explanation: The time constant $TC = RC = 500 \times 0.5\times 10^{-6} = 250 \mu s$. At $t=0$, K is at pos 1, charging from a 20V source. The capacitor voltage at $t = 1TC(-)$ is $V_c = 20(1 - e^{-1}) \approx 20(1 - 0.37) = 12.6$ V. The current at $1TC(-)$ is $I = \frac{20 - V_c}{R} = \frac{20 - 12.6}{500} = 0.0148$ A. At $t = 1TC(+)$, the switch moves to 40V. The capacitor voltage cannot change instantaneously, remaining at 12.6V. The current at $1TC(+)$ is $I = \frac{40 - 12.6}{500} = \frac{27.4}{500} = 0.0548 \approx 0.055$ A.