In the circuit shown, VC is zero at t=0 s. For t>0, the capacitor current IC(t), where t is in second, is ___________(a) 0.50 e^-25t mA(b) 0.25 e^-25t mA(c) 0.50 e^-12.5t mA(d) 0.25 e^-6.25t mAI had been asked this question in an interview.My enquiry is from Advanced Problems on Network Theorems in section Useful Theorems in Circuit Analysis of Network Theory

In the circuit shown, VC is zero at t=0 s. For t>0, the capacitor curr

1.	In the circuit shown, VC is zero at t=0 s. For t>0, the capacitor current IC(t), where t is in second, is ___________(a) 0.50 e^-25t mA(b) 0.25 e^-25t mA(c) 0.50 e^-12.5t mA(d) 0.25 e^-6.25t mAI had been asked this question in an interview.My enquiry is from Advanced Problems on Network Theorems in section Useful Theorems in Circuit Analysis of Network Theory
Answer» Correct choice is (a) 0.50 e^-25T mA Easiest explanation: The CAPACITOR voltage VC (t) = VC (∞) – [VC (∞)-VC (0)]e^-t/RC R = 20 \|\| 20 = \(\FRAC{20×20}{20+20} = \frac{400}{40}\) = 10 kΩ VC (∞) = 10 × \(\frac{20}{20+20}\) = 5 V Given, VC (0) = 0 ∴ VC (t) = 5 – (5-0)e^-t/10×4×10^(-6)×10^3 = 5(1 – e^-25t) IC (t) = C\(\frac{dV_C (t)}{dt} = 4×10^{-6} \frac{d}{dt}5(1-e^{-25t})\) = 4 × 10^-6 × 5 × 25e^-25t ∴ IL (t) = 0.50e^-2.5t mA.

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