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In an RC circuit, a switch S has been closed for a long time so that the capacitor is fully charged. At time t = 0 the switch is opened. The circuit (for t < T) has a capacitor C = 15 microfarad in series with a 100 kilo-ohm resistor (total resistance in loop = 150 kilo-ohm when S is open). The initial voltage across the capacitor is 10 V. At time t = T the switch is closed again, reconnecting an additional 50 kilo-ohm resistor in parallel. Which of the following statements is/are correct?

1. Voltage drop across 100 kohm resistor is 10 * e^(-t/1.5) V for t < T
2. Voltage drop across 100 kohm resistor is (20/3) * e^(-t/1.5) V for t < T
3. Voltage drop across 100 kohm resistor follows a combination of decaying exponentials for t > T
4. The time constant of the circuit is 1.5 s for t > T

Correct answer: Voltage drop across 100 kohm resistor is 10 * e^(-t/1.5) V for t < T

Solution

With switch open, only 100 kohm and C = 15 microfarad form the loop. Time constant tau = 100 * 10³ * 15 * 10⁻⁶ = 1.5 s. Initial voltage across capacitor = 10 V, all of which drives current through the 100 kohm. So V₁₀₀k(t) = 10 * e^(-t/1.5) V for t < T. Option 1 is correct. When S is closed again for t > T, a parallel branch is added; the time constant changes so option 4 (tau = 1.5 s for t > T) is incorrect.

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