In the circuit shown, the switch S has been closed for a long time. At time $t = 0$ , the switch is opened. It remains open for a long time T, after which it is closed again.

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voltage drop across $100 k Ω$ resister is $\frac{20 e^{- t / 1.5}}{3} V$ for t < T

voltage drop across $100 k Ω$ resister is $10 (e^{T - t} - e^{(T - 3 t) / 3})$ for $t > T$

voltage drop across $100 k Ω$ resister is $10 e^{- t / 1.5} V$ for $t < T$

The time constant of the circuit is 1.5sec for t > T

answer is B.

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Detailed Solution

When the switch ‘S’ is opened,
$τ = R C = 1.5 \sec$
Voltage across the capacitor,
$V_{c} = 10 (1 - e^{t / 1.5})$ for t<T
The current, $i = \frac{d Q}{d t} = C \frac{d V_{c}}{d t}$
$= \frac{200}{3} \times 10^{- 6} (e^{- t / 1.5})$ for t<T
Voltage drop across 100 k $μ$ resister,
for t < T
After t = T,