At t = 0,switch S is closed. The charge on th | vedclass

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(A) $Q_0$ is the maximum charge on the capacitor at steady state,i.e.,at $t = \infty$.At $t = \infty$,the capacitor acts as an open circuit,meaning no

Vedclass · Accepted Answer

$\frac{CVR_2}{R_1 + R_2}$

Vedclass · Answer

(A) $Q_0$ is the maximum charge on the capacitor at steady state,i.e.,at $t = \infty$.At $t = \infty$,the capacitor acts as an open circuit,meaning no current flows through it.The circuit simplifies to a series combination of $R_1$ and $R_2$ connected to the voltage source $V$.The potential difference across the capacitor is equal to the potential difference across the resistor $R_2$.Using the voltage divider rule,the voltage across $R_2$ is $V_C = V \left( \frac{R_2}{R_1 + R_2} \right)$.The charge on the capacitor is $Q_0 = C V_C$.Substituting the value of $V_C$,we get $Q_0 = C \left( \frac{V R_2}{R_1 + R_2} \right) = \frac{CVR_2}{R_1 + R_2}$.

At $t = 0$,switch $S$ is closed. The charge on the capacitor is varying with time $t$ as $Q = Q_0(1 - e^{-\alpha t})$. Find the value of $Q_0$.

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