A capacitor of capacitance $C$ is given a charge $Q$. At $t = 0$, it is connected to an uncharged capacitor of equal capacitance through a resistance $R$. The charge on the second capacitor as a function of time is:

Question

An ideal inductor, a resistor of resistance $R$ Ohms and a capacitor with adjustable capacitance are connected in series to an alternating voltage with an effective value of $V$ Volts and frequency of $f$ Hz. The current flowing through the circuit when the capacitance of the capacitor is set to $C_1$ is the same as when the capacitance of the capacitor is set to $C_2$, $C_2 > C_1$. The inductance of the inductor $L$ is given by:

Answer 1

Step 1: Use the effective time constant of two series capacitors. The equivalent capacitance of $C$ and $C$ in series is $C_{eq} = \dfrac{C\cdot C}{C+C} = \dfrac{C}{2}$, so the circuit relaxes with time constant $\tau = R C_{eq} = \dfrac{RC}{2}$, i.e. the decay rate is $2/RC$.

Step 2: Identify the final state. Charge redistributes until both capacitors reach the same voltage. Total charge $Q$ splits equally, so each finally holds $Q/2$; hence $q_{\infty} = Q/2$ for the second capacitor.

Step 3: A first-order relaxation from $q(0)=0$ toward $q_\infty$ has the form
\[q(t) = q_\infty\left(1 - e^{-t/\tau}\right).\]
Step 4: Insert $q_\infty = Q/2$ and $\tau = RC/2$:
\[q(t) = \frac{Q}{2}\left(1 - e^{-2t/RC}\right).\]
\[\boxed{q = \dfrac{Q}{2}\left(1 - e^{-\frac{2t}{RC}}\right)}\]

A capacitor of capacitance \(C\) is given a charge \(Q\). At \(t = 0\), it is connected to an uncharged capacitor of equal capacitance through a resistance \(R\). The charge on the second capacitor as a function of time is:

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The Correct Option is A

Solution and Explanation

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