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Question

A capacitor $C_1 = 1.0 \mu F$ is charged up to a voltage $V = 60$ V by connecting it to battery B through switch (1). Now $C_1$ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $C_2 = 3.0 \mu F$ and $C_3 = 6.0 \mu F$ through switch (2), as shown in the figure. The sum of final charges on $C_2$ and $C_3$ is :

The correct answer is
40 $\mu C$

To solve this problem, we need to calculate the total charge on the capacitors \(C_2\) and \(C_3\) after they are connected to the charged capacitor \(C_1\). Here are the steps to solve the problem:

Step 1: Calculate the initial charge on \(C_1\).

The charge \(Q_1\) on the capacitor \(C_1\) can be determined using the formula:

\(Q = CV\)

Where:

  • \(C = 1.0 \, \mu \text{F}\)
  • \(V = 60 \, \text{V}\)

Thus,

\(Q_1 = 1.0 \times 60 = 60 \, \mu \text{C}\)

Step 2: Determine the equivalent capacitance when \(C_1\) is connected to \(C_2\) and \(C_3\).

The capacitors \(C_2\) and \(C_3\) are connected in series, so the equivalent capacitance \(C_{\text{eq}}\) is given by:

\(\frac{1}{C_{\text{eq}}} = \frac{1}{C_2} + \frac{1}{C_3}\)

Substituting the given values:

\(\frac{1}{C_{\text{eq}}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\)

Thus, \(C_{\text{eq}} = 2 \, \mu \text{F}\)

Step 3: Calculate the final charge distribution.

When \(C_1\) is connected to the equivalent capacitance \(C_{\text{eq}}\), the total charge is conserved. The initial charge on \(C_1\) is now distributed over \(C_1\) and the series combination \(C_2\) and \(C_3\).

The new combined capacitance becomes:

\(C_{\text{total}} = C_1 + C_{\text{eq}} = 1 + 2 = 3 \, \mu \text{F}\)

The voltage across the capacitors is:

\(V_{\text{final}} = \frac{Q_1}{C_{\text{total}}} = \frac{60}{3} = 20 \, \text{V}\)

Hence, the charge on the series combination \((Q_2 = Q_3)\) is:

\(Q_2 = C_{\text{eq}} \times V_{\text{final}} = 2 \times 20 = 40 \, \mu \text{C}\)

Conclusion: The sum of the final charges on \(C_2\) and \(C_3\) is \(40 \, \mu \text{C}\).

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Important Questions from Electricity and Magnetism

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    $\left[ \text{Take } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ Nm}^2\text{/C}^2 \right]$
  3. An inductor stores $16\text{ J}$ of magnetic field energy and dissipates $32\text{ W}$ of thermal energy due to its resistance when an a.c. current of $2\text{ A}$ (rms) and frequency $50\text{ Hz}$ flows through it. The ratio of inductive reactance to its resistance is _________. ($\pi = 3.14$)
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