Question

Consider the circuit shown in the figure. The capacitors C1 and C2 have capacitances \(2\ \mu\text{F}\) and \(8\ \mu\text{F}\), respectively. The switch can connect point X to either Y or Z. Initially XY is connected until the capacitor is fully charged by the battery. Then the switch connects X and Z, and the final charges on C1 and C2 are \(Q_1\) and \(Q_2\), respectively. What is the value of the ratio \(\frac{Q_2}{Q_1 + Q_2}\)?

• A. \(\frac{4}{5}\)
• B. \(\frac{1}{5}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{1}{2}\)

Hint:

When capacitors are connected in parallel, the charge distributes in direct proportion to their capacitances (\(Q \propto C\)).
Thus, the fraction of charge on \(C_2\) is simply \(\frac{C_2}{C_1 + C_2} = \frac{8}{2+8} = \frac{4}{5}\).
You do not need to calculate the actual voltages or intermediate charges.

Answer 1

The Correct Option is A

To find the ratio \(\frac{Q_2}{Q_1 + Q_2}\), follow these steps:

1. Initially, connect the switch to XY. Capacitor \(C_1\) is charged by the battery to its maximum charge, given by: \(Q = C_1 \times V\), where \(V\) is the battery voltage.
2. The charge on \(C_1\) when fully charged is: \(Q_{initial} = 2 \times 10^{-6} \times V = 2V \times 10^{-6} \ \text{Coulombs}\).
3. Now, the switch is moved to XZ, allowing \(C_1\) and \(C_2\) to share the charge. The total charge is conserved: \(Q_{total} = Q_1 + Q_2\).
4. After the charge redistribution, the voltage across each capacitor will be equal, since they are connected in parallel: \(\frac{Q_1}{C_1} = \frac{Q_2}{C_2}\).
5. Using conservation of charge: \(Q_{total} = 2V \times 10^{-6} = Q_1 + Q_2\).
6. Substituting for \(Q_1\) in terms of \(Q_2\): \(Q_1 = \frac{C_1}{C_2} \times Q_2\)
7. Simplifying, we have: \(Q_1 = \frac{2}{8} \times Q_2 = \frac{1}{4} \times Q_2\).
8. Using conservation of charge: \(2V \times 10^{-6} = \frac{1}{4}Q_2 + Q_2\), which simplifies to \(\frac{5}{4}Q_2 = 2V \times 10^{-6}\).
9. Solving for \(Q_2\): \(Q_2 = \frac{4}{5} \times 2V \times 10^{-6}\).
10. Therefore, the ratio is: \(\frac{Q_2}{Q_1+Q_2} = \frac{\frac{4}{5} \times 2V \times 10^{-6}}{2V \times 10^{-6}} = \frac{4}{5}\).

Hence, the value of the ratio \(\frac{Q_2}{Q_1 + Q_2}\) is \(\frac{4}{5}\).