In the circuit shown, the two capacitors are fully charged with the key K open. The charge that flows through the cell when K is closed is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → CV/2
Capacitors in series carry the same charge, and this charge is the charge drawn from the cell. From the figure, the key K is connected across capacitor 2. Closing K joins the two plates of capacitor 2 by a wire, so it discharges and the whole voltage V appears across capacitor 1. The cell then supplies only the extra charge needed by capacitor 1.
With K open, the two capacitors are in series:
\( \begin{aligned} C_{\mathrm{s}} &= \frac{C\times C}{C+C} \\ &= \frac{C}{2} \end{aligned} \)
\( \begin{aligned} Q_1 &= C_{\mathrm{s}}V \\ &= \frac{CV}{2} \end{aligned} \)
With K closed, capacitor 1 alone is across the cell:
\( Q_2=CV \)
The charge that flows through the cell is the increase in charge:
\( \begin{aligned} \Delta Q &= Q_2-Q_1 \\ &= CV-\frac{CV}{2} \\ &= \frac{CV}{2} \end{aligned} \)