Capacitors of \(2\ \mu\mathrm{F}\) and \(4\ \mu\mathrm{F}\) are joined in parallel. This pair is joined in series with a \(3\ \mu\mathrm{F}\) capacitor, and the combination is connected across a 12 V battery. The charge on the \(3\ \mu\mathrm{F}\) capacitor is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(24\ \mu\mathrm{C}\)
The parallel pair acts as one capacitor of \( 2+4=6\ \mu\mathrm{F} \). This pair is in series with the \( 3\ \mu\mathrm{F} \) capacitor. In a series combination, each part carries the same charge as the equivalent capacitor.
\( \begin{aligned} \frac{1}{C} &= \frac{1}{C_p}+\frac{1}{C_3} \\ &= \frac{1}{6}+\frac{1}{3} \\ &= \frac{1}{2}\ \mu\mathrm{F^{-1}} \end{aligned} \)
So \( C=2\ \mu\mathrm{F} \).
\( \begin{aligned} Q &= CV \\ &= 2\ \mu\mathrm{F}\times12\ \mathrm{V} \\ &= 24\ \mu\mathrm{C} \end{aligned} \)
From the above, the series combination carries \( 24\ \mu\mathrm{C} \). Hence, the charge on the \( 3\ \mu\mathrm{F} \) capacitor is \( 24\ \mu\mathrm{C} \).
NCERT (2026-27):
Capacitors in series – “… Thus, in the series combination, charges on the two plates (±Q) are the same on each capacitor. … Now we can regard the combination as an effective capacitor with charge Q and potential difference V.”
Summary – “11. For capacitors in the series combination, the total capacitance C is given by”
\( \displaystyle \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\dots \)
Summary – “In the parallel combination, the total capacitance C is:”
\( C=C_1+C_2+C_3+\dots \)