A network of three capacitors each 9 µF connected in series and the fourth capacitor of 6 µF is supplied 400 V as shown in figure. The ratio of charge on the capacitor \( C_1 \) to the capacitor \( C_4 \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1 : 2
From the figure, \( C_1 \), \( C_2 \) and \( C_3 \) form one branch in series, and this branch is in parallel with \( C_4 \), so the full 400 V is across each branch. Capacitors in series all carry the same charge, which is the charge of their equivalent capacitor.
Series branch
\( \displaystyle \frac{1}{C_s} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{1}{3} \Rightarrow C_s = 3\,\mu\mathrm{F} \)
\( Q_1 = Q_s = C_sV = 3\times10^{-6}\times400 = 1.2\times10^{-3}\,\mathrm{C} \)
Capacitor \( C_4 \)
\( Q_4 = C_4V = 6\times10^{-6}\times400 = 2.4\times10^{-3}\,\mathrm{C} \)
Ratio
\( \displaystyle \frac{Q_1}{Q_4} = \frac{1.2\times10^{-3}}{2.4\times10^{-3}} = \frac{1}{2} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Capacitors in Parallel – …
