Three capacitors of capacitances 3 μF, 6 μF and 12 μF are connected in series. Find potential difference across a 6 μF capacitor, if a battery of 7 V is connected across this combination:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 V
Capacitors in series carry the same charge Q, and the potential difference across each is \( V = Q/C \). First find the equivalent capacitance and the common charge.
\( \displaystyle \frac{1}{C_{\mathrm{eq}}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{4 + 2 + 1}{12} = \frac{7}{12}\,\mu\mathrm{F^{-1}} \Rightarrow C_{\mathrm{eq}} = \frac{12}{7}\,\mu\mathrm{F} \)
\( \displaystyle Q = C_{\mathrm{eq}}V = \frac{12}{7}\,\mu\mathrm{F}\times7\,\mathrm{V} = 12\,\mu\mathrm{C} \)
\( \displaystyle V_{6} = \frac{Q}{6\,\mu\mathrm{F}} = \frac{12\,\mu\mathrm{C}}{6\,\mu\mathrm{F}} = 2\,\mathrm{V} \)
Shortcut: In series, V is proportional to \( 1/C \), so the three voltages are in the ratio \( \frac{1}{3} : \frac{1}{6} : \frac{1}{12} = 4 : 2 : 1 \). The 7 V is shared in 7 parts of 1 V each, and the 6 μF capacitor gets 2 parts, which is 2 V.