Two capacitors marked \( 10\,\mu\mathrm{F},\ 400\,\mathrm{V} \) and \( 4\,\mu\mathrm{F},\ 100\,\mathrm{V} \) are connected in series. What is the maximum safe potential difference that can be applied across the series combination?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 140 V
Capacitors in series carry the same charge. Each capacitor can safely hold at most \( Q=CV \) at its rated voltage. So the capacitor with the smaller safe charge sets the limit for the combination.
The largest safe charge on each capacitor is:
\( \begin{aligned} Q_1 &= C_1V_1 \\ &= 10\,\mu\mathrm{F}\times400\,\mathrm{V} \\ &= 4000\,\mu\mathrm{C} \end{aligned} \)
\( \begin{aligned} Q_2 &= C_2V_2 \\ &= 4\,\mu\mathrm{F}\times100\,\mathrm{V} \\ &= 400\,\mu\mathrm{C} \end{aligned} \)
So the charge on the combination cannot exceed \( 400\,\mu\mathrm{C} \). At this charge the \( 4\,\mu\mathrm{F} \) capacitor has its full 100 V, and the voltage across the \( 10\,\mu\mathrm{F} \) capacitor is:
\( \begin{aligned} V_1' &= \frac{Q_2}{C_1} \\ &= \frac{400\,\mu\mathrm{C}}{10\,\mu\mathrm{F}} \\ &= 40\,\mathrm{V} \end{aligned} \)
The voltages across capacitors in series add:
\( \begin{aligned} V &= V_1'+V_2 \\ &= 40\,\mathrm{V}+100\,\mathrm{V} \\ &= 140\,\mathrm{V} \end{aligned} \)