Two capacitors of capacitances \( 8\,\mu\mathrm{F} \) and \( 20\,\mu\mathrm{F} \) are connected in series with a battery. The voltage across the \( 8\,\mu\mathrm{F} \) capacitor is 5 V. The total battery voltage is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7 V
Capacitors in series carry the same charge, and the battery voltage is the sum of the voltages across them.
\( \displaystyle Q = C_1 V_1 = 8\,\mu\mathrm{F} \times 5\,\mathrm{V} = 40\,\mu\mathrm{C} \)
\( \displaystyle V_2 = \frac{Q}{C_2} = \frac{40\,\mu\mathrm{C}}{20\,\mu\mathrm{F}} = 2\,\mathrm{V} \)
\( \displaystyle V = V_1 + V_2 = 5\,\mathrm{V} + 2\,\mathrm{V} = 7\,\mathrm{V} \)
In series the smaller capacitor takes the larger voltage, so the \( 20\,\mu\mathrm{F} \) capacitor must have less than 5 V across it; that alone rules out 10 V and 12.5 V.