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
When capacitor are connected in series,
$\frac{1}{C_{total}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
$=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}=\frac{7}{12}$
$C_{total}=\frac{12}{7}≃1.71μF$
Charge, $Q=C_{total}×V_{battery}$
$=1.71μF×7V$
$=11.97μC$
$∴V=\frac{Q}{C}=\frac{11.97}{6}≃2V$