Two capacitors of capacitances 3 μF and 6 μF are charged to potentials of 2 V and 3 V, respectively. Now, if they are connected in parallel, the charge on each of the two capacitors, will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 8 μC and 16 μC
Given:
$C_1 = 3\,\mu F,\; V_1 = 2\,V$
$C_2 = 6\,\mu F,\; V_2 = 3\,V$
Initial charges:
$Q_1 = C_1 V_1 = 3 \times 2 = 6\,\mu C$
$Q_2 = C_2 V_2 = 6 \times 3 = 18\,\mu C$
Total charge when connected in parallel:
$Q_{total} = Q_1 + Q_2 = 6 + 18 = 24\,\mu C$
Equivalent capacitance:
$C_{eq} = C_1 + C_2 = 3 + 6 = 9\,\mu F$
Final common potential:
$V_f = \frac{Q_{total}}{C_{eq}} = \frac{24}{9} = \frac{8}{3}\,V \approx 2.67\,V$
Final charge on each capacitor:
$Q_1' = C_1 V_f = 3 \times \frac{8}{3} = 8\,\mu C$
$Q_2' = C_2 V_f = 6 \times \frac{8}{3} = 16\,\mu C$
Final Answer:
$Q_1' = 8\,\mu C,\; Q_2' = 16\,\mu C$