Question

A capacitor of capacity $10\mu F$ is charged to $40V$ and a second capacitor of capacity $15\mu F$ is charged to $30V$. If the capacitors are connected in parallel, the amount of charge that flows from the smaller capacitor to higher capacitor in $\mu C$ is:

• A. $200$
• B. $30$
• C. $60$
• D. $250$

Answer 1

Answer: (C)

Initially, the charge of first capacitance, $Q_1=C_1{V}_1=10\times 40=400\mu C$ and the charge of second one, $Q_2=C_2{V}_2=15\times 30=450\mu C$

When the capacitors are connected in parallel, potential across them will be same. Thus, the common potential, $V=\frac{Q_1+Q_1}{C_1+C_2}=\frac{400+450}{10+15}=34V$

Now charges on capacitors are $Q_1^{\prime }=C_{1}V=10\times 34=340\mu C$ and $Q_2^{\prime }=C_{2}V=15\times 34=510\mu C$

Charge flow $=510-450=60\mu C$