Question

Two capacitors $A(2\, \mu F )$ and $B(5 \,\mu F )$ are connected to two batteries as shown in the figure. Then the potential difference in volts between the plates of $A$ is

• A. 2
• B. 5
• C. 11
• D. 18

Answer 1

Answer: (B)

$(18-11)=\frac{Q}{2 \,\mu F}+\frac{Q}{5\, \mu F}=\frac{7 Q}{10 \,\mu F}$
$Q=10\, \mu C$
$V_{A}=\frac{Q}{C_{A}}=\frac{10 \,\mu C}{2\, \mu F }=5\, V$