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)

$(18-11)=\frac{Q}{2\mu F}+\frac{Q}{5\mu F}=\frac{7Q}{10\mu F}$
$Q=10\mu C$
$V_A=\frac{Q}{C_A}=\frac{10\mu C}{2\mu F}=5V$