Question

In the circuit shown below, the charge on the $60\mu F$ capacitor is

• A. $150\mu C$
• B. $100\mu C$
• C. $50\mu C$
• D. $75\mu C$

Answer 1

Answer: (B)

The circuit shown in the figure is a balanced Wheatstone bridge, so the $80\mu F$ doesn't have any effect on the circuit and it can be removed.
After removing the $80\mu F$ , we see that the $30\mu F$ and $60\mu F$ are in series. Similarly, the $15\mu F$ and $30\mu F$ are also in series.
$C_1=\frac{30\times 60}{30+60}=20\mu F$ , $C_2=\frac{15\times 30}{15+30}=10\mu F$
Hence, the equivalent capacitance of the entire circuit is
$C=C_1+C_2=30\mu F$
$C_{eq}=\frac{C\times 30}{C+30}=15\mu F$
The charge that flows through the battery = $10\times 15=150\mu C$
This charge will be divided into two branches in the ratio of capacitances.
On solving we get that charge on $60\mu F$ = $100\mu C$