Question

For the circuit shown in figure the charge on $4\mu F$ capacitor is

• A. 40 $\mu$ C
• B. 30 $\mu$ C
• C. 24 $\mu$ C
• D. 54 $\mu$ C

Answer 1

Answer: (C)

Combined capacity of $1\mu F$ and $5\mu F$
$=1+5=$ $6\mu F$
Now, $4\mu F$ and $6\mu F$ are in series.
$\therefore \frac{1}{C_s}=\frac{1}{4}+\frac{1}{6}=\frac{3+2}{12}=\frac{5}{12}$
$C_s=\frac{12}{5}\mu F$
Charge in the arm containing $4\mu F$ capacitor is
$q=C_s\times V=\frac{12}{5}\times 10=24\mu C$