Question

In the circuit shown in the figure, the ratio of charges on $5\mu F$ and $4\mu F$ capacitors is :

• A. $4/5$
• B. $3/5$
• C. $3/8$
• D. $1/2$

Answer 1

Answer: (C)

$3/8$

The capacitance 4 and (3,2,5) are in parallel so potential difference 4 and (3,5,1) will be $V=6V$.
$\therefore Q_4=C_{4}V=4\times 6=24\mu C$
The equivalent capacitance of (3,2,5) is $C_{eq}=\frac{3(5+2)}{3+(5+2)}=21/10\mu F$
and $Q_{eq}=C_{eq}V=21/10\times 6=63/5\mu C$
As 3 and (2,5) are in series so charge on 3 and (2,5) is equal to $Q_{eq}$.
Thus potential across (2,5) is $V_{25}=\frac{Q_{eq}}{C_{25}}=\frac{63/5}{5+2}=9/5V$
As 2 and 5 are in parallel so potential difference of each capacitors will be same i.e $9/5V$
Now charge on 5 is is $Q_5=C_5{V}_{25}=5\times 9/5=9\mu C$
$\therefore \frac{Q_5}{Q_4}=\frac{9}{24}=3/8$