Question

Observe the circuit in the figure and match the following (assume $q_1,q_2,$ and $q_3$ to be the charges on the three capacitors $C_1,C_2$ and $C_3$ respectively).

Column I Column II

i.	$q_1\left(\text{in}\mu C\right)$	p.	50
ii.	$q_2\left(\text{in}\mu C\right)$	q.	10/3
iii.	$q_3\left(\text{in}\mu C\right)$	r.	140/3
iv.	Potential difference across the 6$\mu F$ capacitor is (in V)	s.	25/3

• A. i - q ; ii - r ; iii - p ; iv - s
• B. i - p ; ii - r ; iii - s ; iv - q
• C. i - q ; ii - p ; iii - r ; iv - s
• D. i - s ; ii - p ; iii - r ; iv - q

Answer 1

The correct option is A i - q ; ii - r ; iii - p ; iv - s

The charges on three plates that are in contact add to zero, because these plates taken together form an isolated system, which cannot receive charges from the batteries.


Thus, $q_3-q_1$$-q_2=0$ ..........(i)
Applying Kirchhoff's law in loop $ABCFA$ and $CDEFC$ , we get

$-\frac{q_1}{2}-\frac{q_3}{6}+10=0$
$q_3+3q_1=60$ ................(ii)

and
$20-\frac{q_2}{4}-\frac{q_3}{6}=0$
$3q_2+2q_3=240$ ............(iii)

On solving (i), (ii) and (iii)
$q_1=\frac{10}{3}\mu C,q_2=\frac{140}{3}\mu C,q_3=50\mu C$

Potential difference across 6$\mu F$ capacitor is
$$\frac{q_3}{6}=\frac{50\mu C}{6\mu F}=\frac{25}{3}V$$