Question

In the circuit shown in figure
(1) the charge on $C_{2}$ is greater than on $C_{1}$
(2) the charges on $C_{1}$ and $C_{2}$ are same
(3) potential difference across $C_{1}$ and $C_{2}$ are same
(4) potential difference across $C_{1}$ is greater than across $C_{2}$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (C)

Since there is only one path for the charge flow, capacitors $C_{1}$ and $C_{2}$ are in series via the cells of $\operatorname{emf} E_{1}$ and $E_{2}$

The net emf in the circuit
$E=E_{1}-E_{2}=12-6=6 V$
The equivalent capacity
$C =\frac{C_{1} C_{2}}{C_{1}+C_{2}}$
$=\frac{4 \times 8}{4+8}=\frac{8}{3} \mu F$
Charge on each capacitor $\left(q=q_{1}=q_{2}\right)$
$q=C V=\frac{C_{1} C_{2}\left(E_{1}-E_{2}\right)}{C_{1}+C_{2}}=\frac{8}{3} \times 6=16 \mu F$
For a capacitor $q=C V$ or $V=\frac{q}{C}$
$\therefore V_{1}=\frac{q_{1}}{C_{1}}=\frac{q}{C_{1}}=\frac{C_{2}\left(E_{1}-E_{2}\right)}{C_{1}+C_{2}}=\frac{16}{4}=4 \,V $
$V_{2}=\frac{q_{2}}{C_{2}}=\frac{q}{C_{2}}=\frac{C_{1}\left(E_{1}-E_{2}\right)}{C_{1}+C_{2}}=\frac{16}{8}=2 \,V$