Question

‘A’ and ‘B’ are two condensers of capacities 2 $\mu F$ and 4$\mu F$ They are charged to potential differences of 12V and 6V respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is :

• A. 24$\mu$C from A to B
• B. 8$\mu$ C from A to B
• C. 8$\mu$C from B to A
• D. 24$\mu$C from B to A

Answer 1

The correct option is B 8$\mu$ C from A to B

$Q_A=C_A{V}_A$
$=2\times {10}^{-6}\times 12$
$Q_A=24\times {10}^{-6}$
$Q_B=C_B{V}_B$
$=4\times {10}^{-6}\times 6$
$Q_B=24\times {10}^{-6}$
Now when plates are attached total charge $=Q_A+Q_B$
$Q=48\times {10}^{-6}C$
Let x he charge on A then Q-x charge is on B
$\Rightarrow x=2\mu F\times VandQ-x=4\mu F\times V.$
$\Rightarrow \frac{x}{Q-x}=\frac{1}{2}2x=Q-xx=\frac{9}{3}$
$x=16\times {10}^{-6};Q-x=32\times {10}^{-6}$
Now charge flows from A to B is given by,
$=final-initial$
$=32\times {10}^{-6}-20\times {10}^{-6}$
$△Q=8\times {10}^{-6}C$
$△Q=8\mu C$