Question

The capacity of a condenser A is $10\mu F$ and it is charged using a battery of $100V$. The battery is disconnected and the condenser A is connected to a condenser B with common potential as $40V$. The capacity of B is:

• A. $8\mu F$
• B. $15\mu F$
• C. $2\mu F$
• D. $1\mu F$

Answer 1

The correct option is B $15\mu F$

$C_A=10\mu F$
$V_A=100V$
So, charge $q_A=C_A{V}_A$$=10\times {10}^{-6}\times 100$ $={10}^{-3}$

When the battery is disconnected, total charge must be constant.
So, $q_A+q_B={10}^{-3}$

Given common potential $\Delta V=40V$
So, $q_A=C_A\Delta V$$=10\times {10}^{-6}\times 40V=400\times {10}^{-6}V$
And, $q_B=C_B\Delta V$

$\Rightarrow {10}^{-3}=400\times {10}^{-6}+C_B\times 40$
or, $C_B=\frac{0.6\times {10}^{-3}}{40}$$=15\times {10}^{-6}F=15\mu F$