Question

A $4\mu F$ capacitor is charged by a 200V battery. It is then disconnected from the supply and is connected to another uncharged $4\mu F$ capacitor. During this process, Loss of energy (in J) is :

• A. Zero
• B. $5.33\times {10}^{-2}$
• C. $4\times {10}^{-2}$
• D. $2.67\times {10}^{-2}$

Answer 1

The correct option is C $4\times {10}^{-2}$

$V=200V$
$Q=4\times {10}^{-6}\times 200$
$Q=8\times {10}^{-4}C$
Initial energy $u_1=\frac{1}{2}\times 4\times {10}^{-6}\times 200\times 200$
$u_1=8\times {10}^{-2}J$
When it is disconnected from supply total change becomes constant
Let v be the common potential
$x=4\times {10}^{-6}\times u$; $Q-x=4\times {10}^{-6}\times u$
$\frac{x}{Q-x}=1$ $\Rightarrow x=\frac{Q}{2}$
$x=4\times {10}^{-4}C$
Now $v=\frac{x}{c}=\frac{4\times {10}^{-4}}{4\times {10}^{-6}}=100v$
$u_2=\frac{1}{2}\cdot (4\times {10}^{-6})\cdot (100{)}^2+\frac{1}{2}(4\times {10}^{-6}\left)\right(100{)}^2$
$u_2=4\times {10}^{-2}$
Loss in energy $u_2-u_1=4\times {10}^{-2}J$