Question

Two capacitors of 25 $\mu Fand100\mu F$ are connected in series to a source of 120 V. Keeping their
charges unchanged, they are separated and connected in parallel to each other. Find out energy loss in the process.

• A. 5.2 J
• B. 52 J
• C. 50.2 J
• D. 0.052 J

Answer 1

The correct option is D 0.052 J

$\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{25}+\frac{1}{100}=\frac{1}{100}=\frac{1}{20}{C}_s=20\mu F=20\times {10}^{-6}F{U}_1=\frac{1}{2}{C}_s{V}^2=\frac{1}{2}(20\times {10}^{-16})(120{)}^2=144\times {10}^{-3}I$
Charge on each capacitor,
$q_1=q_2=C_s.V=20\times 120=2400\mu C$
In parallel, $C_p=C_1+C_2$
$=25+100=125\mu F=125\times {10}^{-6}F$
$\therefore U_2=\frac{Q^2}{2C_p}={\frac{\left[\right(2400+2400)\times {10}^{-6}]}{2\times 125\times {10}^{-6}}}^2=92.16\times {10}^{-3}J$
Loss of energy $=U_1-U_2=(144-92.16)\times {10}^{-3}J=51.84\times {10}^{-3}J=0.052J$