Question

A 2 $\mu F$ capacitor is charged to a potential of 10 V and another 4 $\mu F$ capacitor is charged to a potential of 20 V. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. The amount of heat evolved in the circuit is

• A. 300 $\mu J$
• B. 600 $\mu J$
• C. 900 $\mu J$
• D. 450 $\mu J$

Answer 1

The correct option is B 600 $\mu J$


Before connection
$Q_1=2\times 10=20\mu C$
$Q_2=4\times 20=80\mu C$
Stored energy $U_i=\frac{1}{2}C{V}^2$
$U_i=\frac{1}{2}2(10{)}^2\times {10}^{-6}+\frac{1}{2}4(20{)}^2\times {10}^{-6}=900\mu J$
After connection,
$Q_{\text{net}}=-20+80$
$=60\mu C$
$V=\frac{60}{2+4}=10volt$
$U=\frac{1}{2}6(10{)}^2=300J$
$|\Delta U|=900-300=600\mu J$