Question

A capacitor of capacitance $C$ is initially charged to a potential difference of $V$ volt. Now it is connected to a battery of $2V$ volt with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be :

• A. $1.75$
• B. $2.25$
• C. $2.5$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

Here $q_i=CV$ and $q_f=2CV$
As battery is connected with opposite polarity so charge flow through the battery is $\Delta q=q_f-(-q_i)=2CV+CV=3CV$.
Initial energy stored ,$U_i=\frac{1}{2}C{V}^2$ and
Final energy stored , $U_f=\frac{1}{2}C(2V{)}^2=2C{V}^2$.
Generated heat, $H=(U_f-U_i)-\Delta q\left(2V\right)=2C{V}^2-\frac{1}{2}C{V}^2-6C{V}^2=-\frac{9}{2}C{V}^2$
Thus,$\frac{H}{U_f}=\frac{9C{V}^2}{2\left(2C{V}^2\right)}=2.25$