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2. Electric Potential and Capacitance
hard
A capacitor of capacitance $C$ is initially charged to a potential difference of $V$ $volt$. Now it is connected to a battery of $2V$ 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
$0.5$
Solution
Here $q_{i}=C V$ and $q_{f}=2 C V$
As battery is connected with opposite polarity so charge flow through the battery is
$\Delta q=q_{f}-\left(-q_{i}\right)=2 C V+C V=3 C V$
Initial energy stored $, U_{i}=\frac{1}{2} C V^{2}$ and
Final energy stored $, U_{f}=\frac{1}{2} C(2 V)^{2}=2 C V^{2}$
Generated heat, $H=\left(U_{f}-U_{i}\right)-\Delta q(2 V)=2 C V^{2}-\frac{1}{2} C V^{2}-6 C V^{2}=-\frac{9}{2} C V^{2}$
Thus, $\frac{H}{U_{f}}=\frac{9 C V^{2}}{2\left(2 C V^{2}\right)}=2.25$
Standard 12
Physics
