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2. Electric Potential and Capacitance
hard
A condenser of capacity ${C_1}$ is charged to a potential ${V_0}$. The electrostatic energy stored in it is ${U_0}$. It is connected to another uncharged condenser of capacity ${C_2}$ in parallel. The energy dissipated in the process is
A
$\frac{{{C_2}}}{{{C_1} + {C_2}}}{U_0}$
B
$\frac{{{C_1}}}{{{C_1} + {C_2}}}{U_0}$
C
$\left( {\frac{{{C_1} - {C_2}}}{{{C_1} + {C_2}}}} \right){U_0}$
D
$\frac{{{C_1}{C_2}}}{{2({C_1} + {C_2})}}{U_0}$
Solution
(a) Loss of energy during sharing $=$ $\frac{{{C_1}{C_2}{{({V_1} – {V_2})}^2}}}{{2({C_1} + {C_2})}}$
In the equation, put ${V_2} = 0,\;{V_1} = {V_0}$
Loss of energy $ = \frac{{{C_1}{C_2}V_0^2}}{{2({C_1} + {C_2})}}$
$ = \frac{{{C_2}{U_0}}}{{{C_1} + {C_2}}}$ $\left[ {\;{U_0} = \frac{1}{2}{C_1}V_0^2} \right]$
Standard 12
Physics
