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2. Electric Potential and Capacitance
hard
સમાન કેપેસિટન્સ $C$ ધરાવતાં કેપેસિટરને $V_1$ અને $V_2$ વોલ્ટેજથી ચાર્જ કરીને સમાંતરમાં જોડતાં તે કેટલી ઊર્જા ગુમાવશે?
A
$\frac{1}{4}C(V_1^2 - V_2^2)$
B
$\frac{1}{4}C(V_1^2 + V_2^2)$
C
$\frac{1}{4}C{\left( {{V_1} - {V_2}} \right)^2}$
D
$\frac{1}{4}C{\left( {{V_1} + {V_2}} \right)^2}$
(IIT-2002)
Solution
(c) Initial energy of the system
${U_i} = \frac{1}{2}C{V_1}^2 + \frac{1}{2}C{V_2}^2$
When the capacitors are joined, common potential $V = \frac{{C{V_1} + C{V_2}}}{{2C}} = \frac{{{V_1} + {V_2}}}{2}$
Final energy of the system
${U_f} = \frac{1}{2}(2C){V^2} = \frac{1}{2}2C\,{\left( {\frac{{{V_1} + {V_2}}}{2}} \right)^2} = \frac{1}{4}C{({V_1} + {V_2})^2}$
Decrease in energy = ${U_i} – {U_f} = \frac{1}{4}C{({V_1} – {V_2})^2}$
Standard 12
Physics
