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પ્રથમ કળ બંધ કરવામાં આવે છે,હવે કળ ખુલ્લી કરીને બંને કેપેસિટરમાં ડાઇઇલેકિટ્રક $3$ ભરતાં તંત્રની પહેલાની અને પછીની ઊર્જાનો ગુણોત્તર કેટલો થાય?
$3:1$
$5:1$
$3:5$
$5:3$
Solution
(c) Initially potential difference across both the capacitor is same hence energy of the system is
${U_1} = \frac{1}{2}C{V^2} + \frac{1}{2}C{V^2} = C{V^2}$$……(i)$
In the second case when key $K$ is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes $3C$, while potential difference across $A$ is $V$ and potential difference across $B$ is $\frac{V}{3}$ hence energy of the system now is
${U_2} = \frac{1}{2}\,(3C){V^2} + \frac{1}{2}\,(3C)\,{\left( {\frac{V}{3}} \right)^2}$$ = \frac{{10}}{6}\,C{V^2}$$……(ii)$
So, $\frac{{{U_1}}}{{{U_2}}} = \frac{3}{5}$
