Capacitors

JEE Main level, numerical answer, medium

128 of 180

Question 128 of 180

A capacitor of capacitance CC and potential VV has energy E.E. It is connected to another capacitor of capacitance 2C2C and potential 2V2V. Then the loss of energy is x3E\dfrac{x}{3}E, where xx is _______.

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Answer: 2

Given, E=12CV2E=\dfrac{1}{2}C{V}^{2}
Required energy loss
=12C1C2C1+C2(V1−V2)2=12C×2CC+2C(V−2V)2=\dfrac{1}{2}\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}{\left({V}_{1}-{V}_{2}\right)}^{2} =\dfrac{1}{2}\dfrac{C\times 2C}{C+2C}{\left(V-2V\right)}^{2}
=26CV2=23⋅E=\dfrac{2}{6}C{V}^{2}=\dfrac{2}{3}\cdot E
∴x=2∴x=2

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