Capacitors

JEE Main level, single correct, medium

152 of 180

Question 152 of 180

Two identical capacitors have same capacitance CC. One of them is charged to the potential VV and other to the potential 2V2V. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

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Answer: (A) 14CV2\dfrac{1}{4}C{V}^{2}

The common potential after connection would be, VC=qnetCnet=CV+C(2V)2C{V}_{C}=\dfrac{{q}_{net}}{{C}_{net}}=\dfrac{CV+C\left(2V\right)}{2C}
⇒VC=3V2\Rightarrow {V}_{C}=\dfrac{3V}{2}
Therefore, loss of energy
=12CV2+12C(2V)2−122C(3V2)2=\dfrac{1}{2}C{V}^{2}+\dfrac{1}{2}C{\left(2V\right)}^{2}-\dfrac{1}{2}2C{\left(\dfrac{3V}{2}\right)}^{2}
=(CV24)=\left(\dfrac{C{V}^{2}}{4}\right)

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