A capacitor of capacitance C and potential V has energy E . It is...

A capacitor of capacitance

C

and potential

V

has energy

E .

It is connected to another capacitor of capacitance

2C

and potential

2V

. Then the loss of energy is

\dfrac{x}{3}E

, where

x

is _______.

Correct answer

Step-by-step solution

Given, $E = \dfrac{1}{2}C{V}^{2}$

Required energy loss

$= \dfrac{1}{2}\dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}{\left({V}_{1} - {V}_{2}\right)}^{2} = \dfrac{1}{2}\dfrac{C \times 2C}{C + 2C}{\left(V - 2V\right)}^{2}$

$= \dfrac{2}{6}C{V}^{2} = \dfrac{2}{3} \cdot E$

$∴ x = 2$

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