Two capacitors of capacitances C and 2, C are charged to potential differences V and 2, V , respecti

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2. Electric Potential and Capacitance

hard

Two capacitors of capacitances $C$ and $2\, C$ are charged to potential differences $V$ and $2\, V$, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is$.....CV^2$

A

$4.5$

B

$4.16$

C

$0$

D

$1.5$

(JEE MAIN-2020)

Solution

$Q _{1}= CV \quad Q _{2}=2 C \times 2 V =4 CV$

$\Rightarrow$ By conservation of charge

$q_{i}=q_{f}$

$Q_{1}+Q_{2}=q_{1}+q_{2}$

$4 CV – CV =( C +2 C ) V _{ C }$

$v _{ C }=\frac{3 CV }{3 C } \Rightarrow V$

$\Rightarrow \frac{1}{2} \times(3 C) \times V_{ c }^{2}$

$=\frac{1}{2} \times 3 C \times V ^{2}=\frac{3}{2} CV ^{2}$

Standard 12

Physics

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