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  • In the circuit shown (Figure), what is the change of total electrical energy stored in the capacitors when the key is pressed? <img alt="" src="https://cdn.entrance360.com/media/uploads/2

In the circuit shown (Figure), what is the change of total electrical energy stored in the capacitors when the key is pressed?

Option: 1

\mathrm{ \frac{C V^2}{12}}


Option: 2

\mathrm{ \frac{7 C V^2}{8}}


Option: 3

\mathrm{ \frac{5 C V^2}{4}}


Option: 4

\mathrm{ \frac{3 C V^2}{8}}


Answers (1)

best_answer

Initially, all three capacitors are in parallel

\mathrm{E_1=\frac{1}{2} \times 3 C V^2=\frac{3}{2} C V^2 }
When key is closed, two capacitors are in series
\begin{aligned} & \mathrm{E_f=\frac{1}{2} C\left(\frac{V}{2}\right)^2 \times 2=\frac{C V^2}{4} }\\ &\mathrm{ \Delta E=E_1-E_f=\frac{5}{4} C V^2 }\end{aligned}

Posted by

Gautam harsolia

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