MHT CET · Physics · Capacitance
Two condensers one of capacity \(\frac{C}{2}\) and other capacity \(\mathrm{C}\) are connected to a battery of voltage \(\mathrm{V}\) as shown. The work done in charging fully both the condensers is
- A \(\frac{1}{2} \mathrm{CV}^2\)
- B \(\frac{3}{4} \mathrm{CV}^2\)
- C \(\frac{3}{2} \mathrm{CV}^2\)
- D \(2 \mathrm{CV}^2\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{4} \mathrm{CV}^2\)
Step-by-step Solution
Detailed explanation
The two condensers are in parallel. Their equivalent capacitance is \(\mathrm{C}_{\mathrm{e}}=\mathrm{C}+\frac{\mathrm{C}}{2}=\frac{3 \mathrm{C}}{2}\)
The formula for work done is
\(\begin{aligned}
& \mathrm{W}=\frac{1}{2} \mathrm{C}_{\mathrm{e}} \mathrm{V}^2 \\
& \mathrm{~W}=\frac{1}{2}\left(\frac{3 \mathrm{C}}{2}\right) \mathrm{V}^2 \\
& \mathrm{~W}=\frac{3}{4} \mathrm{CV}^2
\end{aligned}\)
The formula for work done is
\(\begin{aligned}
& \mathrm{W}=\frac{1}{2} \mathrm{C}_{\mathrm{e}} \mathrm{V}^2 \\
& \mathrm{~W}=\frac{1}{2}\left(\frac{3 \mathrm{C}}{2}\right) \mathrm{V}^2 \\
& \mathrm{~W}=\frac{3}{4} \mathrm{CV}^2
\end{aligned}\)
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