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NEET · Physics · STD 12 - 2. Electric potential and capacitance
Two Identical capacttors \(\mathrm{C}_{1}\) and \(\mathrm{C}_{2}\) of equal capacitance are connected as shown in the circult. Terminals \(a\) and \(b\) of the key \(k\) are connected to charge capacitor \(\mathrm{C}_{1}\) using battery of \(emf \;V\; volt\). Now disconnecting \(a\) and \(b\) the terminals \(b\) and \(c\) are connected. Due to this, what will be the percentage loss of energy?.....\(\%\)

- A \(75\)
- B \(0\)
- C \(50\)
- D \(25\)
Answer & Solution
Correct Answer
(C) \(50\)
Step-by-step Solution
Detailed explanation
\(U_{\text {initial }}=\frac{1}{2} C V^{2},\) Loss \(=\frac{C \cdot C}{2(C+C)}(V-0)^{2}=\frac{1}{4} C V^{2}\) \(\%\) Loss \(=\frac{\frac{1}{4} \mathrm{CV}^{2}}{\frac{1}{2} \mathrm{CV}^{2}} \times 100=50 \%\)
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