AP EAMCET · PHYSICS · Capacitance
A \(4 \mu \mathrm{F}\) capacitor is charged by a \(200 \mathrm{~V}\) battery. It is then disconnected from the supply and is connected to another uncharged \(2 \mu \mathrm{F}\) capacitor. During the process, loss of energy (in \(\mathrm{J}\) ) is
- A \(3.43 \times 10^{-2}\)
- B \(2.67 \times 10^{-2}\)
- C \(2.67 \times 10^{-4}\)
- D \(3.43 \times 10^{-4}\)
Answer & Solution
Correct Answer
(B) \(2.67 \times 10^{-2}\)
Step-by-step Solution
Detailed explanation
Charge stored at the capacitor \(q=C_1 V_1=4 \times 200=800 \mu \mathrm{C}\) When this capacitor is connected with a uncharged capacitor, then common potential on both capacitors \(V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{800+0}{4+2}=\frac{800}{6} \mathrm{~V}\) Loss in energy =…
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