TS EAMCET · Physics · Capacitance
A \(10 \mu \mathrm{~F}\) capacitor is charged by a 100 V battery. It is disconnected from the battery and is connected to another uncharged capacitor of capacitance \(30 \mu \mathrm{~F}\). During this process, the electrostatic energy lost by the first capacitor is
- A \(5 \times 10^{-2} \mathrm{~J}\)
- B \(1.25 \times 10^{-2} \mathrm{~J}\)
- C \(2.75 \times 10^{-2} \mathrm{~J}\)
- D \(3.75 \times 10^{-2} \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(3.75 \times 10^{-2} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{C}_1=10 \mu \mathrm{~F}, \mathrm{~V}_1=100 \mathrm{~V}, \mathrm{C}_2=\mu \mathrm{F}, \mathrm{V}_2=0\) During sharing of capacitors,…
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