TS EAMCET · Physics · Capacitance
The circuit shows two capacitors A and B of capacitances \(\mathrm{C}\) and \(2 \mathrm{C}\) respectively. When they are fully charged, the cell is removed and the capacitors are connected with their plates of opposite polarities touching each other. Then
(a) Charge on \(\mathrm{A}\) is \(\frac{4 \mathrm{CE}}{9}\) (b) Charge on \(\mathrm{B}\) is zero (c) Loss of energy in this process is \(\left(\frac{\mathrm{CE}^2}{3}\right)\) The correct statement \(/ \mathrm{s}\) is/are
- A a and b are correct
- B b and c are correct
- C a, b and c are correct
- D c alone is correct
Answer & Solution
Correct Answer
(B) b and c are correct
Step-by-step Solution
Detailed explanation
\( C_{eq} = \frac{C \cdot 2C}{C+2C} = \frac{2C}{3} \) \( Q_{initial} = C_{eq}E = \frac{2CE}{3} \) \( U_{initial} = \frac{1}{2} C_{eq} E^2 = \frac{1}{2} \left( \frac{2C}{3} \right) E^2 = \frac{CE^2}{3} \)…
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