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Two capacitors of capacitances 3 μF and 6 μF are charged to a potential of 12 V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each capacitor will be
  • a)
    6 V
  • b)
    4 V
  • c)
    3 V
  • d)
    zero
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Two capacitors of capacitances 3 μF and 6 μF are charged to a po...
Explanation:
- Initial Charge on Capacitors:
- The initial charge on a capacitor is given by Q = CV.
- For the 3 μF capacitor: Q1 = 3 μF × 12 V = 36 μC
- For the 6 μF capacitor: Q2 = 6 μF × 12 V = 72 μC
- When Connected in Series:
- When the capacitors are connected in series, the equivalent capacitance is given by:
- 1/Ceq = 1/C1 + 1/C2
- 1/Ceq = 1/3 μF + 1/6 μF
- 1/Ceq = 1/2 μF
- Ceq = 2 μF
- Charge Distribution:
- When capacitors are connected in series, they share the same charge.
- Therefore, the total charge Q is distributed among the capacitors based on their capacitance.
- Q1 = C1 × Veq / (C1 + C2) = 3 μF × Veq / 2 μF = 3/2 Veq
- Q2 = C2 × Veq / (C1 + C2) = 6 μF × Veq / 2 μF = 6/2 Veq
- Final Potential Difference:
- The potential difference across each capacitor in the final state is given by Q/C.
- V1 = Q1 / C1 = (3/2 Veq) / 3 μF = 1/2 Veq
- V2 = Q2 / C2 = (6/2 Veq) / 6 μF = 1/2 Veq
- Conclusion:
- As V1 = V2 = 1/2 Veq, the potential difference across each capacitor will be zero when connected in series.
Therefore, the correct answer is option 'D' (zero).
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Two capacitors of capacitances 3 μF and 6 μF are charged to a potential of 12 V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each capacitor will bea)6 Vb)4 Vc)3 Vd)zeroCorrect answer is option 'D'. Can you explain this answer?
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