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Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?
  • a)
    The charges on the free plated connected together are destroyed.
  • b)
    The energy stored in the system increases.
  • c)
    The potential difference between the free plates is 2 V.
  • d)
    The potential difference remains constant.
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Two identical capacitors are joined in parallel, charged to a potentia...
When the two capacitors charged to same potential are connected in series, then total potential difference V′ = V1 + V2 = V + V = 2V
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Two identical capacitors are joined in parallel, charged to a potentia...
Explanation:
When two identical capacitors are joined in parallel and charged to a potential V, the charges on the plates are equal and have the same magnitude. Let's assume that each capacitor has a charge of Q.

1. Charges on the free plates:
When the capacitors are separated and then connected in series, the positive plate of one capacitor is connected to the negative plate of the other capacitor. This means that the positive charges on one plate will attract the negative charges on the other plate, resulting in a redistribution of charges.

The charges on the free plates connected together are not destroyed. Instead, the positive plate of one capacitor will have a net positive charge and the negative plate of the other capacitor will have a net negative charge. These charges will remain on the plates even after they are connected together.

2. Energy stored in the system:
The energy stored in a capacitor is given by the formula: E = 1/2 * C * V^2, where E is the energy, C is the capacitance, and V is the potential difference.

When the capacitors are joined in parallel, the total capacitance of the system is doubled, but the potential difference remains the same. Therefore, the energy stored in the system increases because the capacitance has increased.

3. Potential difference between the free plates:
When the capacitors are connected in series, the potential difference across each capacitor is the same. In this case, since the capacitors are identical, the potential difference across each capacitor is V.

Since the positive plate of one capacitor is connected to the negative plate of the other capacitor, the potential difference between the free plates is the sum of the potential differences across each capacitor. Therefore, the potential difference between the free plates is 2V.

4. Potential difference remains constant:
The potential difference across each capacitor remains constant when they are connected in series. However, the potential difference between the free plates is not constant in this case. As explained earlier, it is 2V.

Therefore, the correct answer is option 'c' - the potential difference between the free plates is 2V.
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Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?a)The charges on the free plated connected together are destroyed.b)The energy stored in the system increases.c)The potential difference between the free plates is 2 V.d)The potential difference remains constant.Correct answer is option 'C'. Can you explain this answer?
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