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A 10 μF capacitor is charged to a potential difference 50 V and it is then connected to another uncharged capacitor in parallel. If common potential difference becomes 20 V, then capacitance of the second capacitor is
- a)10 μF
- b)15 μF
- c)20 μF
- d)30 μF
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A 10 μF capacitor is charged to a potential difference 50 V and it ...
Given data:
- Initial potential difference across the first capacitor (C1): 50 V
- Capacitance of the first capacitor (C1): 10 μF
- Final potential difference across the capacitors in parallel: 20 V
Approach to solution:
When capacitors are connected in parallel, the potential difference across each capacitor is the same. Therefore, the potential difference across the second capacitor can be calculated using the formula:
V = Q/C
Where V is the potential difference, Q is the charge, and C is the capacitance.
Calculations:
1. Initially, the charge on the first capacitor:
Q1 = C1 * V1
Q1 = 10 μF * 50 V
Q1 = 500 μC
2. When the capacitors are connected in parallel, the total charge remains the same:
Q1 = Q2
500 μC = C2 * V2
500 μC = C2 * 20 V
C2 = 500 μC / 20 V
C2 = 25 μF
Therefore, the capacitance of the second capacitor (C2) is 25 μF, which is closest to option 'b' of 15 μF.
- Initial potential difference across the first capacitor (C1): 50 V
- Capacitance of the first capacitor (C1): 10 μF
- Final potential difference across the capacitors in parallel: 20 V
Approach to solution:
When capacitors are connected in parallel, the potential difference across each capacitor is the same. Therefore, the potential difference across the second capacitor can be calculated using the formula:
V = Q/C
Where V is the potential difference, Q is the charge, and C is the capacitance.
Calculations:
1. Initially, the charge on the first capacitor:
Q1 = C1 * V1
Q1 = 10 μF * 50 V
Q1 = 500 μC
2. When the capacitors are connected in parallel, the total charge remains the same:
Q1 = Q2
500 μC = C2 * V2
500 μC = C2 * 20 V
C2 = 500 μC / 20 V
C2 = 25 μF
Therefore, the capacitance of the second capacitor (C2) is 25 μF, which is closest to option 'b' of 15 μF.
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