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A capacitor of capacitance C1 = 1 µF charged up to a voltage V = 110 V is then connected in parallel to the terminals connected in series and possessing capacitances C2 = 2 µF and C3 = 3 µF. Then, the amount of charge that will flow through the connecting wires is
  • a)
    40 µC
  • b)
    50 µC
  • c)
    60 µC
  • d)
    110 µC
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A capacitor of capacitance C1 = 1 µF charged up to a voltage V = 110 ...
Given information:
Capacitance of C1 = 1 µF
Voltage across C1 = 110 V
Capacitance of C2 = 2 µF
Capacitance of C3 = 3 µF

Analysis:
When C1 is connected in parallel to C2 and C3, the total capacitance (Cp) of the parallel combination can be calculated using the formula:
1/Cp = 1/C2 + 1/C3

Calculation:
1/Cp = 1/2 + 1/3
1/Cp = 3/6 + 2/6
1/Cp = 5/6

Cp = 6/5 µF

The total charge (Q) stored in a capacitor can be calculated using the formula:
Q = CV

Calculation:
Q1 = C1 * V
Q1 = 1 µF * 110 V
Q1 = 110 µC

The charge on the parallel combination of capacitors (Qp) is equal to the charge on C1.

Qp = Q1
Qp = 110 µC

Therefore, the amount of charge that will flow through the connecting wires is 110 µC, which is option (c).
Free Test
Community Answer
A capacitor of capacitance C1 = 1 µF charged up to a voltage V = 110 ...
Initial charge on C1 is Q1 = C1V = 110 µC
Let x charge flow through wires.
where
Solve to get x = 60 µC.
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A capacitor of capacitance C1 = 1 µF charged up to a voltage V = 110 V is then connected in parallel to the terminals connected in series and possessing capacitances C2 = 2 µF and C3 = 3 µF. Then, the amount of charge that will flow through the connecting wires isa)40 µCb)50 µCc)60 µCd)110 µCCorrect answer is option 'C'. Can you explain this answer?
Question Description
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