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Three capacitors C1 = 3 μF, C2 = 6 μF and C3 = 12 μF are joined in series. This series combination is connected to a 14 volt source. The P.D across the plates of capacitor C2 is
  • a)
    2 volt
  • b)
    4 volt
  • c)
    6 volt
  • d)
    8 volt
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Three capacitors C1 = 3 μF, C2 = 6 μF and C3 = 12 μF are joined in se...
Solution:

When capacitors are joined in series, the charge on all of them is the same. Let Q be the charge on each capacitor.

The voltage across each capacitor can be determined using the formula V = Q/C, where V is the voltage across the capacitor and C is the capacitance of the capacitor.

Since the capacitors are in series, the total capacitance is given by:

1/C = 1/C1 + 1/C2 + 1/C3

1/C = 1/3 + 1/6 + 1/12

1/C = 1/2

C = 2 μF

The total charge on the capacitors is given by:

Q = CV

Q = 2 x 14

Q = 28 μC

The voltage across capacitor C2 can be determined using the formula V = Q/C2:

V = 28/6

V = 4 volts

Therefore, the P.D across the plates of capacitor C2 is 4 volts. Hence, the correct answer is option B.
Free Test
Community Answer
Three capacitors C1 = 3 μF, C2 = 6 μF and C3 = 12 μF are joined in se...
Let CS is the effective capacitance.
Charge on C2 = Charge on CS
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