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The figure below shows three capacitors of capacitance 3µF, 2µF, and 6µF. Connected to a 12v supply circuit.
PhycF4T1P2Q14d
Calculate:

  1. The total capacitance of the circuit. 
  2. The total charge stored in the circuit. 
  3. The potential differences across 2µF capacitor.

1 Answer

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  1. CP=2+6=8µf
    CT=3x8 = 24 = 2.18 µf.
          3+8    11
  2.  Q =CV 
    = 2.18 X 10-6 X 12
    = 2.616 X 10-5 C
  3. p.d across 2 µf = 2.616 x 10-6
                                   8 x 10-6
    = 3.27V 
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