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KCET2018PhysicsCapacitance

Two capacitors of ( 3 F ) and ( 6 F ) are connected in series and a potential difference of ( 900 ~V ) is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is

Options

  1. AZero
  2. B( 100 ~V )
  3. C( 200 ~V )
  4. D( 400 ~V )

Correct answer

C. ( 200 ~V )

Step-by-step solution

If ( Q ) is charge on capacitor and ( C ) is capacitance then, potential difference ( V= Q C ) When two capacitors are connected in series, then equivalent capacitance is [ array l 1 C = 1 C₁ + 1 C₂ = 1 3 F + 1 6 F C= 6 3 3+6 =2 F Now, Q =C V Given, V =900 ~V array ] [ Therefore, Q=2 900=1800 C ] When two capacitors are connected in parallel, then equivalent capacitance is [ array l C=C₁+C₂=3 F+6 F=9 F Therefore V= Q C = 1800 C 9 F =200 ~V array ] Thus, potential difference across the combination is ( 200 ~V )

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