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Two capacitors C 1 and C 2 in a circuit are joined as shown in the figure. The potentials of points A and B are V 1 and V 2 respectively. Then the potential of point D will be

Options

  1. AV 1 + V 2 2
  2. BC 2 V 1 + C 1 V 2 C 1 + C 2
  3. CC 1 V 1 + C 2 V 2 C 1 + C 2
  4. DC 2 V 1 - C 1 V 2 C 1 + C 2

Correct answer

C. C 1 V 1 + C 2 V 2 C 1 + C 2

Step-by-step solution

Consider the potential at D be V . The potential drop across C 1 is ( V − V 1 ) and C 2 is ( V 2 − V ) ∴ q 1 = C 1 ( V − V 1 ) ,   q 2 = C 2 ( V 2 − V ) As q 1 = q 2 [capacitors are in series] ∴ C 1 ( V − V 1 ) = C 2 ( V 2 − V ) V = C 1 V 1 + C 2 V 2 C 1 + C 2

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