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KCET2013PhysicsCapacitance

When a potential difference of 10³ ~V is applied between A and B , a charge of 0.75 mC is value of C is (in F )

Options

  1. A1 2
  2. B2
  3. C2.5
  4. D3

Correct answer

B. 2

Step-by-step solution

In the given circuit, 2 F and 2 F capacitors are in series _ S = 2 2 2+2 =1 F So, equivalent circuit will be Now C F and 1 F are in parallel which is in series with 1 F C_ eff = (C+1) 1 (C+1)+1 = C+1 C+2 Given, q=0.75 10⁻³ C =750 10⁻⁶ C , V=10³ ~V So, C_ eff = q V C+1 C+2 = 750 10⁻⁶ 10³ =750 10⁻³ ~F =0.75 F 0.75= C+1 C+2 3 4 = C+1 C+2 3 C+6=4 C+4 C=2 F

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