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AP EAMCET201823 Apr 2018Evening ShiftPhysicsCapacitanceActual

In the circuit shown in figure, if the point R is earthed and point P is given a potential of +1800 ~V , then charges on C₂ and C₃ are respectively

Options

  1. A2.4 10⁻³ C ; 1.2 10⁻³ C
  2. B1.6 10⁻³ C ; 0.8 10⁻³ C
  3. C3.2 10⁻³ C : 1.6 10⁻³ C
  4. D4.8 10⁻³ C ; 2.4 10⁻³ C

Correct answer

A. 2.4 10⁻³ C ; 1.2 10⁻³ C

Step-by-step solution

gathered C_ eq of system = (C₂ parallel C₃ ) series C₁ =1 / ( 1 3 + 1 (4+2) )=2 F gathered So, charge taken from source =q_ eq =C_ eq V=1800 2 10⁻⁶ C =3600 C Potential droop across C₁= q_ C₁ C_ C₁ = 3600 10⁻⁶ 3 10⁻⁶ =1200 ~V So, potential drop across combination of 4 F and 2 F capacitors =1800-1200=600 ~V Hence, aligned q₂ & =C₂ V_ Q R =4 10⁻⁶ 600=2.4 10⁻³ C and q₃ & =C₃ V_ Q R =2 10⁻⁶ 600=1.2 10⁻³ C aligned

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