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JEE Advanced2009PhysicsCurrent ElectricityActual

For the circuit shown in the figure

Options

  1. Athe current I through the battery is 7.5 ~mA
  2. Bthe potential difference across R_L is 18 ~V
  3. Cratio of powers dissipated in R₁ and R₂ is 3
  4. Dif R₁ and R₂ are interchanged, magnitude of the power dissipated in R_L will decrease by a factor of 9.

Correct answer

A. the current I through the battery is 7.5 ~mA

Step-by-step solution

R_ total =2+ 6 1.5 6+1.5 =3.2 k (A) I= 24 ~V 3.2 k =7.5 ~mA =I_ R₁ I_ R₂ = ( R_L R_L+R₂ ) II= 1.5 7.5 7.5=1.5 ~mA (B) V_ R_L = (I_ R_L ) (R_L )=9 ~V (C) P_ R₁ P_ R₂ = (I_ R₁ ^2 ) R₁ (I_ R₂ ^2 ) R₂ = (7.5)^2(2) (1.5)^2(6) = 25 3 (D) When R₁ and R₂ are interchanged, then R₂ R_L R₂+R_L = 2 1.5 3.5 = 6 7 k Now potential difference across R_L will be V_L=24 [ 6 / 7 6+6 / 7 ] Earlier it was 9 ~V Since, P= V^2 R or P V^2 In new situation potential difference has been decreased three times. Therefore, power dissipated will

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