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← IIT-JEE 2010 · Paper 1
PhysicsInteger+3Question 77 of 84

IIT-JEE 2010 · Paper 1, Physics question 77

When two identical batteries of internal resistance 1Ω each are connected in series across a resistor R, the rate of heat produced in R is J1J_1. When the same batteries are connected in parallel across R, the rate is J2J_2. If J1=2.25J2J_1=2.25J_2 then the value of R in Ω is

Answer and worked solution

Official answer: 4

Series. emf 2E2E, internal resistance 2 Ω: J1=4E2R(R+2)2J_1=\dfrac{4E^2R}{(R+2)^2}.

Parallel. emf EE, internal resistance 0.5 Ω: J2=E2R(R+0.5)2J_2=\dfrac{E^2R}{(R+0.5)^2}.

Condition.
J1J2=4(R+0.5)2(R+2)2=2.25  ⟹  2(R+0.5)R+2=1.5  ⟹  R=4 Ω.\frac{J_1}{J_2}=\frac{4(R+0.5)^2}{(R+2)^2}=2.25\implies\frac{2(R+0.5)}{R+2}=1.5\implies R=4\ \Omega.

Answer: 4

Topics:EMF and internal resistanceElectrical power and heating
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