Current Electricity

JEE Main level, single correct, medium

190 of 380

Question 190 of 380

A battery of 3.0V3.0\text{V} is connected to a resistor dissipating 0.5W0.5\text{W} of power. If the terminal voltage of the battery is 2.5V2.5\text{V}, the power dissipated within the internal resistance is:

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Answer: (C) 0.10W0.10\text{W}

Given,
Emf of the battery, E=3VE=3V
Potential difference across resister RR is VR=2.5V{V}_{R}=2.5V
Power dissipation in the resister is P=0.5WP=0.5W
By using Kirchhoff's voltage law is given by,
Vr+VR=E...(1){V}_{r}+{V}_{R}=E...(1)
Vr{V}_{r} is voltage across internal resister.
Using equation (1) and substitute the given values,
⇒Vr+2.5=3\Rightarrow {V}_{r}+2.5=3
⇒Vr=0.5\Rightarrow {V}_{r}=0.5
Now the ratio of potential across RR and rr is given by,
VRVr=IRIr=2.50.5=5\dfrac{{V}_{R}}{{V}_{r}}=\dfrac{IR}{Ir}=\dfrac{2.5}{0.5}=5
⇒Rr=5\Rightarrow \dfrac{R}{r}=5
Now the power dissipation across the internal resistance is given by
PRPr=I2RI2r=Rr\dfrac{{P}_{R}}{{P}_{r}}=\dfrac{{I}^{2}R}{{I}^{2}r}=\dfrac{R}{r}
⇒PRPr=5\Rightarrow \dfrac{{P}_{R}}{{P}_{r}}=5
⇒Pr=PR5\Rightarrow {P}_{r}=\dfrac{{P}_{R}}{5}
⇒Pr=0.55=0.1W\Rightarrow {P}_{r}=\dfrac{0.5}{5}=0.1W

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