JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Current Electricity

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A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is:

(A)0.50 W
(B)0.10 W✓ Correct
(C)0.125 W
(D)0.072 W

Answer: (B)

Solution

The drop across the internal resistance is

ε - V = 3.0 - 2.5 = 0.5

V. The current follows from the external resistor's power and the terminal voltage,

I = \frac{P _{R}}{V} = \frac{0.5}{2.5} = 0.2

A. The power wasted inside the battery is then

P_{r} = I (ε - V) = 0.2 \times 0.5 = 0.1

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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