Question:
A battery of \(3.0\ \mathrm{V}\) is connected to a resistor dissipating \(0.5\ \mathrm{W}\) of power. If the terminal voltage of the battery is \(2.5\ \mathrm{V}\), the power dissipated within the internal resistance is :
A battery of \(3.0\ \mathrm{V}\) is connected to a resistor dissipating \(0.5\ \mathrm{W}\) of power. If the terminal voltage of the battery is \(2.5\ \mathrm{V}\), the power dissipated within the internal resistance is :
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The terminal voltage is the voltage across the external resistor, so current = P/V; the lost voltage (3.0 - 2.5 V) drops across the internal resistance.
Updated On: Nov 5, 2026
- \(0.125\ \mathrm{W}\)
- \(0.50\ \mathrm{W}\)
- \(0.072\ \mathrm{W}\)
- \(0.10\ \mathrm{W}\)
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The Correct Option is D
Solution and Explanation
The resistor is connected across the battery terminals, so the voltage across it is the terminal voltage \(V = 2.5\ \mathrm{V}\).
Current: \(I = P/V = 0.5/2.5 = 0.2\ \mathrm{A}\).
Voltage lost inside the battery: \(3.0 - 2.5 = 0.5\ \mathrm{V}\), so \(r = 0.5/0.2 = 2.5\ \Omega\).
Power dissipated in the internal resistance: \(P_r = I^2 r = (0.2)^2 \times 2.5 = 0.10\ \mathrm{W}\).
Answer: (4) 0.10 W
Current: \(I = P/V = 0.5/2.5 = 0.2\ \mathrm{A}\).
Voltage lost inside the battery: \(3.0 - 2.5 = 0.5\ \mathrm{V}\), so \(r = 0.5/0.2 = 2.5\ \Omega\).
Power dissipated in the internal resistance: \(P_r = I^2 r = (0.2)^2 \times 2.5 = 0.10\ \mathrm{W}\).
Answer: (4) 0.10 W
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