A source with internal resistance does not necessarily deliver its full EMF across an external load. Draw the whole model, including the internal resistance, before applying a circuit equation. That makes the voltage and energy accounting much clearer.
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State the model and find the current
Represent a source by an ideal EMF of 12 V in series with internal resistance r = 1 ohm. Connect an external resistor R = 5 ohms, assuming constant resistances and ideal connecting wires.
The total resistance is 6 ohms, so I = 12/6 = 2 A. The terminal voltage across the load is IR = 10 V, also given by EMF - Ir = 12 - 2 = 10 V.
Follow the power
The ideal source supplies power EI = 12 × 2 = 24 W. The external resistor receives I2R = 20 W, while the internal resistance dissipates I2r = 4 W.
The powers add to 24 W. Counting 12 V across the external resistor would overstate its power and fail this conservation check. EMF is energy supplied per unit charge; despite its name, it is measured in volts, not newtons.
Change the external resistance
If R becomes 11 ohms, the total resistance is 12 ohms and the current is 1 A. The terminal voltage is now 11 V. The smaller internal voltage drop explains why terminal voltage is closer to the EMF.
The ideal source power is 12 W, divided into 11 W in the load and 1 W internally. A larger terminal voltage does not by itself imply a larger load power, because current also changes.
Keep the approximation visible
Real sources may not have constant EMF and internal resistance under all conditions. This simple model is useful within its stated assumptions, not a complete description of every battery.
This is an original calculation exercise for physics preparation, not an experimental instruction or official Olympiad question. Label the source, load and internal loss separately when reviewing your own circuit work.
Continue reading: British Physics Olympiad Mechanics: Momentum vs Kinetic Energy.
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Official information and further reading: Source 1
