A loaded cell model and a changed external network
A theoretical direct-current source is modeled as an ideal electromotive force of 12.0 V in series with a fixed internal resistance of 1.0 ohm. Initially its external load is a 3.0 ohm resistor. Ideal ammeters have negligible resistance and an ideal voltmeter draws no current. A second calculation places a 6.0 ohm resistor in parallel with the original load while leaving the source model unchanged. A student argues that because the source label says 12 V, both resistors must retain 12 V across them in the second calculation. Another assumes that the total current remains the same because no extra source has been added. The task is only a supplied-parameter circuit model: no component power rating or real battery operating approval is implied by the arithmetic.
1. Determine current, terminal voltage and load efficiency with only the 3.0 ohm load.
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Total resistance is 1.0 + 3.0 = 4.0 ohms, so current is 12.0/4.0 = 3.0 A. Terminal voltage is 12.0 - 3.0(1.0) = 9.0 V. External power is I squared R = 27 W and source conversion rate is emf times current = 36 W, giving load efficiency 27/36 = 75 percent in this model.
2. Find the total current and both branch currents after adding the parallel resistor.
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The external equivalent is 3 times 6 divided by 3 + 6 = 2.0 ohms. Total current is 12/(1 + 2) = 4.0 A. Terminal voltage falls to 12 - 4(1) = 8.0 V. Branch currents are 8/3 = 2.667 A and 8/6 = 1.333 A; their sum is 4.0 A. The ideal emf stayed fixed, but terminal voltage did not because internal voltage loss increased.