A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V

Question

A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor known as the "internal resistance", as shown in the figure(Figure 1) . A typical battery has 1.0 ? internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 V, the value of the emf. Suppose the terminals of this battery are connected to a 2.4? resistor.

I got that potential difference between the two terminals is 1.1 V What is the fraction of the battery's power dissapated by the internal resistance?

Explanation / Answer

The internal resistance is effectively in series with the external load of 2.4 ohms.
So the combined resistance is 2.4 + 1 = 3.4 ohms.
Current = v/r = 1.5/3.4 = 0.441 amp
drop across internal resistance = I x R = 0.441 x 1 = 0.441 V
P.D = 1.5 - 0.441 = 1.059V
Power dissipated = I ^2 x R = 0.441^2 x 1 = 0.1945 watts
Power supplied = I ^2 x R (total)
= 0.441^2 x 3.4 = 0.661 W

So 0.1945 / 0.661 = 389/1322 is the fraction of power dissipated in the internal resistance

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