The voltage across the terminals of a real battery will be smaller than its emf

Question

The voltage across the terminals of a real battery will be smaller than its emf due to what is called the internal resistance of the battery. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor, 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.

How much current flows when this battery is connected to the 2.5

What percentage of the battery's power is dissipated by the internal resistance?

Explanation / Answer

when connected to 2.5 ohm.

I = V / Req = V / (R + r)

I = 1.5 / (1 +2.5) = 0.43 A .....Ans

Pr = I^2 r = 0.184 Watt  

P_battery = V I = 1.5 x 0.43 = 0.643 Watt

% = (0.184/0.643) x 100 = 28.6 % ......Ans

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