Lets us consider an example:
Suppose a 10 ohm bulb is connected across the terminals of a 10 V
battery having 2ohm internal resistance.
Then total reistance in series we know, R1 + R2
Thus, R net = 10+ 2 = 12 ohm
The, current across circuit = 10/12= 0.833 A
Now, Power is given by [tex]P = i^{2} R \\ \\ [/tex]
Thus, power dissipated across internal resistance, P = [tex] 0.83^{2} * 2 = 1.37 Watt[/tex]
And, total power dissipated =[tex] 0.83^{2} * 12 = 8.2 watt[/tex]
Thsu, percentage of pwer not avaible = 1.37/8.2 = 16.70%
