Question

What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

Answer 1

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 $P = i^{2} R$

Thus, power dissipated across internal resistance, P = $0.83^{2} * 2 = 1.37 Watt$

And, total power dissipated =$0.83^{2} * 12 = 8.2 watt$

Thsu, percentage of pwer not avaible = 1.37/8.2 = 16.70%