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

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 

Thus, power dissipated across internal resistance, P = 

And, total power dissipated =

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

i'm on my way to history not a mistory.