Question

Initially a battery circuit consists of a 25 ohm load connected to a battery with 3 ohms of internal resistance and the circuit draws 1.5A of current.

What is the terminal voltage and EMF of the batter?

How much (%) power is wasted?

Please help, this doesn't make sense given only V(term) = E - Ir

Answer 1

emf=IRt=1.5(28) volts
terminal voltage=above-1.5*3

waster power: 1.5^2*3

percent=wasted power/EMF*1.5

Answer 2

total R = 28

so i = EMF/28 = 1.5

so EMF = 1.5*28 = 42 volts

voltage drop internal = 1.5*3 = 4.5 volts

so terminal V = 42 -4.5 = 37.5 volts

power waste = i^2R = (1.5)^2 * (3)
= 6.75 Watts