Answer:

Given A battery has an emf of 12.0 V and an internal resistance of 0.05 ohms . Its terminals are connected to a load resistance of 3.00 ohms.

We have to find the current in the circuit and terminal voltage of the battery and also the power delivered.

here the energy is got from battery to load resistor.

we have emf = 12 v, internal resistance r = 0.05 ohms, load resistance R = 3.00 ohms.

current in circuit I = E / R + r = 12 / 3.00 + 0.05

I = 3.93 A

Terminal voltage = Δv = E - Ir

= 12 - 3.93 x 0.05 = 11.8035 v

Δv = IR = 3.93 x 3.00

= 11.8 V

Now power delivered to load resistor PR = I^2 R

= (3.93)^2 x 3.00

= 46.3 W

Power delivered to internal resistance Pr = I^2 r

= (3.93)^2 x 0.05

= 0.772240 W

Power delivered by battery = PR + Pr

= 46.3 + 0.772240 = 46.6722 W