A wire of resistance 6.2 Ω is connected to a battery whose emf ε is 6.0 V and whose internal resistance is 0.45 Ω. In 2.0 min, how much energy is (a) transferred from chemical to electrical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

Answer: a) 648 J; b) 602.64 J c) 43.74 J

Explanation: In order to explain this question we have to take into account the following expressions:

Ohm law V=R*I

Power=V*I=I^2*R

Energy=Power*time

The current in the circuit is:

V / (Rw+Ri) = 6 / (6.2+0.45) = 0.9A

The power in the battery is: I*V=0.9*6=5.4W

Then the energy transferred from chemical to electrical in 2 minutes (120s) is equal:

E=5.4*120=648J

The dissipated thermal energy in the wire is:

Pw=I^2*R=0.9^2*6.2=5.022W

Ew=5.022*120=602.64J

Finally, the thermal energy dissipated in the battery is:

PRi = I^2*Ri=0.9^2*0.45=0.36 W

ERi=0.36*120 = 43.74 J