Question

A car battery has an emf of 12.0 V and an internal resistance of 0.20 Ω. While the car is running, the resistance of the rest of the car (the “load” resistance) is 1.00 Ω.

What is the current through the battery?

What is the terminal voltage of the battery?

What is the power delivered to the rest of the car?

What is the power dissipated as heat by the internal resistance of the battery?

Answer 1

Solution) Emf E = 12 V

Load resistance R = 1 ohm

Internal resistance r = 0.20 ohms

Current through battery i = ?

i = E/(R+r)

i = 12/(1+0.20)

i = 10 A

Terminal voltage V = ?

V = E - ir

V = 12 - (10×0.20)

V= 10 V

Power delivered P = Vi = 10×10 = 100 W

Power dissipated P' = ?

P' = (i^2)(r)

P' = (10^2)(0.20)

P' = 20 W