Question

A 12-V car battery has an internal resistance of 0.020 Ω. When the headlights and stereo are on, a current of 25.0 A is drawn from the battery. What is the value of the terminal voltage when this current is drawn from the battery?

Answer 1

Answer:

11.5 V

Explanation:

Let E be the emf of the battery when it's not supplying current. Let the external resistance (that of the stereo and the headlights) be R. Let the internal resistance of the battery be r and the current be I.

The terminal voltage, V, is, Ohm's law,

$V = IR$

With the load connected, the equivalent resistance, $R_E$, given by

$R_E=\dfrac{E}{I} = \dfrac{12}{25}=0.480\Omega$

Because R and r are in series, $R_E$ is their sum:

0.480 = R + 0.020

R = 0.460

The terminal voltage = 0.460 × 25 = 11.5 V

Answer 2

Answer:

The terminal voltage is 11.5 V

Explanation:

E = I(R + r)

E is the electromotive force (emf) of the car battery = 12 V

I is the current drawn from the battery = 25 A

R is the resistance of the car battery

r is the internal resistance of the car battery = 0.02 ohm

R = (E/I) - r = (12/25) - 0.02 = 0.48 - 0.02 = 0.46 ohm

V = IR = 25 × 0.46 = 11.5 V