When a 29-ω resistor is connected across a 15-v battery, a current of 471 ma flows. what is the internal resistance of the battery?

Remark

You can think of there being 2 resistors in series. The 29W resistor is connected to the external poles of the battery. The battery itself is in series with the internal resistor. She the diagram below. You want to ignore everything but the resistor the battery and the internal resistor of the battery.

Givens

I = 471 ma = 471 / 1000 = 0.471 Amps

R = 29 ohms is what I take that to be.

r = ? it is the internal resistance.

E = 15 volts.

Formula

E = I * R1

R1 = R + r

Sub and Solve

15 V = (29 + r) * 0.471 Divide by 0.471

15 / 0.471 = 29 + r

31.85 = 29 + r Subtract 29

31.85 - 29 = r

2.85 = r

The internal resistance is 2.85.