Question

A voltmeter is connected across the terminals of a 15.0 V battery and a 75 ohm appliances connected across its terminals. If the voltmeter reads 11.3 V How much power is dissipated by the appliance? What is the internal resistance of the battery?

Answer 1

V = 15.0 V
R = 75 Ω
Let the internal resistance of the battery be r.

We know, V = I * R
11.3 = I * 75
I = 11.3/75
I = 0.151 A

Power = I^2 * R = 0.151^2 * 75
P = 1.71 W

(b)
15 - I*r = 11.3
15 - 0.151* r = 11.3
r = 24.5 Ω