A lightbulb with a resistance of 2.9 ohms is operated using a 1.5-volt battery. At what rate is electrical energy transformed in the light bulb?
The Solution
R= 2.90 Ω, V = 1.5 V.
The rate at which electrical energy transformed in the lightbulb i.e. Power: P= V I
Ohm's Law: V = I R, I = V / R
P = V I, P = V²/ R
P = ((1.5)^2)/2.9 = 0.775862
P = 0.78 W
The Answer: 0.78 W