A load resistor RL is connected to a DC source with an EMF of 10 V and an internal resistance of 50 Ω. As RL is varied, what is the maximum power dissipated in the load?

Answer: C. 0.5 W

The load power is maximum when RL equals the internal resistance, 50 Ω (the maximum power transfer theorem). The current is then 10 V/(50 Ω + 50 Ω) = 0.1 A and the load power is (0.1 A)² × 50 Ω = 0.5 W, which can also be written V²/(4R) = 100/200 = 0.5 W. The source delivers 10 V × 0.1 A = 1 W, half of which is lost in the internal resistance, so the efficiency is 50%. RF circuits use this matched condition, whereas power supplies and power transmission favor efficiency and make the load resistance much larger than the internal resistance.

• A. 2 W is V²/R, the power dissipated in the internal resistance with the terminals shorted.
• B. 1 W is the total power from the source, half of which is lost in the internal resistance.
• C. Correct. With RL = 50 Ω, (0.1 A)² × 50 Ω = 0.5 W.