A source Vs(t) = V cos 100πt has an internal impedance of (4 + j3) Ω. If a purely resistive load is connected to this source has to extract the maximum power out of the source, its value in Ohms should be
- A. 3 Ohms
- B. 5 Ohms
- C. 4 Ohms
- D. 7 Ohms
Answer: B. 5 Ohms
Explanation: Source internal impedance Zs = 4 + j3 Ω. To extract maximum average power in a purely resistive load RL, RL = |Zs| = √(4² + 3²) = 5 Ω.