Multiple choice

The average power delivered to an impendence (4 -j3)$\Omega$by a current 5 cos(100$\Omega$t + 100) A is

  1. 44.2 W

  2. 50 W

  3. 62.5 W

  4. 125 W
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Average power = (I_rms)² × R, where I_rms = I_peak/√2. For current 5cos(100t+100), I_rms = 5/√2 A. The impedance is Z = 4-j3 Ω, so |Z| = √(4²+3²) = 5 Ω. Only the real part (resistance) dissipates power. Equivalent resistance R_eq = 4 Ω. P_avg = (5/√2)² × 4 = 25/2 × 4 = 50 W.

Keep practicing — related questions