Discussion
Calculate the average power delivered to impedance of \((4 - j3)\ \Omega\) by a current \(4\cos(100\pi t + 100)\).
Answer: C
The average power is 32 W. The impedance is (4 - j3) Ω, so the resistance is 4 Ω, and the current has a peak value of 4 A. Only the resistive part takes average power, which is the half of the peak current squared times the resistance.
\[ P = \tfrac{1}{2} I_m^2 R \]
\[ = \tfrac{1}{2} \times 4^2 \times 4 = 32\ \text{W} \]
The value 64 W would result from leaving out the factor of one half, and 16 W is the value of the half applied twice, and 48 W does not satisfy the formula. So the third option is correct.
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