AC Power

Electricity and Magnetism Level 1

A sinusoidal AC voltage source is connected across a series combination of a resistor and an inductor. The complex impedances of both elements are shown.

What is the average power dissipated in the resistor?


The answer is 12.

Steven Chase by Steven Chase , 37, USA

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1 solution

The current through the two impedances is I rms = V rms Z R + Z L = 10 ∠ 0 ∘ 3 + j 4 = 10 ∠ 0 ∘ 5 ∠ tan ⁡ − 1 4 3 = 2 ∠ tan ⁡ − 1 4 3 A I_{\text{rms}} = \dfrac {V_{\text{rms}}}{Z_R+Z_L} = \dfrac {10\angle 0^\circ}{3+j4} = \dfrac {10\angle 0^\circ}{5\angle \tan^{-1}\frac 43} = 2\angle \tan^{-1} \frac 43 \text{ A} .

The average power dissipated in the resistor is ∣ I rms ∣ 2 Z R = 2 2 ( 3 ) = 12 W \left|I_{\text{rms}}\right|^2 Z_R = 2^2(3) = \boxed{12} \text{ W} .

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