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Difficulty: MediumAlternating Current (AC) Circuits

An alternating current (AC) circuit consists of a resistor of resistance R=30 ΩR = 30\ \Omega connected in series with a pure inductor across an AC supply of root-mean-square (RMS) voltage 100 V100\ \text{V}. If the average power dissipated in the circuit is 120 W120\ \text{W}, what is the inductive reactance of the inductor?

  1. A
    20 Ω20\ \Omega
  2. 40 Ω40\ \OmegaAnswer
  3. C
    50 Ω50\ \Omega
  4. D
    70 Ω70\ \Omega

Answer

The inductive reactance of the inductor is 40 Ω40\ \Omega.
In an AC circuit containing a resistor and a pure inductor, power is dissipated solely by the resistance. Using P=Irms2RP = I_{\text{rms}}^2 R, the current is Irms=120/30=2 AI_{\text{rms}} = \sqrt{120 / 30} = 2\ \text{A}. The total impedance ZZ is Vrms/Irms=100/2=50 ΩV_{\text{rms}} / I_{\text{rms}} = 100 / 2 = 50\ \Omega. Applying the phasor formula for impedance Z=R2+XL2Z = \sqrt{R^2 + X_L^2}, solving for XLX_L yields XL=502−302=40 ΩX_L = \sqrt{50^2 - 30^2} = 40\ \Omega.

Step-by-Step Solution

1
Calculate the RMS current in the circuit using the average power formula
Irms=2 AI_{\text{rms}} = 2\ \text{A}
In an AC circuit with a resistor and a pure inductor, average power is dissipated only by the resistor: P=Irms2R  ⟹  120=Irms2×30  ⟹  Irms2=4  ⟹  Irms=2 AP = I_{\text{rms}}^2 R \implies 120 = I_{\text{rms}}^2 \times 30 \implies I_{\text{rms}}^2 = 4 \implies I_{\text{rms}} = 2\ \text{A}.
2
Determine the total impedance of the circuit
Z=50 ΩZ = 50\ \Omega
The total impedance is the ratio of RMS voltage to RMS current: Z=VrmsIrms=100 V2 A=50 ΩZ = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{100\ \text{V}}{2\ \text{A}} = 50\ \Omega.
3
Calculate the inductive reactance using the impedance relationship for a series RL circuit
XL=40 ΩX_L = 40\ \Omega
Impedance in a series RL circuit is given by Z=R2+XL2Z = \sqrt{R^2 + X_L^2}. Substituting the known values gives 50=302+XL2  ⟹  2500=900+XL2  ⟹  XL2=1600  ⟹  XL=40 Ω50 = \sqrt{30^2 + X_L^2} \implies 2500 = 900 + X_L^2 \implies X_L^2 = 1600 \implies X_L = 40\ \Omega.

Key Concept

Power Dissipation and Impedance in Series RL AC Circuits
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Topics

Alternating Current (AC) Circuits

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