Power in Alternating-Current Circuits — Question 2

Section 31-9 · Physics pages 5134–5135

Question 2 A wattmeter separates the resistive part from impedance magnitude A series RL load draws 2.00 A rms from 100 V rms at ω = 500 radians per second. A wattmeter reads 120 W, and current lags voltage. P S Qr Schematic; labels, not drawing lengths, specify the data. (a) Infer R, inductive reactance, and L. (b) Find power factor and reactive power.
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Question 2 - Solution A wattmeter separates the resistive part from impedance magnitude P S Qr (a) Method and calculation Z = V/I = 50.0 Ω, R = P/I2 = 30.0 Ω. XL = √ Z2 −R2 = 40.0 Ω, L = XL/ω = .0800 H . (b) Interpretation and further calculation cos ϕ = P/(V I) = 120/200 = .600. Qr = I2XL = 160 var . Physical check Voltage and current alone give only impedance magnitude; real power supplies the missing resistive information.

Question 2

A wattmeter separates the resistive part from impedance magnitude

A series RL load draws 2.00 A rms from 100 V rms at ω=500\omega=500 radians per second. A wattmeter reads 120 W, and current lags voltage.

Diagram shown in the original SVG worksheet.

  1. Infer R, inductive reactance, and L.

  2. Find power factor and reactive power.

Question 2 - Solution

A wattmeter separates the resistive part from impedance magnitude

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Z=V/I=50.0Ω,R=P/I2=30.0Ω.Z=V/I=50.0\,\Omega,\quad R=P/I^2=30.0\,\Omega.XL=Z2−R2=40.0Ω,L=XL/ω=.0800H.\boxed{X_L=\sqrt{Z^2-R^2}=40.0\,\Omega,\quad L=X_L/\omega=.0800\,H}.

(b) Interpretation and further calculation

cos⁡ϕ=P/(VI)=120/200=.600.\cos\phi=P/(VI)=120/200=.600.Qr=I2XL=160var.\boxed{Q_r=I^2X_L=160\,var}.

Physical check

Voltage and current alone give only impedance magnitude; real power supplies the missing resistive information.