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Physics · Ch 4 — Electromagnetic Induction and Alternating Current

Power Factor

4.8.3

Power Factor

The power factor of a circuit can be defined equivalently in any of three ways: (i) as cos⁡ϕ\cos\phi, the cosine of the phase angle of lead or lag between voltage and current; (ii) as the ratio R/Z of resistance to impedance; or (iii) as the ratio Pav/(VRMSIRMS)P_{av}/(V_{RMS}I_{RMS}) of true power to apparent power. All three definitions are algebraically equivalent and give the same numerical value for any given circuit.

Special cases, following directly from Table 4.1: for a PURELY RESISTIVE circuit, the phase angle is zero, so cos⁡ϕ=1\cos\phi=1 and Pav=VRMSIRMSP_{av}=V_{RMS}I_{RMS} -- true power equals apparent power exactly, the best possible case. For a PURELY INDUCTIVE or PURELY CAPACITIVE circuit, the phase angle is ±π/2\pm\pi/2, so cos⁡(±π/2)=0\cos(\pm\pi/2)=0 and Pav=0P_{av}=0 -- despite a real current flowing, no true power is consumed at all (matching the wattless-current result of section 4.8.2). For a general SERIES RLC circuit, ϕ=tan⁡−1[(XL−XC)/R]\phi=\tan^{-1}\left[(X_L-X_C)/R\right], and Pav=VRMSIRMScos⁡ϕP_{av}=V_{RMS}I_{RMS}\cos\phi lies somewhere between these two extremes. Finally, for a series RLC circuit specifically AT RESONANCE, the phase angle is exactly zero (since XL=XCX_L=X_C there), so cos⁡ϕ=1\cos\phi=1 and on …

Misc Example 4.27Impedance, peak current, and power factor (general and at resonance) for a given RLC circuit

Worked out. A series RLC circuit with C=10−4πC=\dfrac{10^{-4}}{\pi} F, L=2πL=\dfrac{2}{\pi} H and R=100 ΩR=100\ \Omega is driven by a 220 V, 50 Hz AC supply, and four quantities are required. First, XL=2πfL=2π(50)(2/π)=200 ΩX_L=2\pi fL=2\pi(50)(2/\pi)=200\ \Omega and XC=1/(2πfC)=1/(2π(50)(10−4/π))=100 ΩX_C=1/(2\pi fC)=1/\left(2\pi(50)(10^{-4}/\pi)\right)=100\ \Omega. (i) Impedance: Z=R2+(XL−XC)2=1002+1002≈141.4 ΩZ=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{100^2+100^2}\approx141.4\ \Omega. (ii) Peak current: Im=Vm/Z=(2202)/141.4≈2.2I_m=V_m/Z=(220\sqrt2)/141.4\approx2.2 A. (iii) Power factor: cos⁡ϕ=R/Z=100/141.4≈0.707\cos\phi=R/Z=100/141.4\approx0.707. (iv) Power factor AT resonance (where XL=XCX_L=X_C so Z=RZ=R): cos⁡ϕ=R/Z=R/R=1\cos\phi=R/Z=R/R=1. The problem strings together impedance, peak current and power factor -- both for the circuit as actually given, and hypothetically at resonance -- into one compreh …