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Zorluk: OrtaAlternating Current (AC) Circuits

A series alternating current (AC) circuit consists of a resistor of resistance R=40 ΩR = 40\ \Omega, an inductor with inductive reactance XL=70 ΩX_L = 70\ \Omega, and a capacitor with capacitive reactance XC=40 ΩX_C = 40\ \Omega, connected to an AC voltage supply of 100 V100\ \text{V}. What is the power factor of the circuit?

Cevap: 0.8

Cevap

The power factor of the circuit is 0.8.
The total impedance ZZ of the series circuit is found using Z=R2+(XL−XC)2=402+(70−40)2=50 ΩZ = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (70 - 40)^2} = 50\ \Omega. The power factor is the ratio of resistance to impedance: cos⁡ϕ=RZ=4050=0.8\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8.

Adım Adım Çözüm

1
Calculate the net reactance of the circuit.
X=XL−XC=70 Ω−40 Ω=30 ΩX = X_L - X_C = 70\ \Omega - 40\ \Omega = 30\ \Omega
In a series RLC circuit, the net reactance is the arithmetic difference between the inductive reactance and the capacitive reactance.
2
Calculate the total impedance of the circuit.
Z=R2+(XL−XC)2=402+302=2500=50 ΩZ = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + 30^2} = \sqrt{2500} = 50\ \Omega
Impedance represents the combined opposition to current flow from resistance and net reactance in quadrature.
3
Determine the power factor of the circuit.
cos⁡ϕ=RZ=4050=0.8\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8
The power factor is equal to the cosine of the phase angle, which is defined as the ratio of resistance to total impedance.

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Power Factor in AC Circuits
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Alternating Current (AC) Circuits

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