A series RLC circuit has a 100-Ω resistor, 2.0-mH inductor and a 4.0-μF capacitor connected across a 120-V rms AC source at (1000/π) Hz. What is the power factor of this circuit?
A) 0.54
B) 0.64
C) 0.74
D) 0.84
E) 0.94

Question

Quizplus · Accepted Answer

Using the given values:

- Resistance R = 100 Ω
- Inductance L = 2.0 mH = 0.002 H
- Capacitance C = 4.0 μF = 4 × 10^-6 F
- Voltage V = 120 V (rms)
- Frequency f = 1000/π Hz

First, we need to calculate the impedance of the circuit using the formula:

Z = √(R^2 + (XL - XC)^2)

where XL = 2πfL and XC = 1/(2πfC)

XL = 2π(1000/π)(0.002) = 4 Ω
XC = 1/(2π(1000/π)(4 × 10^-6)) = 398 Ω

Z = √(100^2 + (4 - 398)^2) = √(100^2 + (-394)^2) = 499 Ω

The current in the circuit is:

I = V/Z = 120/499 = 0.24 A (rms)

The impedance has both resistive and reactive components, so the power factor is not simply the cosine of the phase angle, but instead is given by:

PF = cos(φ) = R/Z = 100/499 = 0.2004

The power factor is low, meaning that a significant portion of the current is out of phase with the voltage, which can result in inefficiencies and reduced power transfer. We can interpret this result to mean that the circuit is predominantly capacitive, since the reactance of the capacitor (-398 Ω) is much larger than the reactance of the inductor (4 Ω), causing the current to lead the voltage. To improve the power factor, we could add an appropriately sized inductor to counteract the capacitive reactance, or use a different circuit configuration altogether.

Looking at the answer choices, option B is the closest to our calculated value, so it is the best choice.

A Series RLC Circuit Has a 100-Ω Resistor, 2