A series ac circuit has a reactance of $11 \mathrm { k } \Omega$ due to its capacitance, a reactance of $3 \mathrm { k } \Omega$ due to its inductance, and a resistance of $29 \mathrm { k } \Omega$. What is the power factor of this circuit?
A) $0.96$
B) $0.48$
C) $0.27$
D) $1.04$

Question

Quizplus · Accepted Answer

The power factor is calculated as the cosine of the phase angle, which is determined by the resistance (R) and the net reactance (X), where $X = X_L - X_C$. In this case, $X = 3k\Omega - 11k\Omega = -8k\Omega$. The impedance (Z) can be calculated using $Z = \sqrt{R^2 + X^2}$, but for power factor (PF), we only need the ratio of R to Z, which simplifies to $PF = R / \sqrt{R^2 + X^2}$. Given $R = 29k\Omega$ and $X = -8k\Omega$, the power factor is $PF = 29 / \sqrt{29^2 + (-8)^2} = 29 / \sqrt{841 + 64} = 29 / \sqrt{905}$. This simplifies to approximately 0.96, indicating that the circuit is mostly resistive with very little phase difference between voltage and current.

A Series Ac Circuit Has a Reactance Of 11kΩ11 \mathrm { k } \Omega11kΩ

A Series Ac Circuit Has a Reactance Of $11 \mathrm { k } \Omega$