Question 53: AC fundamentals

Question

A LCR series circuit driven with $E_{r m s}=90 \mathrm{~V}$ at frequency $f_{\mathrm{d}}=30 \mathrm{~Hz}$ has resistance $R=80 \Omega$, an inductance with inductive reactance $X_L=20.0 \Omega$ and capacitance with capacitive reactance $X_C=80.0 \Omega$. The power factor of the circuit is $\_\_\_\_$ .

Accepted Answer

In an AC circuit, the power factor $(\cos \phi)$ is defined as the ratio of the resistance $(\mathrm{R})$ to the total impedance $(\mathrm{Z})$ of the circuit :
$$ 	ext { Power Factor }=\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}} $$
For a series LCR circuit, the impedance is
$$ \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}ight)^2} $$
The value of resistance is $\mathrm{R}=80 \Omega$, the inductive reactance is $\mathrm{X}_{\mathrm{L}}=20.0 \Omega$, and the capacitive reactance is $\mathrm{X}_{\mathrm{C}}=80.0 \Omega$.
$$ \mathrm{X}_{	ext {net }}=\left|\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}ight|=|20-80|=60 \Omega $$
So, the impedance is,
$$ \mathrm{Z}=\sqrt{80^2+60^2}=\sqrt{6400+3600}=\sqrt{10000}=100 \Omega $$
Thus, the power factor is,
$$ \cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{80}{100}=0.8 $$
Therefore, the correct option is (A).

Answer

0.8

Answer

0.64

Answer

0.9

Answer

0.5

AC fundamentals

A LCR series circuit driven with E_{r m s}=90 \mathrm{~V} at frequency f_{\mathrm{d}}=30 \mathrm{~Hz} has resistance R=80 \Omega, an inductance with inductive reactance X_L=20.0 \Omega and capacitance with capacitive reactance X_C=80.0 \Omega. The power factor of the circuit is \_\_\_\_ .

Choose the correct answer: