Question

In an L-R-C series circuit, R = 300 Ω, X_C = 300 Ω and X_L = 500 Ω. The average electrical power consumed in the resistor is 60.0 W. (a) What is the power factor of the circuit? (b) What is the rms voltage of the source?

Short answer

a) Power factor of the circuit is 0.832

b) The rms value of voltage is 161.17 V.

Step-by-step solution

Concept of impedance, power factor and average power

Impedance of a circuit is the total resistance offered by the circuit.Denoted by Z, the formula for impedance of a LCR circuit is $Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$, where, R is the resistance,$X_L$ is the inductive reactance,role="math" localid="1664194997760" $X_C$ is the capacitive reactance.

Power factor of a circuit is defined as the ratio of the power across the resistor and the power consumed by the circuit. For series combination, the power factor can also be defined as the ratio of the resistance to the impedance of the circuit.

Average power across the resistor in an A/C circuit is given by, $P_{avg}={I_{rms}}^{2}R$

Calculation of impedance of the circuit and power factor

Given that

$$\begin{gathered} R=300\Omega \\ {X}_L=500\Omega \\ {X}_C=300\Omega \end{gathered}$$

Therefore the impedance can be calculated as-

$$\begin{gathered} Z=\sqrt{{300}^2+{\left(500-300\right)}^2} \\ =360.56\Omega \end{gathered}$$

Now, power factor can be calculated as-

$$\begin{gathered} \frac{R}{Z}=\frac{300}{360.56} \\ =0.832 \end{gathered}$$

Hence, power factor of the circuit 0.832

Calculation of rms value of voltage

The average power across the resistor is given by, $P_{avg}={I_{rms}}^{2}R$, therefore,

$$\begin{gathered} I_{rms}=\sqrt{\frac{P_{avg}}{R}} \\ =\sqrt{\frac{60}{300}} \\ =0.447A \end{gathered}$$

Now,$V_{rms}=I_{rms}Z$ , therefore,

$$\begin{gathered} V_{rms}=0.447\times 360.56 \\ =161.17V \end{gathered}$$

Hence, the rms value of voltage is 161.7 V.