Question

A single-phase, $\mathbf{120}\mathbf{}\mathbf{V}\left(\mathrm{rms}\right)\mathbf{,}\mathbf{60}\mathbf{Hz}$source supplies power to a series R-L circuit consisting of$\mathbf{R}\mathbf{=}\mathbf{10}\mathbf{\Omega }$and$\mathbf{L}\mathbf{=}\mathbf{40}\mathbf{mH}$(a) Determine the power factor of the circuit and State whether it is lagging or leading. (b) Determine the real and reactive power absorbed by the load. (c) Calculate the peak magnetic energy${\mathbf{W}}_{\mathbf{int}}$stored in the inductor by using the expression${\mathbf{W}}_{\mathbf{int}}\mathbf{=}\mathbf{L}\left({l_{\mathrm{rms}}}^2\right)$and check whether the reactive power$\mathbf{Q}\mathbf{=}{\mathbf{\omega W}}_{\mathbf{int}}$is satisfied. (Note: The instantaneous magnetic energy storage fluctuates between zero and the peak energy. This energy must be sent twice each cycle to the load from the source by means of reactive power flows.)

Short answer

(a) The power factor of the circuit is $0.552\mathrm{lagging}$.

(b) The real and reactive power absorbed by the load are $439.93\mathrm{KW}$and $662.90\mathrm{VAR}$.

(c) The peak magnetic energy stored in the inductor is $1.76\mathrm{J}$.

Step-by-step solution

Calculate the power factor of the circuit and State whether it is lagging or leading.

The impedance can be calculated as:

$$\begin{gathered} \mathrm{Z}=\mathrm{R}+\mathrm{j\omega L} \\ \mathrm{Z}=\mathrm{R}+\mathrm{j}2\mathrm{\pi fL} \end{gathered}$$

Solve for the impedance as:

$$\begin{gathered} \mathrm{Z}=10+\mathrm{j}2\mathrm{\pi }\times 60\times 40\times {10}^{-3} \\ \mathrm{Z}=10+\mathrm{j}15.07 \end{gathered}$$

Convert the impedance into polar form

$18.08\angle 56.43°\mathrm{\Omega }$

The rms value of the current can be calculated as:

${\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{Z}}$

Substitute the values and solve.

$$\begin{gathered} {\mathrm{I}}_{\mathrm{rms}}=\frac{120}{18.08\angle 56.43°} \\ {\mathrm{I}}_{\mathrm{rms}}=6.63\angle -56.43°\mathrm{A} \end{gathered}$$

(a)

The phase difference between current and voltage can be calculated as:

$$\begin{gathered} \varphi =0°-\left(56.43°\right) \\ \varphi =56.43° \end{gathered}$$

The power factor of the circuit

$$\begin{gathered} p.f=\cos \left(56.43°\right) \\ p.f=0.552 \end{gathered}$$

Hence the power factor of the circuit is$0.552\mathrm{lagging}$.

Calculate the real and reactive power absorbed by the load.

(b)

The real power absorbed by the load can be calculated as:

$$\begin{gathered} P=V_{rms}{I}_{rms}\cos \left(\varphi \right) \\ P=120\times 6.63\cos \left(56.43\right) \\ P=439.93\mathrm{W} \end{gathered}$$

The reactive power absorbed by the load can be calculated as:

$$\begin{gathered} Q=V_{rms}{I}_{rms}\cos \left(\varphi \right) \\ Q=120\times 6.63\sin \left(56.43\right) \\ Q=662.90\mathrm{VAR} \end{gathered}$$

Hence, the real and reactive power absorbed by the load are$439.93\mathrm{KW}$and$662.90\mathrm{VAR}$.

Calculate the peak magnetic energy Wintstored in the inductor by using the expressionWint=L(Irms2).

(c)

The peak magnetic energy can be calculated as:

$$\begin{gathered} \mathrm{W}=\mathrm{I}{\left({\mathrm{I}}_{\mathrm{rms}}\right)}^2 \\ \mathrm{W}=40\times {10}^{-3}\times {\left(6.63\right)}^2 \\ \mathrm{W}=1.76\mathrm{J} \end{gathered}$$

To check whether the reactive power is satisfied substitute the values in the equation $Q=\omega W$and solve.

$$\begin{gathered} Q=2\pi \times 60\times 1.76 \\ Q=663VAR \end{gathered}$$

Hence, the peak magnetic energy stored in the inductor is$1.76\mathrm{J}$.