Chapter 18, Problem 4PP

Assume that the circuit in Figure 18-1 has a power factor of 78%, an apparent power of 374.817 VA, and a frequency of 400 Hz. The inductor has an inductance of 0.0382 H.

$E_T\_\_\_\_\_\_\_\_\_\_\_\_$	$E_R\_\_\_\_\_\_\_\_\_\_\_\_$	$E_L\_\_\_\_\_\_\_\_\_\_\_\_$
$I_T\_\_\_\_\_\_\_\_\_\_\_\_\_$	$I_R\_\_\_\_\_\_\_\_\_\_\_\_$	$I_L\_\_\_\_\_\_\_\_\_\_\_\_\_$
$Z\_\_\_\_\_\_\_\_\_\_\_\_\_$	$R\_\_\_\_\_\_\_\_\_\_\_\_\_$	$X_L\_\_\_\_\_\_\_\_\_\_\_\_$
$VA374.817$	$P\_\_\_\_\_\_\_\_\_\_\_\_\_$	$VA{R}_{S_L}\_\_\_\_\_\_\_\_\_$
$PF78\%$	$\angle \text{θ}\_\_\_\_\_\_\_\_\_\_\_\_$	$L0.0382H$

To determine

The missing values in the table.

Answer to Problem 4PP

ET = 150.062 V	ER = 150.062 V	EL = 150.062 V
IT = 2.498 A	IR = 1.948 A	IL = 1.563 A
Z = 60.073 Ω	R = 77.024 Ω	XL = 96.007 Ω
VA = 374.817	P = 292.357 W	VARSL = 234.55
PF = 78%	$\angle \theta =38.74°$	L = 0.0382 H

Explanation of Solution

Given data:

$\begin{array}{l}\text{ L}=0.0382\text{ H}\\ \text{VA = 374}\text{.817 }\\ \text{ PF}=78\%\\ f=400\text{ Hz}\\ \end{array}$

The inductive reactance X_L is calculated as,

$\begin{array}{l}{X}_L=2\pi fL\\ =2\pi \times 400\times 0.0382\\ =96.007\Omega \end{array}$

Using the Power factor (PF) and Apparent Power(VA), we can calculate True Power (P) as,

$\begin{array}{l}PF=\frac{P}{VA}\times 100\\ \Rightarrow P=\frac{PF\times VA}{100}\\ \Rightarrow P=\frac{78\times 374.817}{100}\\ =292.357\text{W}\end{array}$

The reactive power VAR_SL is calculated as,

$\begin{array}{l}{\text{VAR}}_{\text{SL}}\text{ = }\sqrt{{\text{VA}}^2-{\text{P}}^2}\\ =\sqrt{{374.817}^2-{292.357}^2}\\ =\text{ 234}\text{.55}\end{array}$

The relation between reactive power, voltage and inductive reactance is given as,

$\begin{array}{l}VA{R}_{SL}=\frac{E_L{}^2}{X_L}\\ \Rightarrow E_L{}^2=VA{R}_{SL}\times X_L\\ =234.55\times 96.007\\ =22518.442\\ \\ \Rightarrow E_L=\sqrt{22518.442}=150.062\text{ V}\\ \end{array}$

Since the circuit is parallel R-L,

$E_T=E_R=E_L=150.062\text{ V}$

The current through the inductor is,

$I_L=\frac{E_L}{X_L}=\frac{150.062}{96.007}=1.563\text{ A}$

The relation between true power P, voltage E_R and resistance R is given as,

$\begin{array}{l}P=\frac{E_R{}^2}{R}\\ \Rightarrow R=\frac{E_R{}^2}{P}\\ \Rightarrow R=\frac{{150.062}^2}{292.357}\\ \\ \Rightarrow R=77.024\Omega \end{array}$

The current through the resistor is,

$I_R=\frac{E_R}{R}=\frac{150.062}{77.024}=1.948\text{ A}$

The current through the inductor is,

$I_L=\frac{E_L}{X_L}=\frac{150.062}{96.007}=1.563\text{ A}$

The total current in the circuit

$\begin{array}{l}{I}_T=\sqrt{I_R{}^2+I_L{}^2}\\ =\sqrt{{1.948}^2+{1.563}^2}\\ =2.498\text{ A}\end{array}$

The impedance Z is calculated as,

$Z=\frac{E_T}{I_T}=\frac{150.062}{2.498}=60.073\Omega$

Power factor angle θ will be,

$\begin{array}{l}\cos \theta =PF\\ \cos \theta =0.78\\ \theta ={\cos }^{-1}0.8\\ \theta =38.74°\end{array}$