Chapter 22, Problem 2PP

To determine

The missing values in the table.

Answer to Problem 2PP

ET = 54 V	ER = 54 V	EC = 54 V
IT = 2.5 A	IR = 1.5 A	IC = 2 A
Z = 21.6 Ω	R = 36 Ω	XC = 27 Ω
VA = 135	P = 81 W	VARSC = 108
PF = 60 %	∠θ =53.13 °	C =14.736 µF

Explanation of Solution

Given data:

$\begin{array}{l}{\text{I}}_{\text{C}}=2\text{A}\\ f=400Hz\\ \text{ R = 36 }\Omega \\ \text{ Z = 21}\text{.6 }\Omega \end{array}$

The capacitive reactance X_C is calculated as,

$\begin{array}{l}\frac{1}{Z^2}\text{=}\frac{1}{R^2}\text{+}\frac{1}{X_C^2}\\ \frac{1}{X_C^2}=\frac{1}{Z^2}-\frac{1}{R^2}\\ \frac{1}{X_C^2}=\frac{1}{{21.6}^2}-\frac{1}{{36}^2}\\ \frac{1}{X_C^2}=\frac{1}{466.56}-\frac{1}{1296}\\ \\ \frac{1}{X_C^2}=\frac{1296-466.56}{1296\times 466.56}\\ \\ \frac{1}{X_C^2}=\frac{829.44}{604661.76}\\ \\ \text{Invert the ratio,}\\ X_C^2=\frac{604661.76}{829.44}\\ \\ X_C^2=729\\ \\ \text{Taking square root,}\\ X_C=\sqrt{729}\\ =27\Omega \\ \end{array}$

The value of the capacitor is,

$C=\frac{1}{2\pi f{X}_c}=\frac{1}{2\pi \times 400\times 27}=14.736\mu F$

The voltage drop across the capacitor,

$\begin{array}{l}{E}_C=I_C{X}_C\\ =2\times 27\\ =54\text{V}\end{array}$

Since the circuit is parallel R-C,

$E_C=E_R=E_T=54\text{ V}$

The current flowing through resistance I_R will be,

$I_R=\frac{E_R}{R}=\frac{54}{36}=1.5\text{ A}$

The total current flowing in the parallel R-C network is,

$I_T=\frac{E_T}{Z}=\frac{54}{21.6}=2.5\text{ A}$

The apparent power VA is given as,

$\begin{array}{l}VA=E_T{I}_T\\ =54\times 2.5\\ =135\end{array}$

The true power P is given as,

$\begin{array}{l}P=E_R{I}_R\\ =54\times 1.5\\ =81\text{ W}\end{array}$

The reactive power VAR_SC is calculated as,

$\begin{array}{l}{\text{VAR}}_{\text{SC}}\text{ = }\sqrt{{\text{VA}}^2-{\text{P}}^2}\\ =\sqrt{{135}^2-{81}^2}\\ =\text{ 108}\end{array}$

Power factor (PF) is calculated as,

$\begin{array}{l}PF=\frac{P}{VA}\times 100\\ =\frac{81}{135}\times 100\\ \\ =60\%\end{array}$

Power factor angle θ will be,

$\begin{array}{l}\cos \theta =0.6\\ \theta ={\cos }^{-1}0.6\\ \theta =53.13°\end{array}$