Chapter 24, Problem 3PP

The circuit in Figure 24-2 is connected to a 60-Hz line. The apparent power in the circuit is 48.106 VA. The resistor has a resistance of 12 $\Omega$. The inductor has an inductive reactance of 60 $\Omega$, and the capacitor has a capacitive reactance of 45 $\Omega$.

$\begin{array}{l}\begin{array}{l}{\text{E}}_{\text{T}}\text{ \_\_\_\_\_}\hfill \\ {\text{I}}_{\text{T}}\text{ \_\_\_\_\_}\hfill \\ \text{Z \_\_\_\_\_}\hfill \\ \text{VA 48}\text{.106 }\hfill \\ \text{PF \_\_\_\_\_}\hfill \\ {\text{E}}_{\text{R}}\text{ \_\_\_\_\_}\hfill \\ {\text{I}}_{\text{R}}\text{ \_\_\_\_\_}\hfill \\ \text{R 12 }\Omega \hfill \\ \text{P \_\_\_\_\_}\hfill \\ \angle \text{θ \_\_\_\_\_}\hfill \\ {\text{E}}_{\text{L}}\_\_\_\_\_\hfill \\ {\text{I}}_{\text{L}}\text{ \_\_\_\_\_}\hfill \\ {\text{X}}_{\text{L}}60\Omega \hfill \\ {\text{VARs}}_{\text{L}}\text{ \_\_\_\_\_}\hfill \\ \begin{array}{l}\text{L\_\_\_\_\_}\\ {\text{E}}_{\text{C }}\_\_\_\_\_\end{array}\hfill \end{array}\\ {\text{I}}_{\text{C}}\_\_\_\_\_\\ {\text{X}}_{\text{C}}\text{ 45 }\Omega \\ {\text{VARs}}_{\text{C}}\text{ \_\_\_\_\_}\\ \text{C }\_\_\_\_\_\end{array}$

To determine

The missing values in the table.

Answer to Problem 3PP

ET = 23.99 V	ER = 23.99 V	EL = 23.99 V	EC = 23.99 V
IT = 2.004 A	IR = 1.999 A	IL =0.399 A	IC =0.533 A
Z = 11.973 Ω	R = 12 Ω	XL = 60 Ω	XC = 45 Ω
VA = 48.106	P = 47.96 W	VARSL = 9.575	VARSC = 12.791
PF = 99.725 %	∠θ = 4.25°	L = 0.159 H	C = 58.94 µF

Explanation of Solution

Given data :

$\begin{array}{l}\text{VA}=48.106\\ f=60Hz\\ \text{ R = 12 }\Omega \\ {\text{ X}}_{\text{L}}\text{= 60 }\Omega \\ {\text{ X}}_{\text{C}}\text{= 45 }\Omega \end{array}$

The value of the inductor is given by,

$L\text{=}\frac{X_L}{2\pi f}=\frac{60}{2\pi \times 60}\text{ = 0}\text{.159 H}$

The value of the capacitor is given by,

$C=\frac{1}{2\pi f{X}_C}\text{=}\frac{1}{2\pi \times 60\times 45}=58.94\mu F$

The impedance of the parallel circuit is given as,

$\begin{array}{l}Z=\frac{1}{\sqrt{{\left(\frac{1}{R}\right)}^2+{\left(\frac{1}{X_L}-\frac{1}{X_C}\right)}^2}}\\ Z=\frac{1}{\sqrt{{\left(\frac{1}{12}\right)}^2+{\left(\frac{1}{60}-\frac{1}{45}\right)}^2}}\\ Z=\frac{1}{\sqrt{{\left(\frac{1}{12}\right)}^2+{\left(-\frac{15}{2700}\right)}^2}}\\ Z=\frac{1}{\sqrt{{\left(\frac{1}{12}\right)}^2+{\left(-\frac{1}{180}\right)}^2}}\\ \\ Z=11.973\Omega \end{array}$

To determine the value of total voltage across the parallel R-L-C circuit,

$\begin{array}{l}VA=E_T{I}_T\\ VA=E_T\left(\frac{E_T}{Z}\right)\\ VA=\frac{E_T^2}{Z}\\ \\ \text{Rearrange the terms to get an expression for }{E}_T,\\ E_T=\sqrt{VA\times Z}\\ E_T=\sqrt{48.106\times 11.973}\\ E_T=23.999\text{ V}\end{array}$

Since the circuit is parallel R-L-C circuit,

$E_T=E_R=E_L=E_C=23.999\text{ V}$

The total current flowing through the parallel RLC circuit will be,

$I_T=\frac{E_T}{Z}=\frac{23.999}{11.973}=2.004\text{ A}$

The current flowing through the resistor in the parallel RLC circuit will be,

$I_R=\frac{E_R}{R}=\frac{23.999}{12}=1.999\text{ A}$

The current flowing through the inductor in the parallel RLC circuit will be,

$I_L=\frac{E_L}{X_L}=\frac{23.999}{60}=0.399\text{ A}$

The current flowing through the capacitor in the parallel RLC circuit will be,

$I_C=\frac{E_C}{X_C}=\frac{23.999}{45}=0.533\text{ A}$

The true power in the parallel circuit will be,

$\begin{array}{l}P=E_R{I}_R=23.999\times 1.999=47.974\text{ W}\\ \end{array}$

The reactive power of the inductor is given by,

$VA{R}_{SL}\text{=}{E}_L{I}_L=23.999\times 0.399=9.575$

The reactive power of the capacitor is given by,

$VA{R}_{SC}\text{=}{E}_C{I}_C=23.999\times 0.533=12.791$

The Power factor of the circuit is calculated as,

$\begin{array}{l}PF=\frac{P}{VA}\times 100\\ \\ PF=\frac{47.974}{48.106}\times 100\\ \\ PF=99.725\text{ \%}\\ \end{array}$

Power factor angle θ will be,

$\begin{array}{l}\cos \theta =PF\\ \cos \theta =\frac{99.725}{100}\\ \cos \theta =0.99725\\ \theta ={\cos }^{-1}0.99725\\ \theta =4.25°\end{array}$