Chapter 24, Problem 1PP

To determine

The missing values in the table.

Answer to Problem 1PP

ET = 120 V	ER = 72 V	EL = 200 V	EC = 104 V
IT = 2 A	IR = 2 A	IL = 2 A	IC = 2 A
Z = 60 Ω	R = 36 Ω	XL = 100 Ω	XC = 52 Ω
VA = 240	P = 144 W	VARSL = 400	VARSC = 208
PF = 60 %	∠θ = 53.13 °	L = 0.2652 H	C = 51.01µF

Explanation of Solution

Given data :

$\begin{array}{l}{\text{E}}_{\text{T}}=120\text{ V}\\ f=60Hz\\ \text{ R =36 }\Omega \\ {\text{ X}}_{\text{L}}\text{=100 }\Omega \\ {\text{ X}}_{\text{C}}\text{= 52 }\Omega \end{array}$

The value of the inductor is given by,

$L\text{=}\frac{X_L}{2\pi f}=\frac{100}{2\pi \times 60}\text{ = 0}\text{.2652 H}$

The value of the capacitor is given by,

$C=\frac{1}{2\pi f{X}_C}\text{=}\frac{1}{2\pi \times 60\times 52}=51.01\mu F$

The impedance of the series RLC circuit is calculated as,

$\begin{array}{l}{Z}^2\text{ =}{R}^2+{\left(X_L-X_C\right)}^2\\ Z^2{\text{ = 36}}^2+{\left(100-52\right)}^2\\ Z^2{\text{ = 36}}^2+{48}^2\\ Z^2\text{ = 3600}\\ \\ \text{Taking square root,}\\ Z=\sqrt{\text{3600}}\\ Z=60\Omega \\ \end{array}$

The total current flowing through series RLC circuit will be,

$I_T=\frac{E_T}{Z}=\frac{120}{60}=2\text{A}$

Since the circuit is series R-L-C,

$I_T=I_R=I_L=I_C=2\text{ A}$

The total voltage drop across the resistor,

$\begin{array}{l}{E}_R=I_{R}R\\ =2\times 36\\ =72\text{V}\end{array}$

The total voltage drop across the inductor,

$\begin{array}{l}{E}_L=I_L{X}_L\\ =2\times 100\\ =200\text{V}\end{array}$

The total voltage drop across the capacitor,

$\begin{array}{l}{E}_C=I_C{X}_C\\ =2\times 52\\ =104\text{V}\end{array}$

The apparent power VA is given as,

$\begin{array}{l}VA=E_T{I}_T\\ =120\times 2\\ =240\end{array}$

The true power P is given as,

$\begin{array}{l}P=E_R{I}_R\\ \text{ = 72}\times 2\\ \text{ =}144\text{ W}\end{array}$

The reactive power VAR_SL is calculated as,

$\begin{array}{l}VA{R}_{SL}=E_L{I}_L\\ \text{ = 200}\times 2\\ \text{ = }400\end{array}$

The reactive power VAR_SC is calculated as,

$\begin{array}{l}VA{R}_{SC}=E_C{I}_C\\ \text{ = 104}\times 2\\ \text{ = 208}\end{array}$

Power factor (PF) is calculated as,

$\begin{array}{l}PF=\frac{P}{VA}\times 100\\ =\frac{144}{240}\times 100\\ =60\%\end{array}$

Power factor angle θ will be,

$\begin{array}{l}\cos \theta =PF\\ \cos \theta =\frac{60}{100}\\ \cos \theta =0.6\\ \theta ={\cos }^{-1}0.6\\ \theta =53.13°\end{array}$