Chapter 25, Problem 1RQ

An AC circuit contains a $24-\Omega$ resistor, a 15.9-mH inductor, and a $13.3-\text{μF}$ capacitor connected in parallel. The circuit is connected to a 240-V, 400-Hz power supply. Find the following values.

$\begin{array}{l}{\text{X}}_{\text{L}}=\text{ \_\_\_\_\_\_\_\_\_\_\_\_}\Omega \\ {\text{X}}_{\text{C}}=\text{ \_\_\_\_\_\_\_\_\_\_\_\_}\Omega \\ {\text{I}}_{\text{R}}=\text{ \_\_\_\_\_\_\_\_\_\_\_\_A}\\ {\text{I}}_{\text{L}}=\text{ \_\_\_\_\_\_\_\_\_\_\_\_A}\\ {\text{I}}_{\text{C}}=\text{ \_\_\_\_\_\_\_\_\_\_\_\_A}\\ \text{P}=\text{ \_\_\_\_\_\_\_\_\_\_\_\_W}\\ {\text{VARs}}_{\text{L}}=\text{ \_\_\_\_\_\_\_\_}\\ {\text{VARs}}_{\text{C}}=\text{ \_\_\_\_\_\_\_\_}\\ {\text{I}}_{\text{T}}=\text{ \_\_\_\_\_\_\_\_\_\_\_\_A}\\ \text{VA}=\text{ \_\_\_\_\_\_\_\_\_\_}\\ \text{PF}=\text{ \_\_\_\_\_\_\_\_\_\_\_\%}\\ \angle \text{θ}=\text{ \_\_\_\_\_\_\_\_\_\_\_\_}\circ \end{array}$

To determine

The missing values.

Answer to Problem 1RQ

$\begin{array}{l}{X}_L=39.96\Omega \\ X_C=29.91\Omega \\ I_R=10\text{ A}\\ I_L=6\text{ A}\\ I_C=8.02\text{ A}\\ P=2400\text{ W}\\ VA{R}_{SL}=1440\\ VA{R}_{SC}=1924.8\\ I_T=10.20\text{ A}\\ VA=2448\\ PF=98.03\text{ \%}\\ \angle \theta =11.36°\end{array}$

Explanation of Solution

Given data:

$\begin{array}{l}R=24\Omega \\ L=15.9\text{ mH}\\ C=\text{ 13}\text{.3 μF}\\ {\text{E}}_{\text{T}}=240\text{ V}\\ f=\text{400Hz}\end{array}$

Since the circuit is a parallel RLC,

$E_T=E_R=E_L=E_C=240\text{ V}$

The reactance of the inductor is given by,

$X_L=2\pi fL\text{ = }2\pi \times 400\times 15.9\times {10}^{-3}\text{= 39}\text{.96 }\Omega$

The reactance of the capacitor is given by,

$\begin{array}{l}{X}_C=\frac{1}{2\pi fC}=\frac{1}{2\pi \times 400\times 13.3\times {10}^{-6}}=29.91\Omega \\ \end{array}$

The current flowing in the resistor is,

$I_R=\frac{E_R}{R}=\frac{240}{24}=10\text{ A}$

The current flowing in the inductor is,

$I_L=\frac{E_L}{X_L}=\frac{240}{39.96}=6\text{ A}$

The current flowing in the capacitor is,

$I_C=\frac{E_C}{X_C}=\frac{240}{29.91}=8.02\text{ A}$

The true power in the circuit is,

$P=E_R{I}_R=240\times 10=2400\text{ W}$

The reactive power in the inductor is,

$VA{R}_{SL}=E_L{I}_L=240\times 6=1440$

The reactive power in the capacitor is,

$VA{R}_{SC}=E_C{I}_C=240\times 8.02=1924.8$

The total current in the parallel R-L-C circuit is given,

$\begin{array}{l}{I}_T=\sqrt{I_R^2+{\left(I_L-I_C\right)}^2}\\ I_T=\sqrt{{10}^2+{\left(6-8.02\right)}^2}\\ I_T=\sqrt{{10}^2+{\left(-2.02\right)}^2}\\ I_T=10.20\text{ A}\end{array}$

The apparent power in the circuit is,

$VA=E_T{I}_T=240\times 10.206=2448$

The power factor in the circuit is,

$\begin{array}{l}PF=\frac{P}{VA}\times 100\\ =\frac{2400}{2448}\times 100\\ \\ =98.03\%\\ \end{array}$

The power factor angle is,

$\begin{array}{l}\cos \theta =PF\\ \cos \theta =\frac{98.03}{100}\\ \cos \theta =0.9803\\ \theta ={\cos }^{-1}0.9803\\ \theta =11.36°\end{array}$