Chapter 11, Problem 85P

A regular household system of a single-phase three-wire circuit allows the operation of both 120-V and 240-V, 60-Hz appliances. The household circuit is modeled as shown in Fig. 11.96. Calculate:

1. (a) the currents I₁, I₂, and I_n
2. (b) the total complex power supplied
3. (c) the overall power factor of the circuit

To determine

Find the currents $I_1$, $I_2$, and $I_n$ of the circuit shown in Figure 11.96.

Explanation of Solution

Given data:

Refer to Figure 11.96 in the textbook.

The frequency $f$ is $60\text{Hz}$.

The circuits performs at both the voltages $120\text{V}$ and $240\text{V}$.

The inductance L is $15\text{mH}$.

Formula used:

Write the expression for reactance of an inductor $X_L$.

$X_L=j2\pi fL$ (1)

Here,

$f$ is the frequency, and

L is the inductance.

Calculation:

Refer to Figure 11.96 in the textbook.

Substitute $60\text{Hz}$ for $f$ and $15\text{mH}$ for L in equation (1) to find $X_L$.

$\begin{array}{c}{X}_L=j2\pi \left(60\right)\left(15\times {10}^{-3}\right)\left\{\because 1\text{m}={10}^{-3}\right\}\\ =j5.655\Omega \end{array}$

The modified Figure is shown in Figure 1.

In Figure 1, the current flowing through the lamp $\left(120\Omega \right)$ is calculated by using Ohm’s law as follows.

$\begin{array}{c}{I}_{\text{Lamp}}=\frac{120\text{V}}{120\Omega }\\ =1\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

The current flowing through the refrigerator is calculated by using Ohm’s law as follows.

$\begin{array}{c}{I}_{\text{Refrigerator}}=\frac{120\text{V}}{\left(10+j5.655\right)\Omega }\\ =\frac{120\text{V}}{11.4882\angle 29.488°\Omega }\\ =10.4455\angle -29.488°\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

Convert the equation from polar to rectangular form.

$I_{\text{Refrigerator}}=\left(9.0924-j5.1417\right)\text{A}$

The current flowing through the kitchen range $\left(20\Omega \right)$ is calculated by using Ohm’s law as follows.

$\begin{array}{c}{I}_{\text{Range}}=\frac{240\text{V}}{20\Omega }\\ =12\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

In Figure, apply Kirchhoff’s current law in the circuit. Therefore, the current $I_1$ is,

$\begin{array}{c}{I}_1=I_{\text{Lamp}}+I_{\text{Range}}\\ =1\text{A+12}\text{A}\\ =\text{13}\text{A}\end{array}$

Similarly, the current $I_n$ is,

$\begin{array}{c}{I}_n=I_{\text{Refrigerator}}-I_{\text{Lamp}}\\ =\left(9.0924-j5.1417\right)\text{A}-1\text{A}\\ =\left(8.092-j5.142\right)\text{A}\end{array}$

Convert the equation from rectangular to polar form.

$I_n=9.588\angle -32.43°\text{A}$

Similarly, the current $I_2$ is,

$\begin{array}{c}{I}_2=-I_{\text{Refrigerator}}-I_{\text{Range}}\\ =-\left(9.0924+j5.1417\right)\text{A}-12\text{A}\\ =\left(-21.09+j5.142\right)\text{A}\end{array}$

Convert the equation from rectangular to polar form.

$I_2=21.71\angle 166.3°\text{A}$

Conclusion:

Thus, the currents $I_1$, $I_2$, and $I_n$ flowing in the circuit is $13\text{A}$, $21.71\angle 166.3°\text{A}$, and $9.588\angle -32.43°\text{A}$ respectively.

To determine

Find the total complex power in the circuit of Figure 11.96.

Answer to Problem 85P

The total complex power is $\left(\text{4}\text{.091}+j\text{0}\text{.617}\right)\text{kVA}$.

Explanation of Solution

Given data:

Refer to Figure 11.96 in the textbook.

From part (a),

$I_1=13\text{A}$ and $I_2=21.71\angle 166.3°\text{A}$.

Calculation:

Refer to Figure 1 shown in Part (a).

The complex power delivered by the voltage source 1 is,

$S_1=\left(120\text{V}\right){\left(I_1\right)}^{*}$

Substitute $13\text{A}$ for $I_1$ in the equation to find the complex power $S_1$ in VA.

$\begin{array}{c}{S}_1=\left(120\text{V}\right){\left(13\text{A}\right)}^{*}\\ =\left(120\text{V}\right)\left(13\text{A}\right)\\ =1560\times {10}^3\times {10}^{-3}\text{V}\cdot \text{A}\\ =\text{1}\text{.56}\text{kVA}\left\{\because 1\text{k}={10}^3\right\}\end{array}$

The complex power delivered by voltage source 2 is,

$S_2=\left(120\text{V}\right){\left(-I_2\right)}^{*}$

Substitute $21.71\angle 166.3°\text{A}$ for $I_2$ in the equation to find the complex power $S_2$ in VA.

$\begin{array}{c}{S}_2=\left(120\text{V}\right){\left(-\left(21.71\angle 166.3°\right)\text{A}\right)}^{*}\\ =\left(120\text{V}\right){\left(21.71\angle -13.7°\text{A}\right)}^{*}\\ =\left(120\text{V}\right)\left(21.71\angle 13.7°\text{A}\right)\\ =2605.2\times {10}^3\times {10}^{-3}\angle 13.7°\text{VA}\end{array}$

Simply the equation as follows,

$S_2=2.6052\angle 13.7°\text{kVA}\left\{1\text{k}={10}^3\right\}$

Convert the equation from polar to rectangular form.

$S_2=\left(2.531+j0.617\right)\text{kVA}$

The total complex power is,

$S=S_1+S_2$

Substitute $\text{1}\text{.56}\text{kVA}$ for $S_1$ and $\left(2.531+j0.617\right)\text{kVA}$ for $S_2$ in the equation to find the total complex power in VA.

$S=\text{1}\text{.56}\text{kVA}+\left(2.531+j0.617\right)\text{kVA}$

$S=\left(\text{4}\text{.091}+j\text{0}\text{.617}\right)\text{kVA}$ (2)

Conclusion:

Thus, the total complex power is $\left(\text{4}\text{.091}+j\text{0}\text{.617}\right)\text{kVA}$.

To determine

Find the total power factor of the circuit shown in Figure 11.96.

Answer to Problem 85P

The total power factor of the circuit is 0.9888 (lagging).

Explanation of Solution

Given data:

Refer to Figure 11.96 in the textbook.

Formula used:

Write the expression for complex power.

$S=P+jQ$ (3)

Here,

P is the real power, and

Q is the reactive power.

Write the expression for power factor.

$\text{pf}=\frac{P}{S}$ (4)

Calculation:

On comparing equation (2) and (3), the real power is,

$P=4.091\text{W}$ and $Q=0.617\text{kVAR}$

From equation (2), the apparent power S is,

$\begin{array}{c}S=\left|S\right|\\ =\sqrt{{4.091}^2+{0.617}^2}\\ =4.1373\text{VA}\end{array}$

Substitute $4.1373\text{VA}$ for S and $4.091\text{W}$ for P in equation (4) to find the power factor.

$\begin{array}{c}\text{pf}=\frac{4.091\text{W}}{4.1373\text{VA}}\\ =\frac{4.091\text{W}}{4.1373\text{W}}\left\{\because 1\text{W}=1\text{V}\cdot \text{1}\text{A}\right\}\\ =0.9888\end{array}$

Since, $Q>0$, the load is inductive and the power factor is lagging.

Conclusion:

Thus, the total power factor of the circuit is 0.9888 (lagging).