Chapter 11, Problem 43E

(a)

To determine

Find the apparent power delivered to each load in the circuit in Figure 11.45 in the textbook and power factor of the source.

Answer to Problem 43E

The apparent power delivered to $Z_A$, $Z_B$, $Z_C$, $Z_D$, and the source power factor are $\underset{\_}{3.4\text{kVA}}$, $\underset{\_}{1.48\text{kVA}}$, $\underset{\_}{87.67\text{VA}}$, $\underset{\_}{147\text{VA}}$, and 0.966 leading, respectively.

Explanation of Solution

Given data:

Refer to Figure 11.45 in the textbook for the given circuit.

$V_s=200\angle 0°\text{V}\text{rms}$

The impedance of the loads are given as follows:

$\begin{array}{l}{Z}_A=\left(5-j2\right)\Omega \\ Z_B=3\Omega \\ Z_C=\left(8+j4\right)\Omega \\ Z_D=15\angle -30°\Omega \end{array}$

Formula used:

Write the expression for complex power delivered to the load as follows:

$S=V{I}^{*}$        (1)

Here,

$V$ is the voltage across the load and

$I$ is the current through the load.

Write the expression for voltage in terms of current and impedance as follows:

$V=IZ$        (2)

Write the expression for complex power in the rectangular form as follows:

$S=P+jQ$        (3)

Here,

$P$ is the average power and

$Q$ is the reactive power.

Write the expression for power factor as follows:

$\text{PF}=\cos \left[{\tan }^{-1}\left(\frac{Q}{P}\right)\right]$        (4)

Calculation:

Consider the current through the left-loop as $I_1$ and the current through the right-loop as $I_2$.

Apply KVL to the left-loop as follows:

$-200\angle 0°\text{V}+I_1{Z}_A+\left(I_1-I_2\right)Z_B=0$        (5)

Substitute $\left(5-j2\right)\Omega$ for $Z_A$ and $3\Omega$ for $Z_B$ as follows:

$\begin{array}{l}-200\angle 0°\text{V}+I_1\left[\left(5-j2\right)\Omega \right]+\left(I_1-I_2\right)\left(3\Omega \right)=0\\ \left(8-j2\right)I_1-3I_2=200\\ 3I_2=\left(8-j2\right)I_1-200\end{array}$

$I_2=\frac{\left(8-j2\right)}{3}{I}_1-\frac{200}{3}$        (6)

Apply KVL to the right-loop as follows:

$\left(I_2-I_1\right)Z_B+I_2{Z}_C+I_2{Z}_D=0$        (7)

Substitute $3\Omega$ for $Z_B$, $\left(8+j4\right)\Omega$ for $Z_C$, and $15\angle -30°\Omega$ for $Z_D$ as follows:

$\begin{array}{l}\left(I_2-I_1\right)\left(3\Omega \right)+I_2\left[\left(8+j4\right)\Omega \right]+I_2\left(15\angle -30°\Omega \right)=0\\ 3\left(I_2-I_1\right)+I_2\left(8+j4\right)+I_2\left(12.9903-j7.5\right)=0\\ -3I_1+\left(23.9903-j3.5\right)I_2=0\end{array}$

From Equation (6), substitute $\left[\frac{\left(8-j2\right)}{3}{I}_1-\frac{200}{3}\right]$ for $I_2$ as follows:

$\begin{array}{l}-3I_1+\left(23.9903-j3.5\right)\left[\frac{\left(8-j2\right)}{3}{I}_1-\frac{200}{3}\right]=0\\ -3I_1+\left(61.6408-j25.3268\right)I_1+\left(-1599.3533+j233.3333\right)=0\\ \left(58.6408-j25.3268\right)I_1+\left(-1599.3533+j233.3333\right)=0\end{array}$

Simplify the expression as follows:

$\begin{array}{c}{I}_1=\frac{\left(1599.3533-j233.3333\right)}{\left(58.6408-j25.3268\right)}\text{A}\\ =\left(24.4343+j6.5741\right)\text{A}\\ =25.3032\angle 15.06°\text{A}\end{array}$

Substitute $\left(24.4343+j6.5741\right)\text{A}$ for $I_1$ in Equation (6) to obtain the value of $I_2$.

$\begin{array}{c}{I}_2=\frac{\left(8-j2\right)}{3}\left[\left(24.4343+j6.5741\right)\text{A}\right]-\frac{200}{3}\\ =\left(2.8742+j1.2414\right)\text{A}\\ =3.1308\angle 23.36°\text{A}\end{array}$

Modify the expression in Equation (2) for the voltage across the load $Z_A$.

$V_A=I_1{Z}_A$

Substitute $\left(24.4343+j6.5741\right)\text{A}$ for $I_1$ and $\left(5-j2\right)\Omega$ for $Z_A$ to obtain he voltage across . $Z_A$

$\begin{array}{c}{V}_A=\left[\left(24.4343+j6.5741\right)\text{A}\right]\left[\left(5-j2\right)\Omega \right]\\ =\left(135.3197-j15.9981\right)\text{V}\\ =136.2621\angle -6.74°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the load $Z_A$ as follows:

$S_A=V_A{\left(I_1\right)}^{*}$

Substitute $136.2621\angle -6.74°\text{V}$ for $V_A$ and $25.3032\angle 15.06°\text{A}$ for $I_1$ to obtain the complex power delivered to the load $Z_A$.

$\begin{array}{c}{S}_A=\left(136.2621\angle -6.74°\text{V}\right){\left(25.3032\angle 15.06°\text{A}\right)}^{*}\\ =\left(136.2621\angle -6.74°\text{V}\right)\left(25.3032\angle -15.06°\text{A}\right)\\ =3447.8671\angle -21.8°\text{VA}\\ =3.4478\angle -21.8°\text{kVA}\end{array}$

$S_A\cong 3.4\angle -21.8°\text{kVA}$

Find the apparent power delivered to the load $Z_A$ as follows:

$\begin{array}{c}\left|S_A\right|=\left|3.4\angle -21.8°\text{kVA}\right|\\ =3.4\text{kVA}\end{array}$

Modify the expression in Equation (2) for the voltage across the load $Z_B$.

$V_B=\left(I_1-I_2\right)Z_B$

Substitute $\left(24.4343+j6.5741\right)\text{A}$ for $I_1$, $\left(2.8742+j1.2414\right)\text{A}$ for $I_2$, and $3\Omega$ for $Z_B$

$\begin{array}{c}{V}_B=\left\{\left[\left(24.4343+j6.5741\right)\text{A}\right]-\left[\left(2.8742+j1.2414\right)\text{A}\right]\right\}\left(3\Omega \right)\\ =\left(21.5601+j5.3327\right)\left(3\right)\text{V}\\ =\left(64.6803+15.9981\right)\text{V}\\ =66.6294\angle 13.89°\text{V}\end{array}$

Find the current through the load $Z_B$ as follows:

$I_B=\left(I_1-I_2\right)$

Substitute $\left(24.4343+j6.5741\right)\text{A}$ for $I_1$ and $\left(2.8742+j1.2414\right)\text{A}$ for $I_2$ to obtain the current through the load $Z_B$.

$\begin{array}{c}{I}_B=\left[\left(24.4343+j6.5741\right)\text{A}\right]-\left[\left(2.8742+j1.2414\right)\text{A}\right]\\ =\left(21.5601+j5.3327\right)\text{A}\\ =22.2098\angle 13.89°\text{A}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the load $Z_B$ as follows:

$S_B=V_B{\left(I_B\right)}^{*}$

Substitute $66.6294\angle 13.89°\text{V}$ for $V_B$ and $22.2098\angle 13.89°\text{A}$ for $I_B$ to obtain the complex power delivered to the load $Z_B$.

$\begin{array}{c}{S}_B=\left(66.6294\angle 13.89°\text{V}\right){\left(22.2098\angle 13.89°\text{A}\right)}^{*}\\ =\left(66.6294\angle 13.89°\text{V}\right)\left(22.2098\angle -13.89°\text{A}\right)\\ =1479.8256\angle 0°\text{VA}\\ =1.4798\angle 0°\text{kVA}\end{array}$

$S_B\cong 1.48\angle 0°\text{kVA}$

Find the apparent power delivered to the load $Z_B$ as follows:

$\begin{array}{c}\left|S_B\right|=\left|1.48\angle 0°\text{kVA}\right|\\ =1.48\text{kVA}\end{array}$

Modify the expression in Equation (2) for the voltage across the load $Z_C$.

$V_C=I_2{Z}_C$

Substitute $\left(2.8742+j1.2414\right)\text{A}$ for $I_2$ and $\left(8+j4\right)\Omega$ for $Z_C$ to obtain he voltage across $Z_C$.

$\begin{array}{c}{V}_C=\left[\left(2.8742+j1.2414\right)\text{A}\right]\left[\left(8+j4\right)\Omega \right]\\ =\left(18.028+j21.428\right)\text{V}\\ =28.003\angle 49.93°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the load $Z_C$ as follows:

$S_C=V_C{\left(I_2\right)}^{*}$

Substitute $28.003\angle 49.93°\text{V}$ for $V_C$ and $3.1308\angle 23.36°\text{A}$ for $I_2$ to obtain the complex power delivered to the load $Z_C$.

$\begin{array}{c}{S}_C=\left(28.003\angle 49.93°\text{V}\right){\left(3.1308\angle 23.36°\text{A}\right)}^{*}\\ =\left(28.003\angle 49.93°\text{V}\right)\left(3.1308\angle -23.36°\text{A}\right)\\ =87.6717\angle 26.57°\text{VA}\\ \cong 87.67\angle 26.57°\text{VA}\end{array}$

Find the apparent power delivered to the load $Z_C$ as follows:

$\begin{array}{c}\left|S_C\right|=\left|87.67\angle 26.57°\text{VA}\right|\\ =87.67\text{VA}\end{array}$

Modify the expression in Equation (2) for the voltage across the load $Z_D$.

$V_D=I_2{Z}_D$

Substitute $3.1308\angle 23.36°\text{A}$ for $I_2$ and $15\angle -30°\Omega$ for $Z_D$ to obtain he voltage across $Z_D$.

$\begin{array}{c}{V}_D=\left[3.1308\angle 23.36°\text{A}\right]\left[15\angle -30°\Omega \right]\\ =46.962\angle -6.64°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the load $Z_D$ as follows:

$S_D=V_D{\left(I_2\right)}^{*}$

Substitute $46.962\angle -6.64°\text{V}$ for $V_D$ and $3.1308\angle 23.36°\text{A}$ for $I_2$ to obtain the complex power delivered to the load $Z_D$.

$\begin{array}{c}{S}_D=\left(46.962\angle -6.64°\text{V}\right){\left(3.1308\angle 23.36°\text{A}\right)}^{*}\\ =\left(46.962\angle -6.64°\text{V}\right)\left(3.1308\angle -23.36°\text{A}\right)\\ =147.0286\angle -30°\text{VA}\\ \cong 147\angle -30°\text{VA}\end{array}$

Find the apparent power delivered to the load $Z_D$ as follows:

$\begin{array}{c}\left|S_D\right|=\left|147\angle -30°\text{VA}\right|\\ =147\text{VA}\end{array}$

Modify the expression in Equation (1) for the complex power supplied by the source as follows:

$S_s=V_s{\left(I_1\right)}^{*}$

Substitute $200\angle 0°\text{V}$ for $V_s$ and $25.3032\angle 15.06°\text{A}$ for $I_1$ to obtain the complex power supplied by the source.

$\begin{array}{c}{S}_s=\left(200\angle 0°\text{V}\right){\left(25.3032\angle 15.06°\text{A}\right)}^{*}\\ =\left(200\angle 0°\text{V}\right)\left(25.3032\angle -15.06°\text{A}\right)\\ =5060.6\angle -15.06°\text{VA}\end{array}$

Rewrite the expression for complex power supplied by the source in rectangular form as follows:

$S_s=\left(4886.7899-j1314.8978\right)\text{VA}$

Compare the complex power supplied by the source with the expression in Equation (3) and write the average and reactive power supplied by the source as follows:

$\begin{array}{l}P=4886.7899\text{W}\\ Q=-1314.8978\text{VAR}\end{array}$

Substitute 4886.7899 W for $P$ and $-1314.8978\text{VAR}$ for $Q$ in Equation (4) to obtain the value of source power factor.

$\begin{array}{c}\text{PF}=\cos \left[{\tan }^{-1}\left(\frac{-1314.8978\text{VAR}}{4886.7899\text{W}}\right)\right]\\ =\cos \left(15.06°\right)\\ =0.9656\\ \cong 0.966\end{array}$

If the imaginary part of the complex power (reactive power) is positive value, then the load has lagging power factor. If the imaginary part is negative value, then the load has leading power factor.

As the imaginary part of the given complex power is negative value, the power factor is leading power factor.

$\text{PF}=0.966\text{leading}$

Conclusion:

Thus, the apparent power delivered to $Z_A$, $Z_B$, $Z_C$, $Z_D$, and the source power factor are $\underset{\_}{3.4\text{kVA}}$, $\underset{\_}{1.48\text{kVA}}$, $\underset{\_}{87.67\text{VA}}$, $\underset{\_}{147\text{VA}}$, and 0.966 leading, respectively.

(b)

To determine

Find the apparent power delivered to each load in the circuit in Figure 11.45 in the textbook and power factor of the source.

Answer to Problem 43E

The apparent power delivered to $Z_A$, $Z_B$, $Z_C$, $Z_D$, and the source power factor are $\underset{\_}{9.87\text{kVA}}$, $\underset{\_}{3.86\text{kVA}}$, $\underset{\_}{227.13\text{VA}}$, $\underset{\_}{406.3\text{VA}}$, and 0.85 leading, respectively.

Explanation of Solution

Given data:

The impedance of the loads are given as follows:

$\begin{array}{l}{Z}_A=2\angle -15°\Omega \\ Z_B=1\Omega \\ Z_C=\left(2+j\right)\Omega \\ Z_D=4\angle 45°\Omega \end{array}$

Calculation:

Substitute $2\angle -15°\Omega$ for $Z_A$ and $1\Omega$ for $Z_B$ in Equation (5) as follows:

$\begin{array}{l}-200\angle 0°\text{V}+I_1\left(2\angle -15°\Omega \right)+\left(I_1-I_2\right)\left(1\Omega \right)=0\\ -200+I_1\left(1.9318-j0.5176\right)+\left(I_1-I_2\right)=0\\ \left(2.9318-j0.5176\right)I_1-I_2=200\end{array}$

$I_2=\left(2.9318-j0.5176\right)I_1-200$        (8)

Substitute $1\Omega$ for $Z_B$, $\left(2+j\right)\Omega$ for $Z_C$, and $4\angle 45°\Omega$ for $Z_D$ in Equation (7) as follows:

$\begin{array}{l}\left(I_2-I_1\right)\left(1\Omega \right)+I_2\left[\left(2+j\right)\Omega \right]+I_2\left(4\angle 45°\Omega \right)=0\\ \left(I_2-I_1\right)+I_2\left(2+j\right)+I_2\left(2.8284+j2.8284\right)=0\\ -I_1+\left(5.8284+j3.8284\right)I_2=0\end{array}$

From Equation (5), substitute $\left[\left(2.9318-j0.5176\right)I_1-200\right]$ for $I_2$ as follows:

$\begin{array}{l}-I_1+\left(5.8284+j3.8284\right)\left[\left(2.9318-j0.5176\right)I_1-200\right]=0\\ -I_1+\left(19.0692+j8.2073\right)I_1-\left(1165.68+j765.68\right)=0\end{array}$

Simplify the expression as follows:

$\begin{array}{c}{I}_1=\frac{\left(1165.68+j765.68\right)}{\left(18.0692+j8.2073\right)}\text{A}\\ =\left(69.4342+j10.8368\right)\text{A}\\ =70.2747\angle 8.87°\text{A}\end{array}$

Substitute $\left(69.4342+j10.8368\right)\text{A}$ for $I_1$ in Equation (8) to obtain the value of $I_2$.

$\begin{array}{c}{I}_2=\left(2.9318-j0.5176\right)\left[\left(69.4342+j10.8368\right)\text{A}\right]-200\\ =\left(9.1763-j4.1678\right)\text{A}\\ =10.0784\angle -24.43°\text{A}\end{array}$

Modify the expression in Equation (2) for the voltage across the load $Z_A$.

$V_A=I_1{Z}_A$

Substitute $70.2747\angle 8.87°\text{A}$ for $I_1$ and $2\angle -15°\Omega$ for $Z_A$ to obtain he voltage across . $Z_A$

$\begin{array}{c}{V}_A=\left(70.2747\angle 8.87°\text{A}\right)\left(2\angle -15°\Omega \right)\\ =140.5494\angle -6.13°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the load $Z_A$ as follows:

$S_A=V_A{\left(I_1\right)}^{*}$

Substitute $140.5494\angle -6.13°\text{V}$ for $V_A$ and $70.2747\angle 8.87°\text{A}$ for $I_1$ to obtain the complex power delivered to the load $Z_A$.

$\begin{array}{c}{S}_A=\left(140.5494\angle -6.13°\text{V}\right){\left(70.2747\angle 8.87°\text{A}\right)}^{*}\\ =\left(140.5494\angle -6.13°\text{V}\right)\left(70.2747\angle -8.87°\text{A}\right)\\ =9877.0669\angle -15°\text{VA}\\ =9.8770\angle -15°\text{kVA}\end{array}$

$S_A\cong 9.87\angle -15°\text{kVA}$

Find the apparent power delivered to the load $Z_A$ as follows:

$\begin{array}{c}\left|S_A\right|=\left|9.87\angle -15°\text{kVA}\right|\\ =9.87\text{kVA}\end{array}$

Modify the expression in Equation (2) for the voltage across the load $Z_B$.

$V_B=\left(I_1-I_2\right)Z_B$

Substitute $\left(69.4342+j10.8368\right)\text{A}$ for $I_1$, $\left(9.1763-j4.1678\right)\text{A}$ for $I_2$, and $1\Omega$ for $Z_B$

$\begin{array}{c}{V}_B=\left\{\left[\left(69.4342+j10.8368\right)\text{A}\right]-\left[\left(9.1763-j4.1678\right)\text{A}\right]\right\}\left(1\Omega \right)\\ =\left(60.2579+j15.0046\right)\left(1\right)\text{V}\\ =\left(60.2579+j15.0046\right)\text{V}\\ =62.0979\angle 13.98°\text{V}\end{array}$

Find the current through the load $Z_B$ as follows:

$I_B=\left(I_1-I_2\right)$

Substitute $\left(69.4342+j10.8368\right)\text{A}$ for $I_1$ and $\left(9.1763-j4.1678\right)\text{A}$ for $I_2$ to obtain the current through the load $Z_B$.

$\begin{array}{c}{I}_B=\left[\left(69.4342+j10.8368\right)\text{A}\right]-\left[\left(9.1763-j4.1678\right)\text{A}\right]\\ =\left(60.2579+j15.0046\right)\text{A}\\ =62.0979\angle 13.98°\text{A}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the load $Z_B$ as follows:

$S_B=V_B{\left(I_B\right)}^{*}$

Substitute $62.0979\angle 13.98°\text{V}$ for $V_B$ and $62.0979\angle 13.98°\text{A}$ for $I_B$ to obtain the complex power delivered to the load $Z_B$.

$\begin{array}{c}{S}_B=\left(62.0979\angle 13.98°\text{V}\right){\left(62.0979\angle 13.98°\text{A}\right)}^{*}\\ =\left(62.0979\angle 13.98°\text{V}\right)\left(62.0979\angle -13.98°\text{A}\right)\\ =3856.1491\angle 0°\text{VA}\\ =3.8561\angle 0°\text{kVA}\end{array}$

$S_B\cong 3.86\angle 0°\text{kVA}$

Find the apparent power delivered to the load $Z_B$ as follows:

$\begin{array}{c}\left|S_B\right|=\left|3.86\angle 0°\text{kVA}\right|\\ =3.86\text{kVA}\end{array}$

Modify the expression in Equation (2) for the voltage across the load $Z_C$.

$V_C=I_2{Z}_C$

Substitute $\left(9.1763-j4.1678\right)\text{A}$ for $I_2$ and $\left(2+j\right)\Omega$ for $Z_C$ to obtain he voltage across $Z_C$.

$\begin{array}{c}{V}_C=\left[\left(9.1763-j4.1678\right)\text{A}\right]\left[\left(2+j\right)\Omega \right]\\ =\left(22.5204+j0.8407\right)\text{V}\\ =22.5360\angle 2.14°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the load $Z_C$ as follows:

$S_C=V_C{\left(I_2\right)}^{*}$

Substitute $22.5360\angle 2.14°\text{V}$ for $V_C$ and $10.0784\angle -24.43°\text{A}$ for $I_2$ to obtain the complex power delivered to the load $Z_C$.

$\begin{array}{c}{S}_C=\left(22.5360\angle 2.14°\text{V}\right){\left(10.0784\angle -24.43°\text{A}\right)}^{*}\\ =\left(22.5360\angle 2.14°\text{V}\right)\left(10.0784\angle 24.43°\text{A}\right)\\ =227.1268\angle 26.57°\text{VA}\\ \cong 227.13\angle 26.57°\text{VA}\end{array}$

Find the apparent power delivered to the load $Z_C$ as follows:

$\begin{array}{c}\left|S_C\right|=\left|227.13\angle 26.57°\text{VA}\right|\\ =227.13\text{VA}\end{array}$

Modify the expression in Equation (2) for the voltage across the load $Z_D$.

$V_D=I_2{Z}_D$

Substitute $10.0784\angle -24.43°\text{A}$ for $I_2$ and $4\angle 45°\Omega$ for $Z_D$ to obtain he voltage across $Z_D$.

$\begin{array}{c}{V}_D=\left[10.0784\angle -24.43°\text{A}\right]\left[4\angle 45°\Omega \right]\\ =40.3136\angle 20.57°\text{V}\end{array}$

Modify the expression in Equation (1) for the complex power delivered to the load $Z_D$ as follows:

$S_D=V_D{\left(I_2\right)}^{*}$

Substitute $40.3136\angle 20.57°\text{V}$ for $V_D$ and $10.0784\angle -24.43°\text{A}$ for $I_2$ to obtain the complex power delivered to the load $Z_D$.

$\begin{array}{c}{S}_D=\left(40.3136\angle 20.57°\text{V}\right){\left(10.0784\angle -24.43°\text{A}\right)}^{*}\\ =\left(40.3136\angle 20.57°\text{V}\right)\left(10.0784\angle 24.43°\text{A}\right)\\ =406.2965\angle 45°\text{VA}\\ \cong 406.3\angle 45°\text{VA}\end{array}$

Find the apparent power delivered to the load $Z_D$ as follows:

$\begin{array}{c}\left|S_D\right|=\left|406.3\angle 45°\text{VA}\right|\\ =406.3\text{VA}\end{array}$

Modify the expression in Equation (1) for the complex power supplied by the source as follows:

$S_s=V_s{\left(I_1\right)}^{*}$

Substitute $200\angle 0°\text{V}$ for $V_s$ and $70.2747\angle 8.87°\text{A}$ for $I_1$ to obtain the complex power supplied by the source.

$\begin{array}{c}{S}_s=\left(200\angle 0°\text{V}\right){\left(70.2747\angle 8.87°\text{A}\right)}^{*}\\ =\left(200\angle 0°\text{V}\right)\left(70.2747\angle -8.87°\text{A}\right)\\ =14054.48\angle -8.87°\text{VA}\end{array}$

Rewrite the expression for complex power supplied by the source in rectangular form as follows:

$S_s=\left(-11946.5528-j7403.2616\right)\text{VA}$

Compare the complex power supplied by the source with the expression in Equation (3) and write the average and reactive power supplied by the source as follows:

$\begin{array}{l}P=-11946.5528\text{W}\\ Q=-7403.2616\text{VAR}\end{array}$

Substitute $-11946.5528\text{W}$ for $P$ and $-7403.2616\text{VAR}$ for $Q$ in Equation (4) to obtain the value of source power factor.

$\begin{array}{c}\text{PF}=\cos \left[{\tan }^{-1}\left(\frac{-7403.2616\text{VAR}}{-11946.5528\text{W}}\right)\right]\\ =\cos \left(31.79°\right)\\ =0.8499\\ \cong 0.85\end{array}$

If the imaginary part of the complex power (reactive power) is positive value, then the load has lagging power factor. If the imaginary part is negative value, then the load has leading power factor.

As the imaginary part of the given complex power is negative value, the power factor is leading power factor.

$\text{PF}=0.85\text{leading}$

Conclusion:

Thus, the apparent power delivered to $Z_A$, $Z_B$, $Z_C$, $Z_D$, and the source power factor are $\underset{\_}{9.87\text{kVA}}$, $\underset{\_}{3.86\text{kVA}}$, $\underset{\_}{227.13\text{VA}}$, $\underset{\_}{406.3\text{VA}}$, and 0.85 leading, respectively.