Chapter 12, Problem 87CP

Consider the single-phase three-wire system shown in Fig. 12.78. Find the current in the neutral wire and the complex power supplied by each source. Take V_s as a 220∠0°-V, 60-Hz source.

Figure 12.78

To determine

Calculate the complex power supplied by each source and the current passing through the neutral wire.

Answer to Problem 87CP

The current in the neutral wire is $2.77\angle -176.6°\text{A}$, and the complex power absorbed by each source are $4.581+j2.604\text{ kVA}$ and $3.971+j2.64\text{kVA}$.

Explanation of Solution

Given data:

Refer to Figure 12.78 in the textbook for a single phase three wire system.

The source voltage is $220\angle 0°\text{V}$.

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

Formula used:

Write the expression to find the complex power $S_1$.

$S_1=V_1{I}_1^{*}$        (1)

Here,

$V_1$ is the voltage in the form of complex number,

$I_1^{*}$ is the complex conjugate of the current $I_1$.

Write the expression to find the complex power $S_2$.

$S_2=V_2{I}_2^{*}$        (2)

Here,

$V_2$ is the voltage in the form of complex number,

$I_2^{*}$ is the complex conjugate of the current $I_2$.

Write the expression to find the inductance reactance in ohms.

$X_L=j\omega L$        (3)

Here,

$\omega$ is the angular frequency, and

$L$ is the inductance.

Write the expression to find the angular frequency.

$\omega =2\pi f$        (4)

Here,

$f$ is the supply frequency.

Calculation:

Substitute $60\text{Hz}$ for $f$ in equation (4) to find $\omega$.

$\begin{array}{c}\omega =2\pi \left(60\text{Hz}\right)\\ =376.99\frac{\text{rad}}{\text{s}}\\ \cong 377\frac{\text{rad}}{\text{s}}\end{array}$

Substitute $377\frac{\text{rad}}{\text{s}}$ for $\omega$, and $50\text{mH}$ for $L$ in equation (3) to find the inductance $\left(L\right)$.

$\begin{array}{c}{X}_L=j\left(377\frac{\text{rad}}{\text{s}}\right)\left(50\text{mH}\right)\\ =j\left(377\frac{\text{rad}}{\text{s}}\right)\left(50\times {10}^{-3}\text{H}\right)\left\{\because 1\text{mH=1}\times {\text{10}}^{-3}\text{H}\right\}\\ =18.85\Omega \left\{\because 1\Omega \text{=1}\frac{\text{H}}{\text{s}}\right\}\end{array}$

Modify the given figure with the assumed current directions as shown in Figure.1.

Apply Kirchhoff’s voltage law to loop 1of current $I_1$ in Figure 1.

$\begin{array}{l}-220\angle 0°+I_1+20\left(I_1-I_3\right)+2\left(I_1-I_2\right)=0\\ I_1+20I_1-20I_3+2I_1-2I_2=220\angle 0°\end{array}$

$23I_1-2I_2-20I_3=220$        (5)

Apply Kirchhoff’s voltage law to loop 2 of current $I_2$ in Figure 1.

$\begin{array}{l}-220\angle 0°+2\left(I_2-I_1\right)+30\left(I_2-I_3\right)+I_2=0\\ 2I_2-2I_1+30I_2-30I_3+I_2=220\angle 0°\end{array}$

$-2I_1+33I_2-30I_3=220$        (6)

Apply Kirchhoff’s voltage law to loop 3 of current $I_3$ in Figure 1.

$\begin{array}{l}20\left(I_3-I_1\right)+\left(15+j18.85\right)I_3+30\left(I_3-I_2\right)=0\\ 20I_3-20I_1+15I_3+j18.85I_3+30I_3-30I_2=0\end{array}$

$-20I_1-30I_2+65I_3+j18.85I_3=0$

$-20I_1-30I_2+\left(65+j18.85\right)I_3=0$        (7)

Represent the equation (5) (6) and (7) in matrix form.

$\left[\begin{array}{ccc}23& -2& -20\\ -2& 33& -30\\ -20& -30& 65+j18.85\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ I_3\end{array}\right]=\left[\begin{array}{c}220\\ 220\\ 0\end{array}\right]$

Obtain the value of determinants as follows.

$\begin{array}{c}\Delta =\left|\begin{array}{ccc}23& -2& -20\\ -2& 33& -30\\ -20& -30& 65+j18.85\end{array}\right|\\ =\left[\begin{array}{c}23\left[33\left(65+j18.85\right)-\left(-30\right)\left(-30\right)\right]-\left(-2\right)\left[-2\left(65+j18.85\right)-\left(-20\right)\left(-30\right)\right]\\ -20\left[\left(-2\right)\left(-30\right)-\left(-20\right)\left(33\right)\right]\end{array}\right]\\ \Delta =12,775+j14,231.75\end{array}$

$\begin{array}{c}{\Delta }_1=\left|\begin{array}{ccc}220& -2& -20\\ 220& 33& -30\\ 0& -30& 65+j18.85\end{array}\right|\\ =220\left[33\left(65+j18.85\right)-900\right]-\left(-2\right)\left[220\left(65+j18.85\right)-0\right]-20\left[220\left(-30\right)\right]\\ {\Delta }_1=434500+j145145\end{array}$

$\begin{array}{c}{\Delta }_2=\left|\begin{array}{ccc}23& 220& -20\\ -2& 220& -30\\ -20& 0& 65+j18.85\end{array}\right|\\ =23\left[220\left(65+j18.85\right)-0\right]-220\left[-2\left(65+j18.85\right)-\left(-20\right)\left(-30\right)\right]-20\left(0-220\left(-20\right)\right)\\ {\Delta }_2=401500+j103675\end{array}$

$\begin{array}{c}{\Delta }_3=\left|\begin{array}{ccc}23& -2& 220\\ -2& 33& 220\\ -20& -30& 0\end{array}\right|\\ =23\left[0-220\left(-30\right)\right]-\left(-2\right)\left[0-\left(-20\right)\times \left(220\right)\right]+220\left[\left(-2\right)\left(-30\right)-\left(33\right)\left(-20\right)\right]\\ {\Delta }_3=319000\end{array}$

Write the expression to find the currents, $I_1$, $I_2$ , $I_3$ and $I_n$ using Cramer’s rule.

$I_1=\frac{{\Delta }_1}{\Delta }$        (8)

$I_2=\frac{{\Delta }_2}{\Delta }$        (9)

$I_3=\frac{{\Delta }_3}{\Delta }$        (10)

$I_n=I_2-I_1$        (11)

Substitute $434500+j145145$ for ${\Delta }_1$ and $12,775+j14,231.75$ for $\Delta$ in equation (8) to find $I_1$.

$\begin{array}{c}{I}_1=\frac{434500+j145145}{12,775+j14,231.75}\text{ A}\\ =\frac{458101.867\angle 18.471°}{19124.417\angle 48.087°}\text{A}\\ =23.9538\angle -29.616°\text{A}\end{array}$

$I_1=\text{20}\text{.8245}-j\text{11}\text{.837 A}$

Substitute $401500+j103675$ for ${\Delta }_2$, and $12,775+j14,231.75$ for $\Delta$ in equation (9) to find the current $I_2$.

$\begin{array}{c}{I}_2=\frac{401500+j103675}{12775+j14,231.75}\text{ A}\\ =\frac{414669.4535\angle 14.478°}{19124.417\angle 48.087°}\text{ A}\\ =21.6827\angle -33.609°\text{ A}\end{array}$

$I_2=18.0581-j12.0019\text{ A}$

Substitute $319000$ for ${\Delta }_3$, and $12,775+j14,231.75$ for $\Delta$ in equation (10) to find the current $I_3$.

$\begin{array}{c}{I}_3=\frac{319000}{12775+j14,231.75}\text{A}\\ =\frac{319000}{19124.417\angle 48.087°}\text{A}\\ =16.6802\angle -48.087°\text{A}\\ I_3\text{=11}\text{.1423}-j\text{12}\text{.4128 A}\end{array}$

Substitute $18.0581-j12.0019\text{ A}$ for $I_2$, and $\text{20}\text{.8245}-j\text{11}\text{.837 A}$ for $I_1$ in equation (11) to find the current $I_n$.

$\begin{array}{c}{I}_n=\left(18.0581-j12.0019\text{ A}\right)-\left(\text{20}\text{.8245}-j\text{11}\text{.837 A}\right)\\ I_n=-2.7664-j0.1649\text{}\text{A}\\ I_n=2.7713\angle -176.588°\text{A}\\ I_n\cong 2.77\angle -176.6°\text{A}\end{array}$

Thus, the neutral current is $2.77\angle -176.6°\text{A}$.

Substitute $220\text{V}$ for $V_1$, and $\text{20}\text{.8245}+j\text{11}\text{.837 A}$ for $I_1^{*}$ in equation (1) to find $S_1$.

$\begin{array}{c}{S}_1=\left(220\text{V}\right)\left(\text{20}\text{.8245}+j\text{11}\text{.837 A}\right)\left\{\begin{array}{l}\because I_1=\text{20}\text{.8245}-j\text{11}\text{.837 A}\\ I_1^{*}=\text{20}\text{.8245}+j\text{11}\text{.837 A}\end{array}\right\}\\ =4581.39+j2604.14\text{ VA}\\ S_1=4.581+j2.604\text{ kVA}\end{array}$

Substitute $220\text{V}$ for $V_2$, and $18.0581-j12.0019\text{ A}$ for $I_2^{*}$ in equation (2) to find $S_2$.

$\begin{array}{c}{S}_2=\left(220\text{V}\right)\left(18.0581-j12.0019\text{ A}\right)\\ =3972.782+j2640.418\text{VA}\\ S_2\cong 3.971+j2.64\text{kVA}\end{array}$

Conclusion:

Thus, the current in the neutral wire is $2.77\angle -176.6°\text{A}$, and the complex power absorbed by each source are $4.581+j2.604\text{ kVA}$ and $3.971+j2.64\text{kVA}$.