Chapter 12, Problem 11E

The single-phase three-wire system of Fig. 12.31 has three separate load impedances. If the source is balanced and V_an = 110 +j0 V rms, (a) express V_an and V_bn in phasor notation. (b) Determine the phasor voltage which appears across the impedance Z₃. (c) Determine the average power delivered by the two sources if Z₁ = 50 + j0 Ω, Z₂ = 100 + j45 Ω, and Z₃ = 100 – j90 Ω. (d) Represent load Z₃ by a series connection of two elements, and state their respective values if the sources operate at 60 Hz.

To determine

The expression of phase to neutral voltage of phase $a$ and phase $b$ in phasor notations.

Answer to Problem 11E

The expression of phase to neutral voltage of phase $a$ in phasor notation is $110\angle 0°{\text{V}}_{\text{rms}}$ and of phase $b$ is $110\angle 180°{\text{V}}_{\text{rms}}$.

Explanation of Solution

Given data:

The phase to neutral voltage of phase $a$ $V_{an}$ is $\left(110+j0\right){\text{V}}_{\text{rms}}$.

Calculation:

The given diagram is shown in Figure 1.

The general expression for the phasor notation is given by,

$F=\left|F\right|\angle \theta$        (1)

Here,

$\left|F\right|$ is the magnitude of the $F$.

$\theta$ is the angle measured from the reference.

The magnitude of $F$ is given by,

$\left|F\right|=\sqrt{x^2+y^2}$        (2)

Here,

$x$ is the real part.

$y$ is the imaginary part.

The angle measured from the reference $\theta$ is,

$\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$        (3)

Substitute $110{\text{V}}_{\text{rms}}$ for $x$, $0{\text{V}}_{\text{rms}}$ for $y$ and $V_{an}$ for $F$ in equation (2).

$\begin{array}{c}\left|V_{an}\right|=\sqrt{{\left(110{\text{V}}_{\text{rms}}\right)}^2+{\left(0{\text{V}}_{\text{rms}}\right)}^2}\\ =110{\text{V}}_{\text{rms}}\end{array}$

Substitute $110{\text{V}}_{\text{rms}}$ for $x$ and $0{\text{V}}_{\text{rms}}$ for $y$ in equation (3).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{0{\text{V}}_{\text{rms}}}{110{\text{V}}_{\text{rms}}}\right)\\ =0°\end{array}$

Substitute $110{\text{V}}_{\text{rms}}$ for $\left|F\right|$, $0°$ for $\theta$ and $V_{an}$ for $F$ in equation (1).

$V_{an}=110\angle 0°{\text{V}}_{\text{rms}}$

The voltage $V_{nb}$ is equals to voltage $V_{an}$ since the system is a single phase system.

The voltage $V_{bn}$ using double subscript notation is,

$V_{bn}=-V_{nb}$

Substitute $V_{an}$ for $V_{nb}$ in the above equation.

$V_{bn}=-V_{an}$

Substitute $110\angle 0°{\text{V}}_{\text{rms}}$ for $V_{an}$ in the above equation.

$\begin{array}{c}{V}_{bn}=-110\angle 0°{\text{V}}_{\text{rms}}\\ =110\angle 180°{\text{V}}_{\text{rms}}\end{array}$

Conclusion:

Therefore, the expression of phase to neutral voltage of phase $a$ in phasor notation is $110\angle 0°{\text{V}}_{\text{rms}}$ and of phase $b$ is $110\angle 180°{\text{V}}_{\text{rms}}$.

To determine

The phasor voltage which appears across the impedance $Z_3$.

Answer to Problem 11E

The phasor voltage which appears across the impedance $Z_3$ is $220\angle 0°{\text{V}}_{\text{rms}}$.

Explanation of Solution

Calculation:

The voltage across the impedance $Z_3$ is the sum of the voltages across the impedances $Z_1$ and $Z_2$.

The voltage across the impedance $Z_1$ as per KVL is $V_{an}$.

The voltage across the impedance $Z_2$ as per KVL is $V_{nb}$.

The voltage across the impedance $Z_3$ $V_{Z_3}$ is

$V_{Z_3}=V_{an}+V_{nb}$

Substitute $110\angle 0°{\text{V}}_{\text{rms}}$ for $V_{an}$ and $110\angle 0°{\text{V}}_{\text{rms}}$ for $V_{nb}$ in the above equation.

$\begin{array}{c}{V}_{Z_3}=110\angle 0°{\text{V}}_{\text{rms}}+110\angle 0°{\text{V}}_{\text{rms}}\\ =220\angle 0°{\text{V}}_{\text{rms}}\end{array}$

Conclusion:

Therefore, the phasor voltage which appears across the impedance $Z_3$ is $220\angle 0°{\text{V}}_{\text{rms}}$.

To determine

The average power delivered by the two sources.

Answer to Problem 11E

The average power delivered by source $V_{an}$ is $375.16\text{W}$ and by source $V_{nb}$ is $234.07\text{W}$.

Explanation of Solution

Given data:

The value of the impedance $Z_1$ is $\left(50+j0\right)\Omega$.

The value of the impedance $Z_2$ is $\left(100+j45\right)\Omega$.

The value of the impedance $Z_3$ is $\left(100-j90\right)\Omega$.

Calculation:

The required diagram is shown in Figure 2.

The formula to find current $I_3$ as per KVL in mesh 3 is given by,

$I_3=\frac{V_{Z_3}}{Z_3}$

Substitute $220\angle 0°{\text{V}}_{\text{rms}}$ for $V_{Z_3}$ and $\left(100-j90\right)\Omega$ for $Z_3$ in the above equation.

$\begin{array}{c}{I}_3=\frac{220\angle 0°{\text{V}}_{\text{rms}}}{\left(100-j90\right)\Omega }\\ =\frac{220\angle 0°}{\left(100-j90\right)}\times \frac{100+j90}{100+j90}\text{A}\\ =\frac{22000+j19800}{{100}^2+{90}^2}\text{A}\\ =1.215+j1.093\text{A}\end{array}$

Substitute $1.215\text{A}$ for $x$, $1.093\text{A}$ for $y$ and $I_3$ for $F$ in equation (2).

$\begin{array}{c}\left|I_3\right|=\sqrt{{\left(1.215\text{A}\right)}^2+{\left(1.093\text{A}\right)}^2}\\ =1.63\text{A}\end{array}$

Substitute $1.215\text{A}$ for $x$ and $1.093\text{A}$ for $y$ in equation (3).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1.093\text{A}}{1.215\text{A}}\right)\\ =41.97°\end{array}$

Substitute $1.63\text{A}$ for $\left|F\right|$, $41.97°$ for $\theta$ and $I_3$ for $F$ in equation (1).

$I_3=1.63\angle 41.97°\text{A}$

The formula to find current $I_1$ using KCL and KVL in Figure 2 is given by,

$I_1=\frac{V_{an}}{Z_1}+I_3$

Substitute $\left(50+j0\right)\Omega$ for $Z_1$, $110\angle 0°{\text{V}}_{\text{rms}}$ for $V_{an}$ and $1.63\angle 41.97°\text{A}$ for $I_3$ in the above equation.

$\begin{array}{c}{I}_1=\frac{110\angle 0°{\text{V}}_{\text{rms}}}{\left(50+j0\right)\Omega }+1.63\angle 41.97°\text{A}\\ =\frac{110}{50}+1.215+j1.093\text{A}\\ =3.415+j1.093\text{A}\end{array}$

Substitute $3.415\text{A}$ for $x$, $1.093\text{A}$ for $y$ and $I_1$ for $F$ in equation (2).

$\begin{array}{c}\left|I_1\right|=\sqrt{{\left(3.415\text{A}\right)}^2+{\left(1.093\text{A}\right)}^2}\\ =3.58\text{A}\end{array}$

Substitute $3.415\text{A}$ for $x$ and $1.093\text{A}$ for $y$ in equation (3).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{3.415\text{A}}{1.215\text{A}}\right)\\ =17.7°\end{array}$

Substitute $3.58\text{A}$ for $\left|F\right|$, $17.7°$ for $\theta$ and $I_1$ for $F$ in equation (1).

$I_1=3.58\angle 17.7°\text{A}$

The formula to find current $I_2$ using KVL and KVL in Figure 2 is given by,

$I_2=\frac{V_{nb}}{Z_2}+I_3$

Substitute $\left(100+j45\right)\Omega$ for $Z_2$, $110\angle 0°{\text{V}}_{\text{rms}}$ for $V_{nb}$ and $1.63\angle 41.97°\text{A}$ for $I_3$ in the above equation.

$\begin{array}{c}{I}_2=\frac{110\angle 0°{\text{V}}_{\text{rms}}}{\left(100+j45\right)\Omega }+1.63\angle 41.97°\text{A}\\ =\frac{110\angle 0°}{109.65\angle 24.22°}+1.63\angle 41.97°\text{A}\\ =1.003\angle -24.22°+1.63\angle 41.97°\text{A}\\ =2.233\angle 17.646°\text{A}\end{array}$

The power $P_1$ delivered by the source $V_{an}$ is given by,

$P_1=V_{an}{I}_1\cos \left({\theta }_v-{\theta }_i\right)$

Substitute $110\text{V}$ for $V_{an}$, $3.58\text{A}$ for $I_1$, $0°$ for ${\theta }_v$ and $-17.7°$ for ${\theta }_i$ in the above equation.

$\begin{array}{c}{P}_1=\left(110\text{V}\right)\left(3.58\text{A}\right)\cos \left(0°-\left(-17.7°\right)\right)\\ =375.16\text{W}\end{array}$

The power $P_2$ delivered by the source $V_{nb}$ is given by,

$P_2=V_{nb}{I}_2\cos \left({\theta }_v-{\theta }_i\right)$

Substitute $110\text{V}$ for $V_{nb}$, $2.233\text{A}$ for $I_1$, $0°$ for ${\theta }_v$ and $-17.646°$ for ${\theta }_i$ in the above equation.

$\begin{array}{c}{P}_2=\left(110\text{V}\right)\left(2.233\text{A}\right)\cos \left(0°-\left(-17.646°\right)\right)\\ =234.07\text{W}\end{array}$

Conclusion:

Therefore, the average power delivered by source $V_{an}$ is $375.16\text{W}$ and by source $V_{nb}$ is $234.07\text{W}$.

To determine

The representation of load $Z_3$ for the given condition.

Answer to Problem 11E

The impedance $Z_3$ is the combination of resistance $100\Omega$ and the capacitance $29.5\times {10}^{-6}\text{F}$ in series.

Explanation of Solution

Given data:

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

Calculation:

The value of the impedance $Z_3$ is $\left(100-j90\right)\Omega$.

The real part of $Z_3$ is the resistance.

Hence, the resistance $R$ is $100\Omega$.

The imaginary part of $Z_3$ should be capacitive reactance since series only series capacitive reactance can have negative sign.

Hence, The series capacitive reactance $X_c$ is given by,

$X_c=\frac{-j}{2\pi fC}$

Here,

$f$ is the operating frequency.

$C$ is the capacitance,

Substitute $-j90\Omega$ for $X_c$ and $60\text{Hz}$ for $f$ in the above equation.

$\begin{array}{l}-j90\Omega =\frac{-j}{2\pi \left(60\text{Hz}\right)C}\\ 90=\frac{1}{120\pi C}\\ C=29.5\times {10}^{-6}\text{F}\end{array}$

Conclusion:

Therefore, the impedance $Z_3$ is the combination of resistance $100\Omega$ and the capacitance $29.5\times {10}^{-6}\text{F}$ in series.