Chapter 12, Problem 17E

Assume the system shown in Fig. 12.34 is balanced, R_w = 0, , and a positive phase sequence applies. Calculate all phase and line currents, and all phase and line voltages, if Z_p is equal to (a) 1 kΩ; (b) 100 + j48 Ω; (c) 100 − j48 Ω.

To determine

All the line and phase currents and all the line and phase voltages for the given load impedance.

Answer to Problem 17E

The line and phase currents in sequence are $0.208\angle 0°\text{A}$, $0.208\angle -120°\text{A}$, $0.208\angle -240°\text{A}$, the phase voltages in sequence are $208\angle 0°\text{V}$, $208\angle -120°\text{V}$, $208\angle -240°\text{V}$ and the line voltages in sequence are $360.26\angle 30°\text{V}$, $360.26\angle -90°\text{V}$ and $360.26\angle -210°\text{V}$.

Explanation of Solution

Given data:

The resistance of the wire $R_w$ is $0\Omega$.

The phase to neutral voltage $V_{an}$ for phase $a$ is $208\angle 0°\text{V}$.

The load impedance $Z_P$ is $1\text{k}\Omega$.

Calculation:

The system follows positive sequence hence the phase to neutral voltages will be,

For phase $b$ $V_{bn}$ is,

$V_{bn}=208\angle -120°\text{V}$

For phase $c$ $V_{cn}$ is,

$V_{cn}=208\angle -240°\text{V}$

The required diagram is shown in Figure 1.

The line voltage between phase $a$ and phase $b$ $V_{ab}$ is given by,

$V_{ab}=\sqrt{3}{V}_{an}\angle 30°$

Substitute $208\angle 0°\text{V}$ for $V_{an}$ in the above equation.

$\begin{array}{c}{V}_{ab}=\sqrt{3}\left(208\angle 0°\text{V}\right)\angle 30°\\ =360.26\angle 30°\text{V}\end{array}$

The system follows positive sequence hence the line to line voltages will be,

$\begin{array}{l}{V}_{bc}=360.26\angle -90°\text{V}\\ V_{ca}=360.26\angle -210°\text{V}\end{array}$

All the phase and line voltages are independent of the impedance value and hence these values will remain same for all the values of load impedance $Z_P$.

The conversion from $\text{k}\Omega$ into $\Omega$ is given by,

$\text{1}\text{k}\Omega =1000\Omega$

Hence, the load impedance $Z_P$ is,

$Z_P=1000\Omega$

Since the given system is a star connected system, the phase currents will be equal to the line currents.

The phase $a$ current $I_{AN}$ is,

$I_{AN}=I_{aA}$

Here,

$I_{aA}$ is the line current.

The phase $a$ current $I_{AN}$ is given by,

$I_{AN}=\frac{V_{an}}{Z_P}$

Substitute $208\angle 0°\text{V}$ for $V_{an}$ and $1000\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{AN}=\frac{208\angle 0°\text{V}}{1000\Omega }\\ =0.208\angle 0°\text{A}\end{array}$

The phase $b$ current $I_{BN}$ is,

$I_{BN}=I_{bB}$

Here,

$I_{aA}$ is the line current.

The phase $b$ current $I_{BN}$ is given by,

$I_{BN}=\frac{V_{bn}}{Z_P}$

Substitute $208\angle -120°\text{V}$ for $V_{bn}$ and $1000\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{BN}=\frac{208\angle -120°\text{V}}{1000\Omega }\\ =0.208\angle -120°\text{A}\end{array}$

The phase $c$ current $I_{CN}$ is,

$I_{CN}=I_{cC}$

Here,

$I_{cC}$ is the line current.

The phase $b$ current $I_{CN}$ is given by,

$I_{CN}=\frac{V_{cn}}{Z_P}$

Substitute $208\angle -240°\text{V}$ for $V_{bn}$ and $1000\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{BN}=\frac{208\angle -240°\text{V}}{1000\Omega }\\ =0.208\angle -240°\text{A}\end{array}$

Conclusion:

Therefore, the line and phase currents in sequence are $0.208\angle 0°\text{A}$, $0.208\angle -120°\text{A}$, $0.208\angle -240°\text{A}$, the phase voltages in sequence are $208\angle 0°\text{V}$, $208\angle -120°\text{V}$, $208\angle -240°\text{V}$ and the line voltages in sequence are $360.26\angle 30°\text{V}$, $360.26\angle -90°\text{V}$ and $360.26\angle -210°\text{V}$.

To determine

All the line and phase currents and all the line and phase voltages for the given load impedance.

Answer to Problem 17E

The line and phase currents in sequence are $1.87\angle -25.64°\text{A}$, $1.87\angle -145.64°\text{A}$, $1.87\angle -265.64°\text{A}$, the phase voltages in sequence are $208\angle 0°\text{V}$, $208\angle -120°\text{V}$, $208\angle -240°\text{V}$ and the line voltages in sequence are $360.26\angle 30°\text{V}$, $360.26\angle -90°\text{V}$ and $360.26\angle -210°\text{V}$.

Explanation of Solution

Given data:

The load impedance $Z_P$ is $\left(100+j48\right)\Omega$.

Calculation:

All the phase and line voltages are independent of the impedance value and hence these values will remain same for all the values of load impedance $Z_P$.

The phase $a$ current $I_{AN}$ is,

$I_{AN}=I_{aA}$

Here,

$I_{aA}$ is the line current.

The phase $a$ current $I_{AN}$ is given by,

$I_{AN}=\frac{V_{an}}{Z_P}$

Substitute $208\angle 0°\text{V}$ for $V_{an}$ and $\left(100+j48\right)\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{AN}=\frac{208\angle 0°\text{V}}{\left(100+j48\right)\Omega }\\ =\frac{208\angle 0°}{\left(100+j48\right)}\times \left(\frac{100-j48}{100-j48}\right)\text{A}\\ =\frac{208\times \left(100-j48\right)}{{100}^2+{48}^2}\text{A}\\ =1.87\angle -25.64°\text{A}\end{array}$

The phase $b$ current $I_{BN}$ is,

$I_{BN}=I_{bB}$

Here,

$I_{bB}$ is the line current.

The phase $b$ current $I_{BN}$ is given by,

$I_{BN}=\frac{V_{bn}}{Z_P}$

Substitute $208\angle -120°\text{V}$ for $V_{bn}$ and $\left(100+j48\right)\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{BN}=\frac{208\angle -120°\text{V}}{\left(100+j48\right)\Omega }\\ =\frac{208\angle -120°}{\left(100+j48\right)}\times \left(\frac{100-j48}{100-j48}\right)\text{A}\\ =\frac{208\angle -120°\times \left(100-j48\right)}{{100}^2+{48}^2}\text{A}\\ =1.87\angle -145.64°\text{A}\end{array}$

The phase $c$ current $I_{CN}$ is,

$I_{CN}=I_{cC}$

Here,

$I_{cC}$ is the line current.

The phase $c$ current $I_{CN}$ is given by,

$I_{CN}=\frac{V_{cn}}{Z_P}$

Substitute $208\angle -240°\text{V}$ for $V_{cn}$ and $\left(100+j48\right)\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{CN}=\frac{208\angle -240°\text{V}}{\left(100+j48\right)\Omega }\\ =\frac{208\angle -240°}{\left(100+j48\right)}\times \left(\frac{100-j48}{100-j48}\right)\text{A}\\ =\frac{208\angle -240°\times \left(100-j48\right)}{{100}^2+{48}^2}\text{A}\\ =1.87\angle -265.64°\text{A}\end{array}$

Conclusion:

Therefore, the line and phase currents in sequence are $1.87\angle -25.64°\text{A}$, $1.87\angle -145.64°\text{A}$, $1.87\angle -265.64°\text{A}$, the phase voltages in sequence are $208\angle 0°\text{V}$, $208\angle -120°\text{V}$, $208\angle -240°\text{V}$ and the line voltages in sequence are $360.26\angle 30°\text{V}$, $360.26\angle -90°\text{V}$ and $360.26\angle -210°\text{V}$.

To determine

All the line and phase currents and all the line and phase voltages for the given load impedance.

Answer to Problem 17E

The line and phase currents in sequence are $1.87\angle 25.64°\text{A}$, $1.87\angle -94.36°\text{A}$, $1.87\angle -214.36°\text{A}$, the phase voltages in sequence are $208\angle 0°\text{V}$, $208\angle -120°\text{V}$, $208\angle -240°\text{V}$ and the line voltages in sequence are $360.26\angle 30°\text{V}$, $360.26\angle -90°\text{V}$ and $360.26\angle -210°\text{V}$.

Explanation of Solution

Given data:

The load impedance $Z_P$ is $\left(100-j48\right)\Omega$.

Calculation:

All the phase and line voltages are independent of the impedance value and hence these values will remain same for all the values of load impedance $Z_P$.

The phase $a$ current $I_{AN}$ is,

$I_{AN}=I_{aA}$

Here,

$I_{aA}$ is the line current.

The phase $a$ current $I_{AN}$ is given by,

$I_{AN}=\frac{V_{an}}{Z_P}$

Substitute $208\angle 0°\text{V}$ for $V_{an}$ and $\left(100-j48\right)\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{AN}=\frac{208\angle 0°\text{V}}{\left(100-j48\right)\Omega }\\ =\frac{208\angle 0°}{\left(100-j48\right)}\times \left(\frac{100+j48}{100+j48}\right)\text{A}\\ =\frac{208\times \left(100+j48\right)}{{100}^2+{48}^2}\text{A}\\ =1.87\angle 25.64°\text{A}\end{array}$

The phase $b$ current $I_{BN}$ is,

$I_{BN}=I_{bB}$

Here,

$I_{bB}$ is the line current.

The phase $b$ current $I_{BN}$ is given by,

$I_{BN}=\frac{V_{bn}}{Z_P}$

Substitute $208\angle -120°\text{V}$ for $V_{bn}$ and $\left(100-j48\right)\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{BN}=\frac{208\angle -120°\text{V}}{\left(100-j48\right)\Omega }\\ =\frac{208\angle -120°}{\left(100-j48\right)}\times \left(\frac{100+j48}{100+j48}\right)\text{A}\\ =\frac{208\angle -120°\times \left(100+j48\right)}{{100}^2+{48}^2}\text{A}\\ =1.87\angle -94.36°\text{A}\end{array}$

The phase $c$ current $I_{CN}$ is,

$I_{CN}=I_{cC}$

Here,

$I_{cC}$ is the line current.

The phase $c$ current $I_{CN}$ is given by,

$I_{CN}=\frac{V_{cn}}{Z_P}$

Substitute $208\angle -240°\text{V}$ for $V_{cn}$ and $\left(100-j48\right)\Omega$ for $Z_P$ in the above equation.

$\begin{array}{c}{I}_{BN}=\frac{208\angle -240°\text{V}}{\left(100-j48\right)\Omega }\\ =\frac{208\angle -240°}{\left(100-j48\right)}\times \left(\frac{100+j48}{100+j48}\right)\text{A}\\ =\frac{208\angle -240°\times \left(100+j48\right)}{{100}^2+{48}^2}\text{A}\\ =1.87\angle -214.36°\text{A}\end{array}$

Conclusion:

Therefore, the line and phase currents in sequence are $1.87\angle 25.64°\text{A}$, $1.87\angle -94.36°\text{A}$, $1.87\angle -214.36°\text{A}$, the phase voltages in sequence are $208\angle 0°\text{V}$, $208\angle -120°\text{V}$, $208\angle -240°\text{V}$ and the line voltages in sequence are $360.26\angle 30°\text{V}$, $360.26\angle -90°\text{V}$ and $360.26\angle -210°\text{V}$.