Chapter 12, Problem 51P

Consider the wye-delta system shown in Fig. 12.60. Let , , and . Determine the phase currents, I_AB, I_BC, and I_CA, and the line currents, I_aA, I_bB, and I_cC.

To determine

Find the phase and line currents for the $\text{Y-}\Delta$ system in Figure 12.60.

Answer to Problem 51P

The phase currents $I_{AB}$, $I_{BC}$ and $I_{CA}$ are $2.078\angle 120°\text{A}$, $2.078\angle 90°\text{A}$, and $2.078\angle 150°\text{A}$ respectively.

The line currents $I_{aA}$, $I_{bB}$, and $I_{cC}$ are $1.0759\angle 45°\text{A}$, $\text{1}\text{.0759}\angle \text{15}°\text{A}$, and $\text{2}\text{.078}\angle -\text{150}°\text{A}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 12.60 for the $\text{Y-}\Delta$ system,

In the Y-connected source:

The line to neutral voltage $\left(V_{an}\right)$ is $120\angle 90°\text{V}$.

The line to neutral voltage $\left(V_{bn}\right)$ is $120\angle -30°\text{V}$.

The line to neutral voltage $\left(V_{cn}\right)$ is $120\angle -150°\text{V}$.

In the delta-connected unbalanced load:

The impedance $\left(Z_1\right)$ is $100\text{Ω}$.

The impedance $\left(Z_2\right)$ is $j100\text{Ω}$.

The impedance $\left(Z_3\right)$ is $-j100\text{Ω}$.

Formula used:

Write the expression to find the line to line voltage $V_{AB}$ at the load.

$V_{AB}=V_{an}-V_{bn}$ (1)

Here,

$V_{an}$ and $V_{bn}$ are the line to neutral voltages.

Write the expression to find the line to line voltage $V_{BC}$ at the load.

$V_{BC}=V_{bn}-V_{cn}$ (2)

Here,

$V_{bn}$ and $V_{cn}$ are the line to neutral voltages.

Write the expression to find the line to line voltage $V_{CA}$ at the load.

$V_{CA}=V_{cn}-V_{an}$ (3)

Here,

$V_{cn}$ is the line to neutral voltage.

Write the expression to find the line to line current $I_{AB}$.

$I_{AB}=\frac{V_{AB}}{Z_1}$ (4)

Here,

$V_{AB}$ is the line to line voltage, and

$Z_1$ is the load impedance of the A-B phase.

Write the expression to find the line to line current $I_{BC}$.

$I_{BC}=\frac{V_{BC}}{Z_3}$ (5)

Here,

$V_{BC}$ is the line to line voltage, and

$Z_3$ is the load impedance of the B-C phase.

Write the expression to find the line to line current $I_{CA}$.

$I_{CA}=\frac{V_{CA}}{Z_2}$ (6)

Here,

$V_{CA}$ is the line to line voltage, and

$Z_2$ is the load impedance of the C-A phase.

Write the expression to find the line current $I_{aA}$ for the $\text{Y-}\Delta$ system.

$I_{aA}=I_{AB}-I_{CA}$ (7)

Here,

$I_{AB}$ and $I_{CA}$ are the line to line (phase) currents.

Write the expression to find the line current $I_{bB}$ for the $\text{Y-}\Delta$ system.

$I_{bB}=I_{BC}-I_{AB}$ (8)

Here,

$I_{BC}$ is the line to line (phase) current.

Write the expression to find the line current $I_{cC}$ for the $\text{Y-}\Delta$ system.

$I_{cC}=I_{CA}-I_{BC}$ (9)

Calculation:

Re-draw the given Figure 12.60 as follows.

Substitute $120\angle 90°\text{V}$ for $V_{an}$, and $120\angle -30°\text{V}$ for $V_{bn}$ in equation (1) to find $V_{AB}$.

$\begin{array}{c}{V}_{AB}=\left(120\angle 90°\text{V}\right)-\left(120\angle -30°\text{V}\right)\\ =\left(j120\text{V}\right)-\left(103.923-j60\text{V}\right)\\ =-103.923+j180\text{V}\\ V_{AB}=207.846\angle 120°\text{V}\end{array}$

Substitute $120\angle -30°\text{V}$ for $V_{bn}$ and $120\angle -150°\text{V}$ for $V_{cn}$ in equation (2) to find $V_{BC}$.

$\begin{array}{c}{V}_{BC}=\left(120\angle -30°\text{V}\right)-\left(120\angle -150°\text{V}\right)\\ =\left(103.923-j60\text{V}\right)-\left(-103.923-j60\text{V}\right)\end{array}$

$V_{BC}=207.846\text{V}$

Substitute $120\angle -150°\text{V}$ for $V_{cn}$ and $120\angle 90°\text{V}$ for $V_{an}$ in equation (3) to find $V_{CA}$.

$\begin{array}{c}{V}_{CA}=\left(120\angle -150°\text{V}\right)-\left(120\angle 90°\text{V}\right)\\ =\left(-103.923-60j\text{V}\right)-\left(120j\text{V}\right)\\ =-103.923-180j\text{V}\\ =207.846\angle -119.9°\text{V}\end{array}$

$V_{CA}\cong 207.846\angle -120°\text{V}$

Substitute $207.846\angle 120°\text{V}$ for $V_{AB}$, and $100\text{Ω}$ for $Z_1$ in equation (4) to find $I_{AB}$.

$\begin{array}{c}{I}_{AB}=\frac{207.846\angle 120°\text{V}}{100\text{Ω}}\\ =2.07846\angle 120°\text{A}\left\{\because 1\frac{\text{V}}{\text{Ω}}=1\text{A}\right\}\\ I_{AB}\cong 2.078\angle 120°\text{A}\end{array}$

Substitute $207.846\text{V}$ for $V_{BC}$, and $-j100\text{Ω}$ for $Z_3$ in equation (5) to find $I_{BC}$.

$\begin{array}{c}{I}_{BC}=\frac{207.846\text{V}}{-j100\text{Ω}}\\ =\frac{207.846\text{V}}{100\angle -90°\text{Ω}}\left\{\because 1\frac{\text{V}}{\text{Ω}}=1\text{A}\right\}\\ =2.07846\angle 90°\text{A}\\ I_{BC}\cong 2.078\angle 90°\text{A}\end{array}$

Substitute $207.846\angle -120°\text{V}$ for $V_{CA}$, and $j100\text{Ω}$ for $Z_2$ in equation (6) to find $I_{CA}$.

$\begin{array}{c}{I}_{CA}=\frac{207.846\angle -120°\text{V}}{j100\text{Ω}}\\ =\frac{207.846\angle -120°\text{V}}{100\angle 90°\text{Ω}}\\ =2.07846\left(\angle -120°-\left(90°\right)\right)\text{A}\left\{\because 1\frac{\text{V}}{\text{Ω}}=1\text{A}\right\}\\ I_{CA}=2.078\angle -210°\text{A}\end{array}$

Or

$I_{CA}\cong 2.078\angle 150°\text{A}$

Thus, the phase currents are,

$I_{AB}=2.078\angle 120°\text{A}$

$I_{BC}=2.078\angle 90°\text{A}$

$I_{CA}=2.078\angle 150°\text{A}$

Substitute $2.07846\angle 120°\text{A}$ for $I_{AB}$, and $2.07846\angle 150°\text{A}$ for $I_{CA}$ in equation (7) to find line current $I_{aA}$.

$\begin{array}{c}{I}_{aA}=\left(2.07846\angle 120°\text{A}\right)-\left(2.07846\angle 150°\text{A}\right)\\ =\left(-1.03923+j1.79999\text{A}\right)-\left(-1.79999+j1.03923\text{A}\right)\\ =0.76076+j0.76076\text{A}\\ I_{aA}=1.0759\angle 45°\text{A}\end{array}$

Substitute $2.07846\angle 90°\text{A}$ for $I_{BC}$, and $2.07846\angle 120°\text{A}$ for $I_{AB}$ in equation (8) to find $I_{bB}$.

$\begin{array}{c}{I}_{bB}=\left(2.07846\angle 90°\text{A}\right)-\left(2.07846\angle 120°\text{A}\right)\\ =\left(2.07846j\text{A}\right)-\left(-1.03923+j1.79999\text{A}\right)\\ =1.03923+j0.27847\text{A}\\ =\text{1}\text{.07589}\angle \text{15}°\text{A}\end{array}$

$I_{bB}\cong \text{1}\text{.0759}\angle \text{15}°\text{A}$

Substitute $2.07846\angle 150°\text{A}$ for $I_{CA}$, and $2.07846\angle 90°\text{A}$ for $I_{BC}$ in equation (9) to find $I_{cC}$.

$\begin{array}{c}{I}_{cC}=\left(2.07846\angle 150°\text{A}\right)-\left(2.07846\angle 90°\text{A}\right)\\ =\left(-1.79999+j1.03923\text{A}\right)-\left(j2.07846\text{A}\right)\\ =-1.79999-j1.03923\text{A}\\ =\text{2}\text{.07845}\angle -\text{150}°\text{A}\end{array}$

$I_{cC}\cong \text{2}\text{.078}\angle -\text{150}°\text{A}$

Thus, the line currents are,

$I_{aA}=1.0759\angle 45°\text{A}$

$I_{bB}\cong \text{1}\text{.0759}\angle \text{15}°\text{A}$

$I_{cC}\cong \text{2}\text{.078}\angle -\text{150}°\text{A}$

PSpice simulation:

Assume the angular frequency $\left(\omega \right)$ as $1\frac{\text{rad}}{\text{s}}$.

Write the expression to find the capacitive reactance.

$X_C=\frac{1}{j\omega C}$

Substitute $-j100\Omega$ for $X_C$ and $1\frac{\text{rad}}{\text{s}}$ for $\omega$ to find the capacitance value in above equation.

$\begin{array}{l}-j100\Omega =\frac{1}{j\left(1\frac{\text{rad}}{\text{s}}\right)C}\\ C=\frac{1}{100}\frac{\text{s}}{\text{Ω}}\\ C=0.01\text{F }\left\{\because \frac{\text{s}}{\text{Ω}}=\text{F}\right\}\end{array}$

Write the expression to find the inductive reactance.

$X_L=j\omega L$

Substitute $j100\Omega$ for $X_L$ and $1\frac{\text{rad}}{\text{s}}$ for $\omega$ to find the inductance value in above equation.

$\begin{array}{l}j100\Omega =j\left(1\frac{\text{rad}}{\text{s}}\right)L\\ L=100\text{Ω}\cdot \text{s}\\ L=100\text{H }\left\{\because \text{Ω}\cdot \text{s}=\text{H}\right\}\end{array}$

Draw the circuit as shown in Figure 2 in PSpice, connect the $1\text{ m}\Omega$ resistor in series with the inductor just to avoid the simulation errors.

Provide the Simulation settings as in Figure 2.

Now, run the simulation and check the output file in newly opened PSpice A/D window.

From the PSpice results the phase currents are,

FREQ         IM(V_IAB)    IP(V_IAB)   

   1.5916E-01   2.0785E+00   1.2000E+02

FREQ         IM(V_IBC)    IP(V_IBC)   

   1.5916E-01   2.0785E+00   9.0000E+01

  FREQ         IM(V_ICA)    IP(V_ICA)   

   1.5916E-01   2.0785E+00   1.5000E+02

The phase current $I_{AB}$.

$\begin{array}{l}{I}_{AB}=2.0785\angle 120°\text{A}\\ I_{AB}\cong 2.078\angle 120°\text{A}\end{array}$

The phase current $I_{BC}$.

$\begin{array}{l}{I}_{BC}=2.0785\angle 90°\text{ A}\\ I_{BC}\cong 2.078\angle 90°\text{ A}\end{array}$

The phase current $I_{CA}$.

$\begin{array}{l}{I}_{CA}=2.0785\angle 150°\text{ A}\\ I_{CA}\cong 2.078\angle 150°\text{ A}\end{array}$

From the PSpice results the line currents are,

  FREQ         IM(V_IaA)    IP(V_IaA)   

   1.5916E-01   1.0759E+00   4.5000E+01

  FREQ         IM(V_IbB)    IP(V_IbB)   

   1.5916E-01   1.0759E+00   1.5000E+01

  FREQ         IM(V_IcC)    IP(V_IcC)   

   1.5916E-01   2.0785E+00  -1.5000E+02

The line current $I_{aA}$.

$I_{aA}=1.0759\angle 45°\text{A}$

The line current $I_{bB}$.

$I_{bB}=1.0759\angle 15°\text{A}$

The line current $I_{cC}$.

$\begin{array}{l}{I}_{cC}=2.0785\angle -150°\text{ A}\\ I_{cC}\cong 2.078\angle -150°\text{ A}\end{array}$

Conclusion:

Thus,

The phase currents $I_{AB}$, $I_{BC}$ and $I_{CA}$ are $2.078\angle 120°\text{A}$, $2.078\angle 90°\text{A}$, and $2.078\angle 150°\text{A}$ respectively.

The line currents $I_{aA}$, $I_{bB}$, and $I_{cC}$ are $1.0759\angle 45°\text{A}$, $\text{1}\text{.0759}\angle \text{15}°\text{A}$, and $\text{2}\text{.078}\angle -\text{150}°\text{A}$ respectively.