The phase sequence of the Y-connected generator in Fig. 24.13 is ABC.
a. Find the phase angles $θ_2 and θ_3$ .
b. Find the magnitude of the line voltages.
c. Find the line currents.
d. Verify that, since the load is balanced, $I_N = 0$ .

Question

The phase sequence of the Y-connected generator in Fig. 24.13 is ABC.
a. Find the phase angles [latex]θ_2 and θ_3[/latex].
b. Find the magnitude of the line voltages.
c. Find the line currents.
d. Verify that, since the load is balanced, [latex]I_N = 0[/latex].

Accepted Answer

a. For an ABC phase sequence,

[latex]\theta_2 = −120° \qquad and\qquad \theta_3 = +120°[/latex]

b. [latex]E_L=\sqrt{3}E_\phi =(1.73)(120 V) = 208 V. \enspace Therefore,\[0.5cm] E_{AB} = E_{BC} = E_{CA} = 208 V[/latex]

c. [latex]V_\phi = E_\phi. \enspace Therefore,\[0.5cm] V_{an} = E_{AN}\qquad V_{bn} = E_{BN}\qquad V_{cn} = E_{CN}\[0.5cm] I_{\phi L}=I_{an}=\frac{V_{an}}{Z_{an}}=\frac{120 V ∠0°}{3 Ω + j 4 Ω}=\frac{120 V ∠0°}{5 Ω ∠53.13°}=24 A ∠-53.13°\[0.5cm] I_{bn}=\frac{V_{bn}}{Z_{bn}}=\frac{120 V ∠-120°}{5 Ω ∠53.13°}=24 A ∠-173.13° \[0.5cm] I_{cn}=\frac{V_{cn}}{Z_{cn}} =\frac{120 V ∠+120°}{5 Ω ∠53.13° } =24 A ∠66.87° [/latex]

and, since [latex]I_L = I_{\phi L}[/latex],

[latex]I_{Aa} = I_{an} = 24 A \angle −53.13°\[0.5cm] I_{Bb} = I_{bn} = 24 A \angle −173.13°\[0.5cm] I_{Cc} = I_{cn} = 24 A \angle 66.87°[/latex]

d. Applying Kirchhoff’s current law, we have

[latex]I_N = I_{Aa} + I_{Bb} + I_{Cc}[/latex]

In rectangular form,

[latex]I_{Aa} = 24 A ∠-53.13° = 14.40 A - j 19.20 A\[0.5cm] I_{Bb} = 24 A ∠-173.13° = -22.83 A - j 2.87 A\[0.5cm] I_{Cc} = 24 A ∠66.87° =\underline{9.43 A + j 22.07 A}\[0.5cm]\sum{ (I_{Aa} + I_{Bb} + I_{Cc})} =0 + j 0[/latex]

and [latex]I_N[/latex] is in fact equals to zero, as required for a balanced load.

Question 24.1: The phase sequence of the Y-connected generator in Fig. 24.1...

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