Three-Phase Power

Name	Formula	Related Formulae	Variables/Constants
Balanced three-phase voltages	$v_a=V_m\cos \omega t,v_b=V_m\cos \left(\omega t-120^{\circ }\right),v_c=V_m\cos \left(\omega t+120^{\circ }\right)$	${\mathbf{V}}_a+{\mathbf{V}}_b+{\mathbf{V}}_c=0$	$v_a,v_b,v_c:phasevoltages$ $V_m:peakamplitude$ $\omega :angularfrequency,{\mathbf{V}}_a,{\mathbf{V}}_b$ ${\mathbf{V}}_c:phasors$
Wye (star) connection	$V_L=\sqrt{3}{V}_{ph}$	$I_L=I_{ph}$	$V_L:line-to-linevoltage$ $V_{ph}:phase(line-to-neutral)voltage$ $I_L:linecurrent$ $I_{ph}:phasecurrent$
Delta (mesh) connection	$I_L=\sqrt{3}{I}_{ph}$	$V_L=V_{ph}$	$I_L:linecurrent$ $I_{ph}:currentineachdeltabranch$ $V_L:linevoltage$ $V_{ph}:voltageacrosseachbranch$
Three-phase power	$P=\sqrt{3}{V}_L{I}_L\cos \theta =3V_{ph}{I}_{ph}\cos \theta$	$Q=\sqrt{3}{V}_L{I}_L\sin \theta ,S=\sqrt{3}{V}_L{I}_L$	$P:realpower$ $Q:reactivepower$ $S:apparentpower$ $\begin{array}{r}\theta :anglebetweenphasevoltageandphase\\ current\end{array}$ $\cos \theta :powerfactor$
Wye-delta impedance conversion	$Z_Y=\frac{Z_{\Delta }}{3}$	$Z_{\Delta }=3Z_Y$	$\begin{array}{r}{Z}_Y:equivalentwyeimpedanceperphase,\\ Z_{\Delta }:deltaimpedanceperbranch(balanced\\ load)\end{array}$
Neutral current	${\mathbf{I}}_N={\mathbf{I}}_a+{\mathbf{I}}_b+{\mathbf{I}}_c$	${\mathbf{I}}_N=0(\text{balanced load})$	${\mathbf{I}}_N:neutralcurrentphasor,{\mathbf{I}}_a,{\mathbf{I}}_b$ ${\mathbf{I}}_c:linecurrentphasors$

Important Questions

Ques: Why is the line voltage root three times the phase voltage?

Because it is the difference of two phase voltages that are 120 degrees apart. With ${\mathbf{V}}_a$ at 0 degrees and ${\mathbf{V}}_b$ at -120 degrees, the line voltage ${\mathbf{V}}_{ab}={\mathbf{V}}_a-{\mathbf{V}}_b$ has a magnitude of $2V_{ph}\cos 30^{\circ }=\sqrt{3}{V}_{ph}$ and leads ${\mathbf{V}}_a$ by 30 degrees. The common supply pairs confirm it: 120/208 V and 277/480 V in North America and 230/400 V in Europe and India, all close to a ratio of 1.732 (computed). In a delta connection the same geometry applies to currents, so the line current is $\sqrt{3}$ times the branch current. The 30 degree shift matters when transformers with different connections are run in parallel, because their outputs must be in phase. Reference: C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 6th ed. (McGraw-Hill Education, 2017); J. D. Glover, T. J. Overbye and M. S. Sarma, Power System Analysis and Design, 6th ed. (Cengage Learning, 2017).

Ques: Are root three times VL IL cos theta and three times Vph Iph cos theta the same power?

Yes, they are the same quantity written with different voltages and currents. In a wye connection $V_L=\sqrt{3}{V}_{ph}$ and $I_L=I_{ph}$, so $3V_{ph}{I}_{ph}=3(V_L/\sqrt{3})I_L=\sqrt{3}{V}_L{I}_L$; in a delta connection the current carries the $\sqrt{3}$ instead, with the same result. The line form is preferred because line voltage and line current are what is usually measured and quoted. A 400 V, 10 A load at a power factor of 0.85 draws about 5.9 kW (computed). In both forms, $\theta$ is the angle between each phase voltage and its own phase current, and the formula applies to balanced loads. For unbalanced loads, the three per-phase powers are calculated separately and added. Reference: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill, 2012); C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 6th ed. (McGraw-Hill Education, 2017).

Ques: Why is three-phase power used instead of single-phase?

For three reasons. First, the total instantaneous power of a balanced three-phase system is constant, because the cosine-squared terms of the three phases add to a constant, whereas single-phase power pulses at twice the line frequency; constant power means smooth torque in motors and generators. Second, three-phase lines carry the same power with less conductor: for equal line voltage and losses, three wires of half the cross-section replace two full-size ones, about 75 per cent of the copper (computed). Third, three-phase currents create a rotating magnetic field directly, which makes induction motors self-starting. Balanced loads also need no neutral current. Reference: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill, 2012); J. D. Glover, T. J. Overbye and M. S. Sarma, Power System Analysis and Design, 6th ed. (Cengage Learning, 2017).

Ques: How is a delta-connected load analysed?

By converting it to an equivalent wye and solving one phase. For a balanced load $Z_Y=Z_{\Delta }/3$, the phase voltage is $V_L/\sqrt{3}$, and the per-phase power is one third of the total. A 30 $\Omega$ delta load is equivalent to a 10 $\Omega$ wye load; at 400 V line-to-line each phase sees 231 V and draws 23.1 A, 16 kW in total (computed). Inside the delta, each 30 $\Omega$ branch sees the full 400 V and carries 13.3 A, and $\sqrt{3}$ times that is 23.1 A, matching the line current (computed). The per-phase method applies only to balanced systems; unbalanced loads are solved phase by phase or with symmetrical components. Reference: J. D. Glover, T. J. Overbye and M. S. Sarma, Power System Analysis and Design, 6th ed. (Cengage Learning, 2017); C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 6th ed. (McGraw-Hill Education, 2017).

References

1. C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 6th ed. New York: McGraw-Hill Education, 2017.
2. J. D. Glover, T. J. Overbye and M. S. Sarma, Power System Analysis and Design, 6th ed. Boston: Cengage Learning, 2017.
3. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. New York: McGraw-Hill, 2012.