Three-phase phasor diagrams

A three-phase generator does not produce one alternating voltage — it produces three, each identical in size and frequency but staggered 120∘120^\circ apart in time. Trying to reason about such a system with sine-wave graphs quickly turns messy: three overlapping waveforms, six sign changes, and a headache. Phasor diagrams cut through that mess. By freezing each sinusoid as a rotating arrow at one instant, the whole three-phase source collapses into three arrows spaced 120∘120^\circ apart, and every relationship you need — line voltage from phase voltage in a star connection, line current from phase current in a delta connection, load current shifted by the impedance angle — becomes a picture you can draw with a ruler and protractor rather than an equation you must re-derive each time. This article builds that picture, shows the one geometric rule (3\sqrt3 magnitude, 30∘30^\circ rotation) that governs both star and delta, and works it through numerically so the pattern sticks.

The core insight

A three-phase phasor diagram represents balanced phase quantities as three RMS-magnitude arrows 120∘120^\circ apart, and subtracting any two of them — the operation behind both star line voltages and delta line currents — always produces a resultant 3\sqrt3 times as large, rotated 30∘30^\circ from the phasor being subtracted from.

The core idea

A phasor diagram plots the phase quantities — voltages or currents — as vectors of fixed RMS magnitude, rotating together at the supply frequency but drawn frozen at one instant so only their relative angles matter. For a balanced source with phase sequence R-Y-B, the three phase-voltage phasors sit 120∘120^\circ apart:

V⃗R=Vph∠0∘,V⃗Y=Vph∠−120∘,V⃗B=Vph∠−240∘\vec V_R = V_{ph}\angle0^\circ,\qquad \vec V_Y = V_{ph}\angle{-120}^\circ,\qquad \vec V_B = V_{ph}\angle{-240}^\circ

where: VphV_{ph} is the RMS phase voltage (the same for all three, by the balanced assumption), and the angles are measured from V⃗R\vec V_R taken as the reference. Because the three phasors are equal in magnitude and evenly spaced, they sum to zero — V⃗R+V⃗Y+V⃗B=0\vec V_R+\vec V_Y+\vec V_B=0 — which is the phasor statement of "no neutral current flows in a balanced star load."

Star connection — line voltage from phase voltage. The voltage between any two line terminals is the phasor difference of the two phase voltages meeting there, e.g. V⃗RY=V⃗R−V⃗Y\vec V_{RY}=\vec V_R-\vec V_Y. Graphically this means reversing V⃗Y\vec V_Y (rotate it 180∘180^\circ) and adding it head-to-tail onto V⃗R\vec V_R; the resultant lies 30∘30^\circ ahead of V⃗R\vec V_R and is 3\sqrt3 times as long. Repeating the construction for V⃗YB=V⃗Y−V⃗B\vec V_{YB}=\vec V_Y-\vec V_B and V⃗BR=V⃗B−V⃗R\vec V_{BR}=\vec V_B-\vec V_R produces a second, larger equilateral triangle of line voltages, rotated 30∘30^\circ ahead of the phase-voltage star.

Figure
Phasor star of three balanced phase voltages 120 degrees apart with the line voltage drawn as the diagonal joining two phase-voltage tips
Phase voltages 120 degrees apart, and one line voltage V_R (0 deg) V_Y (-120 deg) V_B (-240 deg) V_RY joins the tips of V_R and V_Y magnitude root3 times V_ph, leads V_R by 30 degrees

Delta connection — line current from phase current. The same subtraction rule applies to currents in a delta-connected source or load. With phase (winding) currents I⃗RY,I⃗YB,I⃗BR\vec I_{RY},\vec I_{YB},\vec I_{BR} each 120∘120^\circ apart, the current in a line conductor is the phasor difference of the two winding currents meeting at that junction, e.g. I⃗R=I⃗RY−I⃗BR\vec I_R = \vec I_{RY}-\vec I_{BR}. By the identical geometry, each line current has magnitude 3 Iph\sqrt3\,I_{ph} and lags the corresponding phase current by 30∘30^\circ.

Load current phasors. For a balanced load of impedance angle θ\theta (lagging for an inductive load, leading for a capacitive one), each phase current is displaced by θ\theta from its own phase voltage: I⃗R\vec I_R lags or leads V⃗R\vec V_R by θ\theta, and likewise for Y and B, with the 120∘120^\circ mutual spacing between the three currents preserved. This is simply the familiar single-phase R-L-C phasor relationship applied independently, phase by phase. The one rule to memorise covers both cases: subtracting two equal-magnitude phasors that are 120∘120^\circ apart always yields a resultant of magnitude 3\sqrt3 times the original, rotated 30∘30^\circ from the phasor being subtracted from. In star, this rule acts on voltages (phase voltage to line voltage); in delta, it acts on currents (phase current to line current) — the two connections are mirror images of each other in this respect, and mixing them up is the single most common exam slip.

Derivation APhase-to-line transformation: root-3 magnitude, 30-degree rotation

The goal is to show, starting only from the three balanced phase-voltage phasors written in complex (rectangular) form, that any line voltage formed by subtracting two adjacent phase voltages has magnitude exactly 3\sqrt3 times the phase voltage and leads the "from" phasor by exactly 30∘30^\circ — the single fact that generates every star line-voltage and delta line-current relationship in this topic.

Step 1. Write the two phase voltages involved in complex form, taking V⃗R\vec V_R as reference: V⃗R=Vph(cos⁡0∘+jsin⁡0∘)=Vph\vec V_R = V_{ph}(\cos0^\circ + j\sin0^\circ) = V_{ph}, and V⃗Y=Vph(cos⁡(−120∘)+jsin⁡(−120∘))=Vph(−0.5−j0.866)\vec V_Y = V_{ph}(\cos(-120^\circ)+j\sin(-120^\circ)) = V_{ph}(-0.5 - j0.866).

Step 2. Subtract as instructed by the subscript order, V⃗RY=V⃗R−V⃗Y\vec V_{RY}=\vec V_R-\vec V_Y. Substituting, V⃗RY=Vph[1−(−0.5−j0.866)]=Vph(1.5+j0.866)\vec V_{RY} = V_{ph}\big[1-(-0.5-j0.866)\big] = V_{ph}(1.5+j0.866). The real part has grown to 1.5 Vph1.5\,V_{ph} and an imaginary part of 0.866 Vph0.866\,V_{ph} has appeared purely from the geometry of the 120∘120^\circ spacing.

Step 3. Convert this rectangular result to polar form to read off magnitude and angle. The magnitude is ∣V⃗RY∣=Vph1.52+0.8662=Vph2.25+0.75=Vph3|\vec V_{RY}| = V_{ph}\sqrt{1.5^2+0.866^2} = V_{ph}\sqrt{2.25+0.75} = V_{ph}\sqrt3, and the angle is tan⁡−1(0.866/1.5)=30.0∘\tan^{-1}(0.866/1.5) = 30.0^\circ. Both numbers are independent of VphV_{ph} — they depend only on the 120∘120^\circ spacing, which is why the 3\sqrt3-and-30∘30^\circ rule is universal for any balanced three-phase set, not just this particular pair.

Step 4. Generalise by symmetry. The identical algebra applied to V⃗YB=V⃗Y−V⃗B\vec V_{YB}=\vec V_Y-\vec V_B and V⃗BR=V⃗B−V⃗R\vec V_{BR}=\vec V_B-\vec V_R gives magnitude 3 Vph\sqrt3\,V_{ph} in each case, each leading its own "from" phasor by 30∘30^\circ; the same steps applied to delta winding currents (replace VphV_{ph} with IphI_{ph} and note the subtraction direction that defines a line current) give magnitude 3 Iph\sqrt3\,I_{ph}, this time lagging by 30∘30^\circ because the line current is defined as (incoming winding current) minus (outgoing winding current) rather than the other way round. The single complex-number calculation above is therefore the engine behind VL=3 VphV_L=\sqrt3\,V_{ph} in star and IL=3 IphI_L=\sqrt3\,I_{ph} in delta.

Worked example

A balanced star source has Vph=230 VV_{ph}=230\ \text{V} RMS, phase sequence R-Y-B, with V⃗R\vec V_R taken as reference. Find all three line voltages V⃗RY\vec V_{RY}, V⃗YB\vec V_{YB}, V⃗BR\vec V_{BR} using phasor subtraction, and confirm they sum to zero.

Step 1. Write the phase voltages in rectangular form. V⃗R=230∠0∘=230+j0\vec V_R = 230\angle0^\circ = 230+j0, V⃗Y=230∠−120∘=−115−j199.2\vec V_Y = 230\angle{-120}^\circ = -115-j199.2, and V⃗B=230∠−240∘=−115+j199.2\vec V_B = 230\angle{-240}^\circ = -115+j199.2.

Step 2. Compute V⃗RY=V⃗R−V⃗Y=(230−(−115))+j(0−(−199.2))=345+j199.2\vec V_{RY}=\vec V_R-\vec V_Y = (230-(-115))+j(0-(-199.2)) = 345+j199.2, giving ∣V⃗RY∣=3452+199.22=398.4 V|\vec V_{RY}|=\sqrt{345^2+199.2^2}=398.4\ \text{V} at tan⁡−1(199.2/345)=30.0∘\tan^{-1}(199.2/345)=30.0^\circ — exactly the 3×230=398.4 V\sqrt3\times230=398.4\ \text{V} predicted by the derivation above.

Step 3. Compute V⃗YB=V⃗Y−V⃗B=(−115−(−115))+j(−199.2−199.2)=0−j398.4\vec V_{YB}=\vec V_Y-\vec V_B = (-115-(-115))+j(-199.2-199.2) = 0-j398.4, which is 398.4∠−90∘ V398.4\angle{-90}^\circ\ \text{V} — again magnitude 398.4 V398.4\ \text{V}, and 30∘30^\circ ahead of V⃗Y\vec V_Y (at −120∘-120^\circ), since −120∘+30∘=−90∘-120^\circ+30^\circ=-90^\circ.

Step 4. Compute V⃗BR=V⃗B−V⃗R=(−115−230)+j(199.2−0)=−345+j199.2\vec V_{BR}=\vec V_B-\vec V_R = (-115-230)+j(199.2-0) = -345+j199.2, which is 398.4∠150∘ V398.4\angle150^\circ\ \text{V}, again 30∘30^\circ ahead of V⃗B\vec V_B (at −240∘≡120∘-240^\circ\equiv120^\circ), since 120∘+30∘=150∘120^\circ+30^\circ=150^\circ. Adding all three: (345+0−345)+j(199.2−398.4+199.2)=0+j0(345+0-345)+j(199.2-398.4+199.2)=0+j0. The three line voltages, each of magnitude 398.4 V\mathbf{398.4\ V} at 30∘,−90∘,150∘\mathbf{30^\circ, -90^\circ, 150^\circ}, sum to zero — confirming they form a closed equilateral triangle, exactly as three balanced phasors must.

Competitive stretch

For the same source feeding a balanced star-connected inductive load with θ=30∘\theta=30^\circ lagging and Iph=10 AI_{ph}=10\ \text{A}, the load current is I⃗R=10∠−30∘ A\vec I_R = 10\angle{-30}^\circ\ \text{A}. Comparing this to V⃗RY=398.4∠30∘ V\vec V_{RY}=398.4\angle30^\circ\ \text{V} from Step 2, the angle between the line voltage V⃗RY\vec V_{RY} and the phase current I⃗R\vec I_R is 30∘−(−30∘)=60∘=θ+30∘30^\circ-(-30^\circ)=60^\circ=\theta+30^\circ — this (θ+30∘)(\theta+30^\circ) relationship, rather than the plain θ\theta between phase voltage and phase current, is what the two-wattmeter power-measurement method relies on.

Common traps

  • Drawing the line-voltage triangle as a simple "stretched" copy of the phase-voltage star without the 30∘30^\circ rotation — both the 3\sqrt3 scaling and the 30∘30^\circ shift must appear together.
  • Getting the subtraction order backwards: V⃗RY=V⃗R−V⃗Y\vec V_{RY}=\vec V_R-\vec V_Y, not V⃗Y−V⃗R\vec V_Y-\vec V_R; the subscript order fixes which phasor is reversed and added.
  • Assuming the load current always sits 30∘30^\circ from the phase voltage — that 30∘30^\circ belongs to the star/delta geometric transformation; the load's own current-to-phase-voltage angle is the impedance angle θ\theta, which depends entirely on the load.
  • Mixing up which quantity picks up the 3\sqrt3/30∘30^\circ transformation in each connection — voltages transform this way in star (phase voltage to line voltage); currents transform this way in delta (phase current to line current).

Ask the tutor: want to practice constructing the full phasor diagram for a specific lagging or leading power-factor load, step by step?

Why does the line voltage end up root 3 times the phase voltage instead of just twice as much?

Because the two phase voltages being subtracted are not in phase or exactly opposite — they are 120∘120^\circ apart. If they were opposite (180∘180^\circ apart) the difference would indeed be twice as large; if in phase, the difference would be zero. At 120∘120^\circ apart, the vector-triangle geometry gives a resultant of 3≈1.732\sqrt3\approx1.732 times either phasor — a number that falls straight out of the rectangular subtraction shown in the derivation, specifically from 1.52+0.8662\sqrt{1.5^2+0.866^2}.

Does the same root-3, 30-degree rule apply if the phase sequence is reversed, say R-B-Y instead of R-Y-B?

The magnitudes are unchanged — you still get 3 Vph\sqrt3\,V_{ph} and 3 Iph\sqrt3\,I_{ph} — but the sign of every angle flips, since reversing the sequence mirrors the whole phasor star. A line voltage that led its phase voltage by 30∘30^\circ under R-Y-B will instead lag by 30∘30^\circ under R-B-Y. Always redraw the phasor star for the sequence actually specified rather than assuming R-Y-B by default.

Is this root-3 rule specific to voltages, or does it apply to currents too?

It is a pure geometry result about subtracting two equal-magnitude phasors 120∘120^\circ apart, so it applies to any such pair — voltages or currents. In a star connection it happens to relate phase and line voltages, because that is where the subtraction naturally occurs (line voltage is a difference of two phase voltages). In a delta connection the same subtraction occurs between winding currents, so it relates phase and line currents instead. The underlying algebra in the derivation above never assumed "voltage" — it works identically with IphI_{ph} in place of VphV_{ph}.

Before this

Where this leads