Wye (Y) Phasor Calculations — Nigeria Case Study (230/400 V)

This post converts the WYE diagram in your screenshot into clear, accurate phasor math using **Nigeria standard voltages**. You’ll learn: how to compute line-to-line magnitude from phase voltages, perform phasor subtraction (digit-by-digit), compute currents for balanced Y loads, and calculate kVA/kW. A 40-question timed CBT quiz is included.

Quick orientation — voltage systems

Two common low/medium voltage three-phase standards you may encounter:

US-style 120/208 V Wye: Phase (line-to-neutral) ≈ 120 V, Line-to-line ≈ 208 V (because 120 × √3 ≈ 208).
Nigeria / many countries: 230/400 V Wye: Phase = 230 V (line-to-neutral), Line-to-line = 230 × √3 ≈ 399–400 V (commonly called 400 V).

In your screenshot the sketch used a 120/208 framework. Below we rework the same geometry using 230 V phase and show the math step-by-step so you can reproduce for any nominal voltage.

Phasor basics — short reminder

For balanced three-phase Wye:

Phase voltages (to neutral) are 120° apart as phasors: V_A = V_ph ∠0°, V_B = V_ph ∠−120°, V_C = V_ph ∠+120°.
Line-to-line voltage is the difference between two phase phasors: e.g. V_AB = V_A − V_B.
Magnitude |V_AB| = √3 × V_ph (for balanced system). We’ll show this with actual complex arithmetic too.

Case study (Nigeria): numbers we use

Line-to-neutral (phase) voltage: V_ph = 230.00 V
Line-to-line nominal: V_LL ≈ √3 × 230 = ? → compute step-by-step below
We'll use √3 ≈ 1.732 (carry enough digits for accuracy).

Step 1 — compute nominal line-to-line magnitude

Formula: V_LL = √3 × V_ph

Show the digit-by-digit arithmetic

√3 ≈ 1.732
V_ph = 230.00 V

Multiply:
1.732 × 230.00

Step by step:
1.732 × 200 = 346.4
1.732 × 30  = 51.96
Add: 346.4 + 51.96 = 398.36 V

Rounding to typical nominal: ≈ 398.4 V → industry rounds to ≈ 400 V

So V_LL ≈ 398.36 V (often called 400 V nominal)

Conclusion:

For Nigeria: V_ph = 230 V → V_LL ≈ 398.36 V (≈400 V).

Step 2 — phasor check using complex subtraction (digit-by-digit)

We will compute V_AB = V_A − V_B using complex numbers to show the √3 factor emerges.

Define phasors (rectangular form):

V_A = 230 ∠ 0°  = 230 × (cos 0° + j sin 0°) = 230 + j0
V_B = 230 ∠ −120° = 230 × (cos(−120°) + j sin(−120°))

We need cos(−120°) and sin(−120°):
cos(−120°) = cos(120°) = −0.5
sin(−120°) = −sin(120°) = −0.86602540378 (≈ −0.8660)

Compute V_B components step-by-step

V_B (real part) = 230 × (−0.5) = −115.000
V_B (imag part) = 230 × (−0.86602540378) ≈ −199.38584287 ≈ −199.3858

So:
V_B ≈ −115.000  − j199.3858
V_A = 230.000 + j0.0000

Now subtract: V_AB = V_A − V_B

Show subtraction digit-by-digit

V_AB = (230.000 + j0.0000) − (−115.000 − j199.3858)

Real part: 230.000 − (−115.000) = 230.000 + 115.000 = 345.000
Imag part: 0.000 − (−199.3858) = +199.3858

So V_AB (rectangular) ≈ 345.000 + j199.3858

Now magnitude |V_AB| = sqrt(345.000^2 + 199.3858^2)

Compute squares:
345.000^2 = 119,025.000
199.3858^2 ≈ 39,754.476  (199.3858 × 199.3858 ≈ 39,754.476)

Sum = 119,025.000 + 39,754.476 = 158,779.476

Square root:
√158,779.476 ≈ 398.471... V

Which matches our earlier multiplication result (rounding differences). We reported V_LL ≈ 398.36 by multiplication; here precise phasor math ≈ 398.47 V.

Takeaway

Phasor subtraction reproduces the √3 relationship. In practice we use the simpler rule: |V_LL| = √3 × V_ph → for Nigeria: 1.732 × 230 ≈ 398–400 V.

Step 3 — Balanced Y-connected resistive load example (currents & power)

Suppose a balanced Y load has phase impedance Z_ph = 10.0 Ω (purely resistive) on each phase. Compute phase currents, line currents and total power.

Compute phase current (I_ph)

I_ph = V_ph / Z_ph = 230.00 V / 10.0 Ω = 23.000 A
(digit-by-digit: 230 ÷ 10 = 23.0)

For a Y connection, line currents equal phase currents (I_line = I_ph). So:

I_line = 23.000 A

Compute per-phase real power (P_phase)

P_phase = V_ph × I_ph × cosφ

For resistive load, cosφ = 1.

P_phase = 230.00 × 23.000 = 5,290.00 W
(230 × 23 = 230 × (20 + 3) = 4,600 + 690 = 5,290)

Total real power (P_total)

P_total = 3 × P_phase = 3 × 5,290.00 = 15,870.00 W ≈ 15.87 kW

Total apparent power (S_total) using line values

Use S_total = √3 × V_LL × I_line (kVA in VA)

We will compute with V_LL ≈ 398.36 V and I_line = 23.000 A:

First compute √3 × V_LL × I_line exactly:
√3 ≈ 1.732
V_LL ≈ 398.36 V
I_line = 23.000 A

Step 1: √3 × V_LL = 1.732 × 398.36 ≈ (1.732 × 400) − (1.732 × 1.64)
         1.732 × 400 = 692.8
         1.732 × 1.64 ≈ 2.841
         difference ≈ 692.8 − 2.841 ≈ 689.959 (but simpler: we can use 1.732×398.36 ≈ 689.96)

Step 2: Multiply by current:
S_total (VA) ≈ 689.96 × 23.000 ≈ 15,869.08 VA

Which matches P_total = 15,870 W (difference due to rounding), so S ≈ 15.87 kVA, power factor = P/S ≈ 1 (resistive).

Summary of this worked case

Phase voltage: 230 V
Line voltage: ≈ 398–400 V
Phase current: 23.0 A (for 10 Ω resistive per phase)
Total real power ≈ 15.87 kW
Total apparent power ≈ 15.87 kVA (PF ≈ 1)

Step 4 — Example with inductive load (power factor & kVA conversion)

Suppose same volts but each phase has an impedance with PF = 0.8 lagging and phase real power per phase we want: P_phase = 5,000 W (for example).

Compute kVA per phase and currents

Given P_phase = 5,000 W and PF = 0.8

Apparent power per phase S_ph = P_phase / PF = 5,000 / 0.8 = 6,250 VA

Phase current:
I_ph = S_ph / V_ph = 6,250 / 230 ≈ 27.1739 A

Check total:
S_total = 3 × 6,250 = 18,750 VA => 18.75 kVA
P_total = 3 × 5,000 = 15,000 W => 15.00 kW
PF_total = P_total / S_total = 15,000 / 18,750 = 0.8 (lags)

Common on-site quick checks (cheat sheet)

To estimate kVA needed from measured kW: kVA ≈ kW / PF. If PF unknown, assume 0.8 for mixed loads.
To get line current from kVA (3φ, 400 V): I ≈ (kVA × 1000) / (√3 × 400). Shortcut: I ≈ kVA × 1.443 / 400? — easier: I(A) ≈ kVA × 1.443 (when using kVA and 400 V). Example: 10 kVA at 400 V → I ≈ 10 × 1.443 ≈ 14.43 A.
To convert single-phase kVA to amps at 230 V: I ≈ (kVA × 1000) / 230 ≈ kVA × 4.348.

Embedded videos — quick visual recap

40-question CBT quiz (5s start, 15 minutes)

Start the quiz, wait the 5-second countdown, then answer 40 multiple-choice questions in 15 minutes. The quiz grades locally and you may email your score to edwinogielibrary@gmail.com.

FAQ & quick notes

Q: Why is Nigeria's line-to-line called 400 V if calculation gave 398.36 V?

Nominal values are standardized & rounded (230/400). Small rounding is normal and equipment is designed for that tolerance.

Q: When do we use phasor subtraction (Va−Vb) instead of √3 shortcut?

Use phasor subtraction when phases are unbalanced or when angles are not exactly 120°; the √3 shortcut only applies for balanced systems with perfect 120° separation.

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