Balanced three-phase circuits
Keep line and phase values straight
Compare star and delta using the same load impedances, calculate total power, and find a common √3 mistake.
What you’ll learn
- Choose the correct line-to-phase relationship for star or delta.
- Calculate three-phase power and check it by adding the phase powers.
Before you start: Know RMS quantities, impedance magnitude and power factor. This guide assumes a balanced sinusoidal source and three identical loads.
In this guide 6 sections
1. Name the voltage before using it
A three-phase calculation often goes wrong before any arithmetic: the source voltage is line-to-line, but a student treats it as the voltage across one load. Begin by writing two pairs of symbols. VL and IL describe the supply lines; Vph and Iph describe one load branch.
Our example uses a balanced sinusoidal 400 V RMS line-to-line supply. Each of the three identical load impedances is 16 + j12 Ω. Its magnitude is 20 Ω and its power factor is 16 / 20 = 0.8 lagging. Assume ideal connecting wires and neglect source impedance.
| Connection | Branch voltage | Line current |
|---|---|---|
| Star / Y | Vph = VL / √3 | IL = Iph |
| Delta / Δ | Vph = VL | IL = √3 Iph |
These relationships concern magnitudes. The √3 factor comes from combining phasors separated in angle; it is not a multiplier to attach to every three-phase number.
2. Solve one branch in star
With the loads connected in star, each branch receives 400 / √3 ≈ 230.94 V RMS. Divide that voltage by the branch impedance magnitude: Iph = 230.94 / 20 ≈ 11.55 A RMS. A line feeds one branch directly, so the line current is also 11.55 A.
P = √3 VLIL cos φ
Using unrounded values: √3 × 400 × (400 / (√3 × 20)) × 0.8 = 6400 W.
If you use rounded current, your answer may be a few watts away. Keep full calculator precision until the final result. In a perfectly balanced star system the three phase-current phasors sum to zero, so the neutral carries zero current. This conclusion depends on the balance assumption.
3. Reconnect the same impedances in delta
Now each impedance sits directly between two lines, so each branch receives 400 V RMS. Its current is 400 / 20 = 20 A RMS. A supply line meets two branches, whose current phasors combine to give IL = √3 × 20 ≈ 34.64 A RMS.
The total real power becomes √3 × 400 × (√3 × 20) × 0.8 = 19,200 W. For this specific comparison, delta draws three times the line current and consumes three times the power of star. The supply voltage and the three impedances were held fixed; this is not a statement that any delta-connected machine always consumes three times as much power.
4. Check by adding the three branch powers
For star, each branch dissipates Iph²R = (400 / (√3 × 20))² × 16 ≈ 2133.33 W. Multiply by three to recover 6400 W. For delta, each branch dissipates 20² × 16 = 6400 W, giving 19,200 W in total.
| Result | Star | Delta |
|---|---|---|
| Branch voltage | 230.94 V | 400 V |
| Branch current | 11.55 A | 20 A |
| Line current | 11.55 A | 34.64 A |
| Total real power | 6.4 kW | 19.2 kW |
Try two checks of your own
A. Work backwards from power
A balanced load takes 9 kW from a 400 V RMS line-to-line supply at 0.75 power factor. Find its line current. Do you need to know whether the load is star or delta?
Show the working
IL = 9000 / (√3 × 400 × 0.75) ≈ 17.32 A. You do not need the connection to calculate line current from total real power, line voltage and power factor. You would need it to find the individual branch voltage and current.
B. Find the extra factor
A student calculates the star branch current correctly as 11.55 A, then multiplies by √3 and reports 20 A as the line current. Which relationship was used incorrectly?
Show the explanation
The student used the delta current relationship for a star load. In star, each line and its branch are in series, so their current magnitudes are equal. The √3 factor for star belongs in the line-to-phase voltage relationship.
Check the assumptions before generalizing
Unequal impedances, a floating star point, harmonics or non-sinusoidal currents require a more detailed analysis. Do not assume zero neutral current or use one branch to represent all three when the load is unbalanced. Treat star–delta reconnection here as a paper calculation; actual machines have winding ratings and switching requirements.
Further study: MIT OpenCourseWare, Power Electronics, Lecture 21: Three-Phase Systems, Part 1. The comparison values and practice questions above were prepared for Padhnus.