Three-phase balancing exercises
Installations with all the load concentrated on a single phase. Redistribute loads across R/S/T, watch the imbalance in real time and get validation per I7-2011 Art. 3.1.5.8.
Imbalance & neutral current calculator
Power on each phase → imbalance, neutral current, V43 verdict
Imbalance
150 %
Neutral current
15.7 A
✗ Imbalance
imbalance% = (5.0 − 1.0) / 2.67 × 100 = 150 %
I_N = √(I_R² + I_S² + I_T² − I_R·I_S − I_S·I_T − I_T·I_R) · I_phase = P / (230 × cos φ)
V43 thresholds (engineering): ≤ 15% balanced · 16–30% caution · > 30% error. Art. 3.1.5.8 only requires “as balanced a load as possible”.
Phase distributor (interactive exercise)
Move the consumers across R/S/T until you bring the imbalance below 15%
All consumers start on phase R — maximum imbalance. Press the phase button on each one (R → S → T) to move it, or use “Auto-balance”. Target: imbalance below 15%.
Heat pump
3.5 kW
Ceramic hob
3.0 kW
Electric oven
2.5 kW
Washing machine
2.2 kW
Electric water heater
2.0 kW
Air conditioning
1.5 kW
Lighting + sockets
1.0 kW
Target: bring the imbalance below 15% by moving the consumers across the phases.
Calculation assumptions and scope
- Convention: imbalance = (P_max − P_min) / P_avg × 100, rounded to the nearest integer; V43 verdict (identical to the validator): balanced at ≤ 15%, caution at 16–30%, error above 30% (engineering thresholds, not numeric limits from I7-2011)
- The neutral current is estimated for linear loads at the same cos φ (currents 120° out of phase); 3rd-order harmonics (LED, switched-mode supplies) add up in the neutral and can increase it further
- Phase currents are calculated at 230 V (single-phase consumers): I = P / (230 × cos φ). A balanced three-phase circuit (RST) contributes equally across the three phases
- The check is done at installation level (all boards combined), not per board — a single-phase board is locally “unbalanced”, but globally it may be balanced
Dig deeper on the blog
Three-phase balancing — step by step
From the definition of imbalance to the neutral current, the regulatory basis and the balancing method.
A. What three-phase imbalance is
Three phases 120° apart
A three-phase network delivers three alternating voltages 120° apart: phases R, S and T. Single-phase consumers (sockets, lighting, appliances) connect one at a time to a single phase.
Equal loads → unloaded neutral
When the power is distributed equally across the three phases, the line currents cancel vectorially and the current through the neutral (N) conductor is zero or minimal. Unequal distribution loads the neutral and distorts the voltages.
B. The imbalance formula
Relate the deviation to the average
imbalance% = (P_max − P_min) / P_avg × 100
P_max and P_min are the powers of the most loaded and least loaded phases; P_avg = (R + S + T) / 3.
Example
R = 5 kW, S = 3 kW, T = 1 kW → average = 3 kW → imbalance = (5 − 1) / 3 × 100 = 133%.
≤ 15%
balanced
16–30%
caution
> 30%
error
The 15% / 30% thresholds are engineering conventions (rule V43), not numeric limits from I7-2011.
C. The current through the neutral conductor
Vector sum of the three currents
I_N = √(I_R² + I_S² + I_T² − I_R·I_S − I_S·I_T − I_T·I_R)
The formula is valid for linear loads at the same cos φ (currents 120° out of phase). At I_R = I_S = I_T it gives I_N = 0.
3rd-order harmonics increase the neutral
Non-linear loads (LED, switched-mode supplies) inject 3rd-order harmonics that add up in the neutral instead of cancelling. That is why Art. 5.2.4.6.3 requires a full-section neutral when the harmonics exceed 15%.
D. The regulatory basis — I7-2011
Art. 3.1.5.8 — I7-2011
“The distribution across phases and across the supply circuits of the electrical receivers must be carried out so as to ensure as balanced a load of the phases as possible.”
Art. 5.2.4.6.1 (d) — neutral cross-sectional area
“The cross-sectional area of the neutral conductor ... must be equal to the cross-sectional area of the phase conductors ... d) in normal operation, balancing between the phases and the neutral is not ensured (for example lighting and socket boards).”
Art. 5.2.4.6.3 (a) — when the neutral can be smaller
“... the cross-sectional area of the neutral conductor may be smaller than that of the phase conductors, if ... a) the load ... is distributed in a balanced way across the phases and the level of 3rd-order harmonics ... does not exceed 15% in the phase conductor.”
The 15% and 30% thresholds used by rule V43 come from engineering practice; they are not numeric limits from the standard — Art. 3.1.5.8 only requires “as balanced a load as possible”.
E. How you balance in practice
Distribute the large consumers one by one
Consumers ≥ 3 kW are placed one per phase, rotating R → S → T. The “largest on the least loaded phase” (LPT) algorithm minimises the imbalance — exactly what the “Auto-balance” button does.
Check at installation level
Add up the loads on the same phase across all the boards (main + sub-boards). A single-phase board is locally “unbalanced”, but globally it may be perfectly balanced.
Document the phase of each circuit
Note the phase assigned to each circuit in the detailed schematic handed over at commissioning, so the check is reproducible.
F. The demand factor (what you are actually balancing)
Installed power vs absorbed power
Pa = Pi × k_u × k_s
k_s (demand) = the power running simultaneously / the installed power; k_u (utilisation) = the actual power / the installed power of a consumer. Not all consumers run at maximum at the same time.
You balance the SIMULTANEOUS load, not the installed one
Only the effective (absorbed) power flows through the conductors and the neutral. That is why the distribution across phases is done on Pa, not on the sum of the nameplates. E.g.: Pi = 12 kW, k_s = 0.5 → Pa = 6 kW to balance across R/S/T.
Art. 3.2.1.2 / 3.2.2.1 — I7-2011
“To determine the absorbed power ... the demand factor (k_s) and the utilisation factor (k_u) must be taken into account. ... Pa = Pi · k_u · k_s.” (indicative values in Tables 3.3–3.5)
Details and value tables (k_u, k_s by dwelling type): the “Demand factors” article on the blog.
Exercises — pick the correct option
You go through each step by choosing the correct answer. Wrong? Try again.
Imbalance — flat
A flat has R = 4 kW, S = 2 kW, T = 1 kW (single-phase consumers across the three phases).
Step 1: What is the average power per phase?
Redistribution — the optimal phase
The board has R = 3 kW, S = 3 kW, T = 1 kW. You add a new 2 kW consumer.
Step 1: On which phase do you put it to balance things out?
Neutral current — balanced installation
Perfectly equal, resistive loads on the three phases: R = S = T = 3 kW.
Step 1: What is the current through the neutral conductor?
Demand — power to balance
Flat with installed power Pi = 12 kW and demand factor k_s = 0.5.
Step 1: What is the simultaneous (absorbed) power Pa = Pi × k_s?
Exercises — you calculate each step
Now you enter the value for each step yourself. You get hints if you go wrong.
Imbalance — you calculate
A board has R = 5 kW, S = 3 kW, T = 1 kW.
Step 1: Average power per phase = ?
Neutral current — three unequal currents
Phase currents (resistive loads): I_R = 10 A, I_S = 6 A, I_T = 2 A.
Step 1: Current through the neutral I_N = ?
Target balancing — perfect distribution
Installation with 12 kW of single-phase consumers, distributed perfectly across the three phases (cos φ = 1).
Step 1: Power on each phase = ?
Demand — absorbed power
House with Pi = 20 kW, utilisation factor k_u = 0.6 and demand factor k_s = 0.83.
Step 1: Absorbed power Pa = Pi × k_u × k_s = ?
Exercises — direct answer
You solve it on your own and give the final answer. The step-by-step solution is available if needed.
Imbalance, direct
R = 6 kW, S = 4 kW, T = 2 kW. What is the imbalance (percentage)?
Neutral current — one phase missing
I_R = 16 A, I_S = 16 A, I_T = 0 A. What is the current through the neutral (A)?
Neutral current — three loads
I_R = 20 A, I_S = 12 A, I_T = 8 A. What is the current through the neutral (A)?
Direct simultaneous power
Pi = 15 kW installed, demand factor k_s = 0.6. What is the absorbed simultaneous power (kW)?
Discussion
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