Lighting Circuit Voltage Drop

Overview

Calculates voltage drop for single-phase parallel-connected lighting circuits with sequential cable runs per IEC 60364, BS 7671, and AS/NZS 3000. Each light fixture taps off a shared cable that runs from the supply through each light in series. The cable section between any two lights carries the cumulative current of all downstream lights, producing a different voltage drop at each fixture location.

Input Parameters

Circuit Parameters

Parameter	Type	Default	Range	Description
`supply_voltage`	Number	230 V	100 – 400 V	Nominal single-phase supply voltage
`power_factor`	Number	1.0	0.5 – 1.0	Load power factor (typically 0.95–1.0 for LED)
`frequency`	Enum: 50 / 60 Hz	50	—	System frequency
`temperature`	Number	30 °C	−10 – 90 °C	Cable operating temperature

Conductor Parameters

Parameter	Type	Default	Range	Description
`conductor_material`	Enum: copper / aluminium	copper	—	Conductor material
`conductor_csa`	Enum	1.5 mm²	0.75 – 500 mm²	Conductor cross-sectional area

Light String Configuration

Parameter	Type	Default	Range	Description
`lights`	Array of objects	4 lights	1 – 50 items	Sequential light fixtures with cable lengths and power ratings
`lights[n].length`	Number	10 m	0.1 – 1000 m	Cable length from previous light (or supply for first light)
`lights[n].power`	Number	20 W	0.1 – 5000 W	Light fixture power consumption

Compliance Settings

Parameter	Type	Default	Range	Description
`voltage_drop_limit`	Number	3%	0.5 – 10%	Maximum allowable voltage drop percentage

Output Results

Summary Outputs

Output	Unit	Description
`total_power`	W	Sum of all light fixture powers
`cable_resistance`	Ω/km	Conductor resistance at operating temperature
`supply_current`	A	Total current drawn at supply point (first section current)
`max_voltage_drop`	V	Maximum cumulative voltage drop at any light
`max_voltage_drop_percent`	%	Maximum voltage drop as percentage of supply
`max_voltage_drop_location`	Integer	Light number where maximum voltage drop occurs
`min_voltage`	V	Minimum voltage at any light fixture
`compliant`	Boolean	Whether \(V_{d\%,max} \leq\) voltage drop limit
`warnings`	List	Warning messages (over-current, excessive length, etc.)

Per-Light Outputs (`light_details` array)

Output	Unit	Description
`section_current`	A	Current through cable section feeding this light
`section_voltage_drop`	V	Voltage drop across this cable section only
`cumulative_voltage_drop`	V	Total voltage drop from supply to this light
`voltage_at_light`	V	Actual voltage at this light fixture
`voltage_drop_percent`	%	Cumulative voltage drop as percentage of supply

Formulas

Conductor Resistance at Temperature

\[ R(T) = R_{20} \times \bigl(1 + \alpha \times (T - 20)\bigr) \quad (\Omega/\text{km}) \]

Where \(R_{20}\) is the conductor resistance at 20 °C (from lookup table), \(\alpha\) is the temperature coefficient, and \(T\) is the operating temperature in °C.

Material	\(\alpha\) (/°C)
Copper	0.00393
Aluminium	0.00403

Section Current

\[ I_{\text{section},i} = \frac{\displaystyle\sum_{k=i}^{n} P_k}{V_{\text{supply}} \times \cos\phi} \]

Each cable section carries the cumulative current of all downstream lights. Section 1 (supply to first light) carries the total circuit current. Working backwards from the last light, each section adds the power of the next upstream light.

Section Voltage Drop

\[ \Delta V_i = \frac{2 \times I_{\text{section},i} \times L_i \times R(T)}{1000} \]

Where the factor of 2 accounts for active and neutral conductors (go-and-return path), \(L_i\) is the cable section length in meters, and \(R(T)\) is in Ω/km. Division by 1000 converts the length–resistance product from m·Ω/km to Ω.

Cumulative Voltage Drop

\[ \Delta V_{\text{cum},n} = \sum_{i=1}^{n} \Delta V_i \]

The total voltage drop at light \(n\) is the sum of all section voltage drops from the supply to that light.

Voltage at Light

\[ V_{\text{light},n} = V_{\text{supply}} - \Delta V_{\text{cum},n} \]

Voltage Drop Percentage

\[ V_{d\%,n} = \frac{\Delta V_{\text{cum},n}}{V_{\text{supply}}} \times 100 \]

Compliance Check

\[ \max(V_{d\%,n}) \leq V_{d,\text{limit}\%} \implies \text{PASS} \]

Key Lookup Tables

Copper Resistance at 20 °C (Ω/km)

CSA (mm²)	R₂₀ (Ω/km)
0.75	24.5
1.0	18.1
1.5	12.1
2.5	7.41
4.0	4.61
6.0	3.08
10	1.83
16	1.15
25	0.727
35	0.524
50	0.387
70	0.268
95	0.193
120	0.153
150	0.124
185	0.0991
240	0.0754
300	0.0601
400	0.0470
500	0.0366

Aluminium Resistance at 20 °C (Ω/km)

IEC 60228 lists aluminium conductors from 10 mm²; the 2.5, 4 and 6 mm² aluminium values are the component's own (about 1.64× the copper value).

CSA (mm²)	R₂₀ (Ω/km)
2.5	12.2
4.0	7.56
6.0	5.08
10	3.08
16	1.91
25	1.20
35	0.868
50	0.641
70	0.443
95	0.320
120	0.253
150	0.206
185	0.164
240	0.125
300	0.100
400	0.0778
500	0.0605

Worked Examples

The 6 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_voltage_drop; worked examples for every calculation).

#	Case	Cable	Lights / load	R(T) (Ω/km)	Worst light	Max VD	Total vs limit	Result
1	Four LED downlights on 1.5 mm² Cu	1.5 mm² Cu, 230 V	4 / 95 W	12.5755	4 (229.7 V)	0.297 V (0.129 %)	0.129 / 3 %	PASS
2	Ten 100 W lights at 75 °C, pf 0.9	2.5 mm² Cu, 230 V	10 / 1,000 W	9.0117	10 (225.2 V)	4.789 V (2.082 %)	2.082 / 3 %	PASS
3	Warehouse high-bays, 5 × 250 W on 2.5 mm²	2.5 mm² Cu, 230 V	5 / 1,250 W	8.1380	5 (219.8 V)	10.242 V (4.453 %)	4.453 / 3 %	FAIL
4	Aluminium 4 mm² at 50 °C, three 100 W lights	4 mm² Al, 230 V	3 / 300 W	8.4740	3 (229.2 V)	0.774 V (0.336 %)	0.336 / 3 %	PASS
5	Street lighting, 12 × 150 W on 16 mm² Al, 1.5 % upstream	16 mm² Al, 240 V	12 / 1,800 W	2.2179	12 (230.4 V)	9.611 V (4.005 %)	5.505 / 5 %	FAIL
6	First light at the supply, ten 58 W fittings on 4 mm²	4 mm² Cu, 230 V	10 / 580 W	4.9723	10 (229.0 V)	0.979 V (0.426 %)	0.426 / 3 %	PASS

Example 1 — Four LED downlights on 1.5 mm² Cu

The help page's basic example: unequal spacing and loads, worst drop at the far light.

Given: 4 lights fed in sequence on 1.5 mm² Cu (active + neutral), 230 V single-phase, cos φ = 1, conductor at 30 °C; (section length, light power) from the supply: (15 m, 20 W), (8 m, 30 W), (10 m, 25 W), (12 m, 20 W); voltage-drop limit 3 % from the point of supply (tool default). Engine c0015.

Resistance R(T) = R₂₀ × (1 + α(T − 20)) = 12.1 × (1 + 0.00393 × (30 − 20)) = 12.57553 Ω/km

Load ΣP = 95 W; supply current I₁ = ΣP / (V × cos φ) = 95 / (230 × 1) = 0.4130 A

Sections

I_k = ΣP_downstream / (V × cos φ); ΔV_k = 2 × I_k × L_k × R / 1000; cumulative from the supply:

k	L (m)	ΣP (W)	I_k (A)	ΔV_k (V)	ΣΔV (V)	V (V)	VD %
1	15	95	0.4130	2 × 0.4130 × 15 × 12.5755/1000 = 0.1558	0.1558	229.84	0.068
2	8	75	0.3261	2 × 0.3261 × 8 × 12.5755/1000 = 0.0656	0.2214	229.78	0.096
3	10	45	0.1957	2 × 0.1957 × 10 × 12.5755/1000 = 0.0492	0.2706	229.73	0.118
4	12	20	0.0870	2 × 0.0870 × 12 × 12.5755/1000 = 0.0262	0.2969	229.70	0.129

Worst light Light 4: ΣΔV = 0.2969 V → 0.2969 / 230 × 100 = 0.129 %; minimum voltage 229.70 V

Voltage drop check 0.129 % ≤ 3 %? → 0.129 vs limit 3.000 % PASS

Headroom Load at the limit = (limit − upstream) / max VD = 3.00 / 0.129 = 23.24 × the present load

Result 1/1 checks pass → PASS; worst light 4 at 0.129 % total; governing: voltage drop

Reference Hand calculation: R(30 °C) = 12.1 × (1 + 0.00393 × 10) = 12.5755 Ω/km (IEC 60228 class 2 R20); section currents 95/75/45/20 W ÷ 230 V = 0.41304/0.32609/0.19565/0.08696 A; ΣΔV = 2 × 12.5755/1000 × (0.41304 × 15 + 0.32609 × 8 + 0.19565 × 10 + 0.08696 × 12) = 0.2969 V = 0.129 %; V = 229.70 V (validation case c0015-4-lights-1mm5-230v)

Example 2 — Ten 100 W lights at 75 °C, pf 0.9

Uniformly spaced equal lights (tapered current) with a conservative 75 °C conductor temperature.

Given: 10 lights fed in sequence on 2.5 mm² Cu (active + neutral), 230 V single-phase, cos φ = 0.9, conductor at 75 °C; (section length, light power) from the supply: 10 × (10 m, 100 W); voltage-drop limit 3 % from the point of supply (tool default). Engine c0015.

Resistance R(T) = R₂₀ × (1 + α(T − 20)) = 7.41 × (1 + 0.00393 × (75 − 20)) = 9.01167 Ω/km

Load ΣP = 1,000 W; supply current I₁ = ΣP / (V × cos φ) = 1,000 / (230 × 0.9) = 4.8309 A

Sections

I_k = ΣP_downstream / (V × cos φ); ΔV_k = 2 × I_k × L_k × R / 1000; cumulative from the supply:

k	L (m)	ΣP (W)	I_k (A)	ΔV_k (V)	ΣΔV (V)	V (V)	VD %
1	10	1,000	4.8309	2 × 4.8309 × 10 × 9.0117/1000 = 0.8707	0.8707	229.13	0.379
2	10	900	4.3478	2 × 4.3478 × 10 × 9.0117/1000 = 0.7836	1.6543	228.35	0.719
3	10	800	3.8647	2 × 3.8647 × 10 × 9.0117/1000 = 0.6966	2.3509	227.65	1.022
4	10	700	3.3816	2 × 3.3816 × 10 × 9.0117/1000 = 0.6095	2.9604	227.04	1.287
5	10	600	2.8986	2 × 2.8986 × 10 × 9.0117/1000 = 0.5224	3.4828	226.52	1.514
6	10	500	2.4155	2 × 2.4155 × 10 × 9.0117/1000 = 0.4353	3.9181	226.08	1.704
7	10	400	1.9324	2 × 1.9324 × 10 × 9.0117/1000 = 0.3483	4.2664	225.73	1.855
8	10	300	1.4493	2 × 1.4493 × 10 × 9.0117/1000 = 0.2612	4.5276	225.47	1.969
9	10	200	0.9662	2 × 0.9662 × 10 × 9.0117/1000 = 0.1741	4.7017	225.30	2.044
10	10	100	0.4831	2 × 0.4831 × 10 × 9.0117/1000 = 0.0871	4.7888	225.21	2.082

Worst light Light 10: ΣΔV = 4.7888 V → 4.7888 / 230 × 100 = 2.082 %; minimum voltage 225.21 V

Voltage drop check 2.082 % ≤ 3 %? → 2.082 vs limit 3.000 % PASS

Headroom Load at the limit = (limit − upstream) / max VD = 3.00 / 2.082 = 1.44 × the present load

Result 1/1 checks pass → PASS; worst light 10 at 2.082 % total; governing: voltage drop

Reference Hand calculation: R = 7.41 × (1 + 0.00393 × 55) = 9.01167 Ω/km; I_k = (11 − k) × 100 ÷ (230 × 0.9); ΣI = 5500 ÷ 207 = 26.570 A; ΔV = 2 × 10 × 9.01167/1000 × 26.570 = 4.7888 V = 2.082 %; load at the 3 % limit 3 ÷ 2.082 = 1.44× (validation case c0015-10-lights-2mm5-75c-pf09)

Example 3 — Warehouse high-bays, 5 × 250 W on 2.5 mm²

The help page's long run: the drop passes 3 % from the third light on, so the circuit fails.

Given: 5 lights fed in sequence on 2.5 mm² Cu (active + neutral), 230 V single-phase, cos φ = 0.95, conductor at 45 °C; (section length, light power) from the supply: (50 m, 250 W), 4 × (30 m, 250 W); voltage-drop limit 3 % from the point of supply (tool default). Engine c0015.

Resistance R(T) = R₂₀ × (1 + α(T − 20)) = 7.41 × (1 + 0.00393 × (45 − 20)) = 8.13803 Ω/km

Load ΣP = 1,250 W; supply current I₁ = ΣP / (V × cos φ) = 1,250 / (230 × 0.95) = 5.7208 A

Sections

I_k = ΣP_downstream / (V × cos φ); ΔV_k = 2 × I_k × L_k × R / 1000; cumulative from the supply:

k	L (m)	ΣP (W)	I_k (A)	ΔV_k (V)	ΣΔV (V)	V (V)	VD %
1	50	1,250	5.7208	2 × 5.7208 × 50 × 8.1380/1000 = 4.6556	4.6556	225.34	2.024
2	30	1,000	4.5767	2 × 4.5767 × 30 × 8.1380/1000 = 2.2347	6.8903	223.11	2.996
3	30	750	3.4325	2 × 3.4325 × 30 × 8.1380/1000 = 1.6760	8.5663	221.43	3.724
4	30	500	2.2883	2 × 2.2883 × 30 × 8.1380/1000 = 1.1174	9.6837	220.32	4.210
5	30	250	1.1442	2 × 1.1442 × 30 × 8.1380/1000 = 0.5587	10.2424	219.76	4.453

Worst light Light 5: ΣΔV = 10.2424 V → 10.2424 / 230 × 100 = 4.453 %; minimum voltage 219.76 V

Voltage drop check 4.453 % ≤ 3 %? → 4.453 vs limit 3.000 % FAIL

Headroom Load at the limit = (limit − upstream) / max VD = 3.00 / 4.453 = 0.67 × the present load

Result 0/1 checks pass → FAIL; worst light 5 at 4.453 % total; governing: voltage drop. Voltage drop of 4.45% exceeds limit of 3.0%

Reference Hand calculation: R(45 °C) = 7.41 × 1.09825 = 8.13803 Ω/km; I = 1250/1000/750/500/250 W ÷ (230 × 0.95) = 5.72082/4.57666/3.43249/2.28833/1.14416 A; ΔV = 2 × 8.13803/1000 × (5.72082 × 50 + (4.57666 + 3.43249 + 2.28833 + 1.14416) × 30) = 4.6556 + 5.5868 = 10.2424 V = 4.453 % > 3 % (validation case c0015-warehouse-5x250w-2mm5-fail)

Example 4 — Aluminium 4 mm² at 50 °C, three 100 W lights

Aluminium conductor resistance and its own temperature coefficient (0.00403 /°C).

Given: 3 lights fed in sequence on 4 mm² Al (active + neutral), 230 V single-phase, cos φ = 1, conductor at 50 °C; (section length, light power) from the supply: (20 m, 100 W), 2 × (15 m, 100 W); voltage-drop limit 3 % from the point of supply (tool default). Engine c0015.

Resistance R(T) = R₂₀ × (1 + α(T − 20)) = 7.56 × (1 + 0.00403 × (50 − 20)) = 8.47400 Ω/km

Load ΣP = 300 W; supply current I₁ = ΣP / (V × cos φ) = 300 / (230 × 1) = 1.3043 A

Sections

I_k = ΣP_downstream / (V × cos φ); ΔV_k = 2 × I_k × L_k × R / 1000; cumulative from the supply:

k	L (m)	ΣP (W)	I_k (A)	ΔV_k (V)	ΣΔV (V)	V (V)	VD %
1	20	300	1.3043	2 × 1.3043 × 20 × 8.4740/1000 = 0.4421	0.4421	229.56	0.192
2	15	200	0.8696	2 × 0.8696 × 15 × 8.4740/1000 = 0.2211	0.6632	229.34	0.288
3	15	100	0.4348	2 × 0.4348 × 15 × 8.4740/1000 = 0.1105	0.7737	229.23	0.336

Worst light Light 3: ΣΔV = 0.7737 V → 0.7737 / 230 × 100 = 0.336 %; minimum voltage 229.23 V

Voltage drop check 0.336 % ≤ 3 %? → 0.336 vs limit 3.000 % PASS

Headroom Load at the limit = (limit − upstream) / max VD = 3.00 / 0.336 = 8.92 × the present load

Result 1/1 checks pass → PASS; worst light 3 at 0.336 % total; governing: voltage drop

Reference Hand calculation: R(50 °C) = 7.56 × (1 + 0.00403 × 30) = 8.4740 Ω/km (component R20 for 4 mm² Al); I = 300/200/100 W ÷ 230 V = 1.30435/0.86957/0.43478 A; ΔV = 2 × 8.4740/1000 × (1.30435 × 20 + 0.86957 × 15 + 0.43478 × 15) = 0.4421 + 0.2211 + 0.1105 = 0.7737 V = 0.336 % (validation case c0015-al-4mm2-50c-3x100w)

Example 5 — Street lighting, 12 × 150 W on 16 mm² Al, 1.5 % upstream

A 5 % limit from the point of supply with 1.5 % already used upstream: the run itself is 4.0 %, so the total fails.

Given: 12 lights fed in sequence on 16 mm² Al (active + neutral), 240 V single-phase, cos φ = 0.9, conductor at 60 °C; (section length, light power) from the supply: 12 × (40 m, 150 W); voltage-drop limit 5 % from the point of supply, 1.5 % already used upstream. Engine c0015.

Resistance R(T) = R₂₀ × (1 + α(T − 20)) = 1.91 × (1 + 0.00403 × (60 − 20)) = 2.21789 Ω/km

Load ΣP = 1,800 W; supply current I₁ = ΣP / (V × cos φ) = 1,800 / (240 × 0.9) = 8.3333 A

Sections

I_k = ΣP_downstream / (V × cos φ); ΔV_k = 2 × I_k × L_k × R / 1000; cumulative from the supply:

k	L (m)	ΣP (W)	I_k (A)	ΔV_k (V)	ΣΔV (V)	V (V)	VD %
1	40	1,800	8.3333	2 × 8.3333 × 40 × 2.2179/1000 = 1.4786	1.4786	238.52	0.616
2	40	1,650	7.6389	2 × 7.6389 × 40 × 2.2179/1000 = 1.3554	2.8340	237.17	1.181
3	40	1,500	6.9444	2 × 6.9444 × 40 × 2.2179/1000 = 1.2322	4.0661	235.93	1.694
4	40	1,350	6.2500	2 × 6.2500 × 40 × 2.2179/1000 = 1.1089	5.1751	234.82	2.156
5	40	1,200	5.5556	2 × 5.5556 × 40 × 2.2179/1000 = 0.9857	6.1608	233.84	2.567
6	40	1,050	4.8611	2 × 4.8611 × 40 × 2.2179/1000 = 0.8625	7.0233	232.98	2.926
7	40	900	4.1667	2 × 4.1667 × 40 × 2.2179/1000 = 0.7393	7.7626	232.24	3.234
8	40	750	3.4722	2 × 3.4722 × 40 × 2.2179/1000 = 0.6161	8.3787	231.62	3.491
9	40	600	2.7778	2 × 2.7778 × 40 × 2.2179/1000 = 0.4929	8.8716	231.13	3.696
10	40	450	2.0833	2 × 2.0833 × 40 × 2.2179/1000 = 0.3696	9.2412	230.76	3.851
11	40	300	1.3889	2 × 1.3889 × 40 × 2.2179/1000 = 0.2464	9.4876	230.51	3.953
12	40	150	0.6944	2 × 0.6944 × 40 × 2.2179/1000 = 0.1232	9.6109	230.39	4.005

Worst light Light 12: ΣΔV = 9.6109 V → 9.6109 / 240 × 100 = 4.005 %; minimum voltage 230.39 V

Point of supply Total = upstream + this circuit = 1.5 + 4.005 = 5.505 %

Voltage drop check 5.505 % ≤ 5 %? → 5.505 vs limit 5.000 % FAIL

Headroom Load at the limit = (limit − upstream) / max VD = 3.50 / 4.005 = 0.87 × the present load

Result 0/1 checks pass → FAIL; worst light 12 at 5.505 % total; governing: voltage drop. Voltage drop of 4.00% exceeds limit of 3.5%

Reference Hand calculation: R(60 °C) = 1.91 × (1 + 0.00403 × 40) = 2.21789 Ω/km; I_k = (13 − k) × 150 ÷ (240 × 0.9), ΣI = 78 × 150 ÷ 216 = 54.1667 A; ΔV = 2 × 40 × 2.21789/1000 × 54.1667 = 9.6109 V = 4.005 %; total 1.5 + 4.005 = 5.505 % > 5 % (validation case c0015-street-al16-240v-upstream-fail)

Example 6 — First light at the supply, ten 58 W fittings on 4 mm²

A zero-length first section (light mounted at the board) and tapered spacing — only the c0015 engine allows it.

Given: 10 lights fed in sequence on 4 mm² Cu (active + neutral), 230 V single-phase, cos φ = 0.95, conductor at 40 °C; (section length, light power) from the supply: (0 m, 58 W), 2 × (6 m, 58 W), 2 × (8 m, 58 W), 2 × (10 m, 58 W), 2 × (12 m, 58 W), (15 m, 58 W); voltage-drop limit 3 % from the point of supply (tool default). Engine c0015.

Resistance R(T) = R₂₀ × (1 + α(T − 20)) = 4.61 × (1 + 0.00393 × (40 − 20)) = 4.97235 Ω/km

Load ΣP = 580 W; supply current I₁ = ΣP / (V × cos φ) = 580 / (230 × 0.95) = 2.6545 A

Sections

I_k = ΣP_downstream / (V × cos φ); ΔV_k = 2 × I_k × L_k × R / 1000; cumulative from the supply:

k	L (m)	ΣP (W)	I_k (A)	ΔV_k (V)	ΣΔV (V)	V (V)	VD %
1	0	580	2.6545	2 × 2.6545 × 0 × 4.9723/1000 = 0.0000	0.0000	230.00	0.000
2	6	522	2.3890	2 × 2.3890 × 6 × 4.9723/1000 = 0.1425	0.1425	229.86	0.062
3	6	464	2.1236	2 × 2.1236 × 6 × 4.9723/1000 = 0.1267	0.2693	229.73	0.117
4	8	406	1.8581	2 × 1.8581 × 8 × 4.9723/1000 = 0.1478	0.4171	229.58	0.181
5	8	348	1.5927	2 × 1.5927 × 8 × 4.9723/1000 = 0.1267	0.5438	229.46	0.236
6	10	290	1.3272	2 × 1.3272 × 10 × 4.9723/1000 = 0.1320	0.6758	229.32	0.294
7	10	232	1.0618	2 × 1.0618 × 10 × 4.9723/1000 = 0.1056	0.7814	229.22	0.340
8	12	174	0.7963	2 × 0.7963 × 12 × 4.9723/1000 = 0.0950	0.8764	229.12	0.381
9	12	116	0.5309	2 × 0.5309 × 12 × 4.9723/1000 = 0.0634	0.9398	229.06	0.409
10	15	58	0.2654	2 × 0.2654 × 15 × 4.9723/1000 = 0.0396	0.9794	229.02	0.426

Worst light Light 10: ΣΔV = 0.9794 V → 0.9794 / 230 × 100 = 0.426 %; minimum voltage 229.02 V

Voltage drop check 0.426 % ≤ 3 %? → 0.426 vs limit 3.000 % PASS

Headroom Load at the limit = (limit − upstream) / max VD = 3.00 / 0.426 = 7.05 × the present load

Result 1/1 checks pass → PASS; worst light 10 at 0.426 % total; governing: voltage drop

Reference Hand calculation: R(40 °C) = 4.61 × (1 + 0.00393 × 20) = 4.97235 Ω/km; I_k = (11 − k) × 58 ÷ (230 × 0.95); section 1 has L = 0 so ΔV1 = 0; ΣI·L = (9 × 6 + 8 × 6 + 7 × 8 + 6 × 8 + 5 × 10 + 4 × 10 + 3 × 12 + 2 × 12 + 1 × 15) × 0.265446 = 371 × 0.265446 = 98.4805 A·m; ΔV = 2 × 4.97235/1000 × 98.4805 = 0.9794 V = 0.426 % (validation case c0015-first-light-at-supply-4mm2)

Standards & References

IEC 60364-5-52 — Low-voltage electrical installations — Selection and erection of electrical equipment — Wiring systems
BS 7671 (IET Wiring Regulations) — Requirements for electrical installations, Appendix 12 (voltage drop)
AS/NZS 3000:2018 — Electrical installations (Wiring Rules), Clause 3.6 (voltage drop)
NEC (NFPA 70) — National Electrical Code, Articles 210.19(A) and 215.2(A) (voltage drop recommendations)
IEC 60228 — Conductors of insulated cables (resistance values at 20 °C)
Standard voltage drop limit for lighting: typically 3% (configurable per local regulations)

Tips & Limitations

Tips

The default 3% limit is typical for lighting sub-circuits. Some standards allow up to 5% total (main + sub-circuit combined).
The worst voltage drop always occurs at the last (most distant) light because cumulative drops accumulate along the cable run.
LED drivers with wide input voltage ranges (e.g., 180–265 V) tolerate higher voltage drops than fluorescent or incandescent loads.
For long runs, increasing cable CSA by one standard size often eliminates voltage drop issues cost-effectively.
Power factor for modern LED luminaires is typically 0.95–1.0. Older magnetic ballast fluorescents may be 0.5–0.6 without correction.
Higher ambient temperatures increase cable resistance — account for ceiling void, roof space, or enclosed luminaire temperatures.

Limitations

Single-phase only — three-phase lighting circuits are not supported by this calculator.
Resistive model only — cable reactance is not included. For small lighting conductors (≤ 16 mm²) reactance is negligible; for larger conductors this is slightly non-conservative.
No inrush current analysis — LED driver inrush currents at switch-on are not modelled. This calculator is for steady-state voltage drop only.
Uniform cable — the same conductor material and CSA is assumed for all sections. Mixed cable sizes are not supported.
Simplified current capacity check — the built-in over-current warning uses an approximate table. Always verify current capacity separately using the full cable sizing calculator (c0030 / c0031).
Balanced loading assumed — each light draws its rated power at the supply voltage. Voltage-dependent load variation is not modelled.

Developer Guide

Architecture

File layout:

components/c0015_lighting_voltage_drop/component.py — Component registration, parameter schema, summary cards builder
components/c0015_lighting_voltage_drop/calculations.py — Pure math functions and resistance lookup tables (no Django dependencies)
components/c0015_lighting_voltage_drop/equations.py — MathJax equation builder for report generation
components/c0015_lighting_voltage_drop/excel_generator.py — Excel report generation with detailed light-by-light breakdown

The calculator class is LightingVoltageDropCalculator, extending ElectricalCalculator. The lights array input is decomposed into cable_lengths and light_powers lists before being passed to the pure calculation function calculate_lighting_voltage_drop().

Lookup tables are defined as Python dicts:

COPPER_RESISTANCE_20C — Cu R at 20 °C (Ω/km), keyed by CSA (0.75–500 mm²)
ALUMINIUM_RESISTANCE_20C — Al R at 20 °C (Ω/km), keyed by CSA (2.5–500 mm²)

Temperature constants:

COPPER_TEMP_COEFF = 0.00393
ALUMINIUM_TEMP_COEFF = 0.00403

Code Paths

Key functions in calculations.py:

Function	Purpose
`calculate_conductor_resistance(csa, material, temperature)`	Lookup R₂₀ from table, apply temperature correction, return Ω/km
`calculate_lighting_voltage_drop(data)`	Main entry point — computes section currents, voltage drops, cumulative drops, compliance, and warnings. Returns complete results dict.

Key functions in component.py:

Function / Class	Purpose
`LightingVoltageDropCalculator.validate_parameters()`	Validates supply voltage, CSA, lights array, and power factor
`LightingVoltageDropCalculator.calculate()`	Orchestrates: extract lights array, decompose into lengths/powers, call calculation engine, merge results with parameter echoes
`equation_builder(calculation_results)`	Delegates to `equations.build_equations()` for report articles
`summary_cards_builder(results)`	Returns 4 summary cards: max VD%, min voltage, total power, supply current
`register_component(registry)`	Registers calc type `'lighting_voltage_drop'` with schema, builders, and sidebar template

Calculation flow:

Calculate R(T) from lookup table + temperature correction
Sum downstream powers for each section (reverse iteration)
Divide cumulative power by \(V \times \cos\phi\) for section current
Compute \(\Delta V = 2 \times I \times L \times R / 1000\) per section
Accumulate voltage drops, derive voltage at each light
Check max VD% against limit, generate warnings

Testing

Running tests:

pytest calcs/components/c0015_lighting_voltage_drop/ -v
pytest calcs/test_c0015* -v

How to extend:

To add new conductor materials, define a new resistance table dict (e.g., TINNED_COPPER_RESISTANCE_20C) and temperature coefficient, then update calculate_conductor_resistance() and the conductor_material enum in PARAMETER_SCHEMA.
To support mixed cable sizes per section, refactor calculate_lighting_voltage_drop() to accept a per-section CSA array and call calculate_conductor_resistance() once per section.
To add reactance for larger conductors, introduce an XC_TABLE and modify the section voltage drop formula to use impedance \(Z = \sqrt{R^2 + X^2}\).
Write tests covering edge cases: single light, 50 lights (max), very long runs, aluminium with high temperature, PF < 1.

Overview

Input Parameters

Circuit Parameters

Conductor Parameters

Light String Configuration

Compliance Settings

Output Results

Summary Outputs

Per-Light Outputs (light_details array)

Formulas

Conductor Resistance at Temperature

Section Current

Section Voltage Drop

Cumulative Voltage Drop

Voltage at Light

Voltage Drop Percentage

Compliance Check

Key Lookup Tables

Copper Resistance at 20 °C (Ω/km)

Aluminium Resistance at 20 °C (Ω/km)

Worked Examples

Example 1 — Four LED downlights on 1.5 mm² Cu

Example 2 — Ten 100 W lights at 75 °C, pf 0.9

Example 3 — Warehouse high-bays, 5 × 250 W on 2.5 mm²

Example 4 — Aluminium 4 mm² at 50 °C, three 100 W lights

Example 5 — Street lighting, 12 × 150 W on 16 mm² Al, 1.5 % upstream

Example 6 — First light at the supply, ten 58 W fittings on 4 mm²

Standards & References

Tips & Limitations

Tips

Limitations

Developer Guide

Architecture

Code Paths

Testing

Per-Light Outputs (`light_details` array)