From Tables to Answers

Voltage Drop, Diversity & Cable Sizing Calculations

Voltage Drop & Cable Sizing · Diversity · Overcurrent Protection · Standard Circuits · Trunking Calculations · OSG Appendix Revision

⚡ Voltage Drop & Cable Sizing

Where Does This Fit? — The 9-Step Design Process

Before you can size a cable or check voltage drop, you need to understand where these calculations sit in the overall design procedure. BS 7671, Chapter 13, Section 132 provides the framework. The designer follows these steps for every circuit:

Design from First Principles — the 9-step checklist
  1. Design current (Ib) — calculate the load: I = P ÷ V. Apply diversity where appropriate.
  2. Protective device (In) — pick a standard rating ≥ Ib → On-Site Guide Appendix B
  3. Installation method — identify the reference method (A, B, C, E…) → OSG Table 7.1(ii)
  4. Correction factors & cable formula (It) — apply Ca, Cg, Ci, Cf → OSG Appendix F ← this page covers steps 4–7
  5. Select cable size — find a cable whose tabulated rating ≥ It → OSG Table F4/F6
  6. Voltage drop formula — VD = (mV/A/m) × Ib × L ÷ 1000
  7. Confirm voltage drop — within 3% (6.9 V lighting) / 5% (11.5 V power)
  8. Confirm shock protection — measured or calculated Zs ≤ tabulated maximum → OSG Appendix B
  9. Thermal constraint — S = √(I²t) ÷ k → BS 7671, Chapter 54
Steps 4–7 (highlighted above) are what this resource focuses on. But remember: a cable that passes the current-rating and voltage-drop checks may still fail on shock protection or thermal constraint. Every step must be checked independently.

1. Core Principles of Circuit Design

To design a circuit safely, we follow this regulatory logic (from the 9-step checklist above):

Ib — Design Current
The actual load current the circuit must carry.
In — Nominal Rating
The breaker / fuse size (In ≥ Ib).
It — Tabulated Current
Min. cable rating after applying Correction Factors.
Vd — Voltage Drop
The loss of voltage over the cable length.
Statutory Voltage Drop Limits (230 V Supply)
🔦 Lighting: 3% = max 6.9 V
🔌 Power / Other: 5% = max 11.5 V

2. Calculating It — Why We De-rate and How to Use the OSG Tables

A cable's current-carrying capacity is published for ideal conditions — a single cable at 30 °C, not touching insulation, not protected by a BS 3036 fuse. Real installations are rarely ideal, so we apply correction factors to work out how much current the cable really needs to handle. Each factor is less than 1, which increases It, meaning we may need a bigger cable.

It ≥ In / (Ca × Cg × Ci × Cf)

It = the minimum tabulated cable rating you need to find in the OSG table. In = the rating of the protective device you chose in Step 2. The correction factors make It bigger than In to compensate for the real-world conditions.

FactorWhat It RepresentsWhere to Find ItWhen to Apply
CaAmbient temperature — a hotter environment means the cable can carry less current before its insulation overheats.OSG Table F1 — look up the ambient temperature (left column) against the cable insulation type (top row).Only if ambient temp > 30 °C. At 30 °C the factor is 1.0 (no correction needed). At 35 °C → 0.94. At 40 °C → 0.87.
CgGrouping — cables bunched together trap each other's heat, reducing the capacity of each one.OSG Table F3 — look up the number of circuits grouped (left column) against the installation method.When two or more circuits are bunched, clipped or enclosed together. A single cable run on its own needs no grouping factor.
CiThermal insulation — insulation around the cable prevents heat from escaping, so the cable runs hotter.OSG Table F2 — look up the length of cable surrounded by insulation. Over 0.5 m → 0.5 (the worst case — half capacity).When the cable passes through or is surrounded by thermal insulation (e.g. loft insulation). If the cable is clear of insulation, Ci = 1.
CfSemi-enclosed fuse — a BS 3036 rewirable fuse is less accurate than a CB or cartridge fuse, so the cable needs extra headroom.Always 0.725 (a constant from Appendix F of the OSG).Only when the protective device is a BS 3036 semi-enclosed (rewirable) fuse. For MCBs, HBC fuses and cartridge fuses, Cf = 1.
If a correction factor does not apply, treat it as 1. Dividing by 1 has no effect. For example, if the ambient temperature is 30 °C (Ca = 1), the cable is not grouped (Cg = 1), it is clear of insulation (Ci = 1), and the device is an MCB (Cf = 1), then It = In ÷ 1 = In — no de-rating needed.

3. Selecting the Cable — How to Use the OSG Tables

Once you have calculated It, you need to find a cable whose tabulated current rating is ≥ It. The table you use depends on the installation method:

Which OSG table do I use for cable ratings?

Table F4 — for cables in conduit or trunking (Methods A and B), or clipped direct (Method C). This is the most commonly used table.

Table F6 — for PVC flat-profile (twin & earth) cables in insulated walls or surrounded by insulation (Method 100/101). Use this for domestic loft/wall situations.

In both tables, find the column for your installation method and read down to find the first cable size whose rating is ≥ It. That is your minimum cable size.

Important: compare the current rating values across different methods in the quick-reference table below. A 2.5 mm² cable carries 27 A when clipped direct (Method C) but only 20 A in an insulated wall (Method A). The installation method directly determines whether a cable size is adequate — same cable, different rating.

4. Calculating Voltage Drop (Vd) — the Second Check

Selecting a cable that carries enough current is only half the job. The designer must also check that the voltage arriving at the load is close enough to 230 V. Over a long cable run, resistance causes a voltage loss — this is the voltage drop.

Vd = (mV/A/m × Ib × L) / 1000

mV/A/m = millivolt drop per amp per metre — a physical constant of the cable size, found in the last column of OSG Table F4 or F6 (the same row as the cable you selected).

Ib = the design current (the actual load, not the device rating In).

L = the cable length in metres (the actual route, including vertical runs).

We divide by 1000 because the mV/A/m value is in millivolts, but we need the answer in Volts.

Key point: the mV/A/m value is a property of the cable size — it is not affected by correction factors. Correction factors change the required current rating (It) and therefore may force you to pick a larger cable, but the mV/A/m you use in the voltage drop formula always comes from the cable you have actually selected.

Quick Reference — Twin & Earth Cable Data (from OSG Tables F4 & F6)

This table summarises the key data you need for cable selection and voltage drop. The current-carrying capacity columns show how the same cable has different ratings depending on how it is installed. The mV/A/m column is the value you put into the voltage drop formula.

Cable Size (mm²)Method C — Clipped Direct (Amps)Method B — In Conduit/Trunking (Amps)Method A — In Insulated Wall (Amps)mV/A/m (Voltage Drop Factor)
1.0161311.544
1.52016.514.529
2.527232018
4.037302611
6.04738327.3
10.06450344.4
16.08568462.8
As the cross-sectional area increases, the mV/A/m value decreases — thicker cables have lower resistance and therefore less voltage drop. This is why increasing the cable size is the standard fix when the voltage drop calculation fails. A cable that passes the current-rating check may still need to be upsized to pass the voltage drop check — both criteria must be satisfied before the cable is accepted.
📝 Voltage Drop — Worked Examples
📐 Example 1: Method C — Clipped Direct (Ambient Temp.)

Question: A 230 V radial supplies a 16 A load (Ib) using a 20 A MCB (In). The 20 m cable is Clipped Direct (Method C) at 40 °C.

1
Identify correction factors: The ambient temperature is above 30 °C, so we need Ca. From OSG Table F1, look up 40 °C for 70 °C PVC → Ca = 0.87. No grouping, no insulation, MCB not BS 3036, so Cg = Ci = Cf = 1.
2
Calculate It:
It = 20 / 0.87 = 22.98 A
This means the cable must have a tabulated rating of at least 22.98 A under Method C conditions.
3
Select Cable from OSG Table F4 — find the Method C column: 1.5 mm² = 20 A (too small — less than 22.98 A). Select 2.5 mm² = 27 A (first cable ≥ It).
4
Calculate Vd — from the same table row, the mV/A/m for 2.5 mm² is 18. Use Ib (the actual load, not In):
Vd = (18 × 16 × 20) / 1000 = 5.76 V
✓ PASS — 5.76 V is within the 11.5 V limit (5% for power)
📐 Example 2: Method B — In Trunking (Grouping)

Question: A 32 A Ring (Ib = 25 A) is in Trunking (Method B) grouped with two other circuits. Length is 25 m.

1
Identify correction factors: 3 circuits grouped together. From OSG Table F3, look up 3 circuits → Cg = 0.70. Ambient is 30 °C (Ca = 1), no insulation (Ci = 1), MCB not BS 3036 (Cf = 1).
2
Calculate It:
It = 32 / 0.70 = 45.71 A
The grouping factor has pushed It well above In — the cables are trapping each other's heat.
3
Select Cable from OSG Table F4 — find the Method B column: 6 mm² = 38 A (too small). Select 10 mm² = 50 A (first cable ≥ 45.71 A).
4
Calculate Vd — mV/A/m for 10 mm² = 4.4. Use Ib = 25 A (the actual load):
Vd = (4.4 × 25 × 25) / 1000 = 2.75 V
✓ PASS — 2.75 V is well within the 11.5 V limit
Notice that grouping forced us from 6 mm² up to 10 mm². The 10 mm² cable has a much lower mV/A/m (4.4 vs 7.3), so the voltage drop is easily met. Sometimes a correction factor that forces a larger cable actually solves the voltage drop problem at the same time.
📐 Example 3: Method 101 — In Thermal Insulation

Question: A 3 A lighting load (Ib) uses a 6 A MCB (In). The 30 m cable is in a loft, touching a ceiling, and is totally surrounded by 150 mm of insulation (Method 101).

1
Identify correction factors: The cable is surrounded by insulation for more than 0.5 m. From OSG Table F2, the factor for cable totally surrounded by insulation > 0.5 m → Ci = 0.50. This is the worst case — the cable's capacity is halved.
2
Calculate It:
It = 6 / 0.50 = 12 A
Even though the load is only 3 A and the MCB is 6 A, the cable needs to be rated for at least 12 A because the insulation traps heat.
3
Select Cable from OSG Table F6 — this table is specifically for PVC flat-profile (twin & earth) cables in insulated walls or loft insulation: 1.5 mm² = 10.5 A (too small). Select 2.5 mm² = 13.5 A.
4
Calculate Vd — mV/A/m for 2.5 mm² = 18. This is a lighting circuit, so the limit is 6.9 V (3%):
Vd = (18 × 3 × 30) / 1000 = 1.62 V
✓ PASS — 1.62 V is well within the 6.9 V limit (3% for lighting)
📐 Example 4: Method C — Rewirable Fuse (BS 3036)

Question: A 10 A heater (Ib) uses a 15 A BS 3036 fuse. The 40 m cable is Clipped Direct (Method C).

1
Calculate It (BS 3036 → Cf = 0.725):
It = 15 / 0.725 = 20.68 A
2
Select Cable — Table F4 (Method C): Select 2.5 mm² (handles 27 A).
3
Calculate Vd:
Vd = (18 × 10 × 40) / 1000 = 7.2 V
✓ PASS — 7.2 V is within the 11.5 V limit
📐 Example 5: Method A — Multiple Factors (Ca + Cg)

Question: A 32 A cooker (Ib = 28 A) is in Conduit in an insulated wall (Method A). Ambient temp is 35 °C and it is grouped with one other circuit for 15 m.

1
Calculate It (Ca = 0.94 at 35 °C, Cg = 0.80 for 2 circuits):
It = 32 / (0.94 × 0.80) = 32 / 0.752 = 42.55 A
2
Select Cable — OSG Table F4 (Method A): 10 mm² handles 34 A (too small). Select 16 mm² (handles 43 A).
3
Calculate Vd:
Vd = (2.8 × 28 × 15) / 1000 = 1.18 V
✓ PASS — 1.18 V is well within the 11.5 V limit
If your calculated voltage drop is too high, the most common solution is to increase the cross-sectional area of the cable (e.g. jump from 2.5 mm² to 4 mm²). This lowers the mV/A/m value and therefore reduces the voltage drop.
🔒 Step 8 — Shock Protection (Earth Fault Loop Impedance)

What Is Shock Protection?

Shock protection ensures that if a fault occurs — for example a live conductor touches the metal casing of an appliance — the protective device disconnects the supply quickly enough to prevent a lethal electric shock. The key question the designer must answer is: is the earth fault loop impedance low enough for the device to trip within the required time?

The Principle in Plain English

When a fault occurs, the fault current flows in a loop — from the transformer, through the line conductor (R1), through the fault, back through the CPC (R2), through the main earthing terminal, and back to the transformer via the supply earth (Ze). The total impedance of this loop is Zs. The lower Zs is, the higher the fault current, and the faster the protective device trips.

The Formula — Zs

Zs = Ze + (R1 + R2)

Ze = external earth fault loop impedance (the supply side — obtained from the electricity distributor or by measurement).

R1 = resistance of the line conductor for the cable length of the circuit.

R2 = resistance of the circuit protective conductor (CPC) for the cable length of the circuit.

Where Does Each Value Come From?

ValueSourceHow to Find It
ZeElectricity distributor or measurementAsk the DNO, or use the typical worst-case values: TN-S = 0.8 Ω, TN-C-S (PME) = 0.35 Ω. For TT systems the value depends on the earth electrode.
R1 + R2 per metreOSG Table I1Look up the line conductor size (left column) against the CPC size (top row). The table gives the combined resistance in mΩ/m (milliohms per metre) at 20 °C. Multiply by the cable length to get the total.
Temperature correctionOSG Table I3Under fault conditions the cable heats up, increasing its resistance. Multiply the (R1 + R2) value by the factor from Table I3 — for 70 °C PVC cable this factor is 1.20.
Maximum permitted ZsOSG Appendix BLook up the type and rating of the protective device. For Type B MCBs use Table B6. The table gives the maximum Zs value — your calculated Zs must be equal to or less than this value.
OSG vs BS 7671: The maximum Zs values in the On-Site Guide and Guidance Note 3 can be read directly — they already include a safety margin. If you use the tables in BS 7671 instead (Tables 41.2 and 41.3), you must multiply the tabulated value by 0.8 (i.e. your measured Zs must be ≤ 80% of the BS 7671 figure).

Maximum Disconnection Times (Reg 411.3.2.2)

Circuit TypeTN System (230 V)TT System (230 V)
Final circuit ≤ 32 A0.4 s0.2 s
Final circuit > 32 A / Distribution5 s1 s

Maximum Zs Values for Type B MCBs (OSG Table B6)

MCB Rating (A)61016202532404550
Max Zs (Ω)7.284.372.732.191.751.371.090.880.87

Key R1 + R2 Values at 20 °C (OSG Table I1 — extract)

Line Conductor (mm²)CPC (mm²)R1 + R2 (mΩ/m)
1.01.036.20
1.51.030.20
1.51.524.20
2.51.025.51
2.51.519.51
2.52.514.82
4.01.516.71
4.02.512.02
6.02.510.49
10.04.06.44
16.06.04.23
PVC flat-profile (twin & earth) cable has a CPC that is smaller than the line conductor — for example, 2.5 mm² cable comes with a 1.5 mm² CPC. You must use the correct line/CPC combination from Table I1, not assume they are the same size. The CPC size for each cable can be found in OSG Table 7.1(i).

Worked Examples — Shock Protection

📐 Example 1: 20 A Radial Socket Circuit (TN-C-S Supply)

Scenario: A 20 A Type B MCB protects a radial socket circuit wired in 2.5 mm² PVC flat cable (which incorporates a 1.5 mm² CPC). The cable length is 30 m. The supply is TN-C-S (PME), so Ze = 0.35 Ω.

1
Find R1 + R2 per metre — from OSG Table I1, look up 2.5 mm² line with 1.5 mm² CPC → 19.51 mΩ/m.
2
Calculate R1 + R2 for the cable length:
19.51 × 10⁻³ × 30 = 0.585 Ω
3
Apply the temperature correction — from OSG Table I3, for 70 °C PVC the factor is 1.20:
0.585 × 1.20 = 0.702 Ω
This accounts for the cable heating up during a fault.
4
Calculate Zs:
Zs = 0.35 + 0.702 = 1.052 Ω
5
Check against the maximum permitted value — from OSG Table B6, the maximum Zs for a 20 A Type B MCB is 2.19 Ω.
✓ PASS — 1.052 Ω is less than 2.19 Ω. The protective device will disconnect within 0.4 s.
📐 Example 2: 45 A Kiln Circuit — Shock Protection FAILS

Scenario: A 10 kW kiln (Ib = 43.47 A) is protected by a 45 A Type B MCB. The cable is 10 mm² PVC flat cable (with a 4 mm² CPC), run for 15 m. The supply is TN-C-S and Zs has been measured on site at 0.81 Ω.

1
Find the maximum permitted Zs — from OSG Table B6, for a 45 A Type B MCB: 0.88 Ω.
2
Compare — measured Zs = 0.81 Ω. Is 0.81 ≤ 0.88?
0.81 Ω ≤ 0.88 Ω ✓ — just within the limit
✓ PASS — 0.81 Ω is just within the 0.88 Ω maximum. But note how tight this is.
What if Zs had been 0.91 Ω instead? Then 0.91 > 0.88 and shock protection would FAIL. In that case the designer would need to: shorten the cable route, increase the CPC size (to lower R2), or install an RCD to achieve the required disconnection time by a different means. This is exactly the scenario in the textbook's kiln exercise — it shows that passing voltage drop does not guarantee passing shock protection.
📐 Example 3: Lighting Circuit on a TN-S Supply

Scenario: A 6 A Type B MCB protects a lighting circuit wired in 1.5 mm² PVC flat cable (with a 1.0 mm² CPC). The cable length is 25 m. The supply is TN-S, so Ze = 0.8 Ω.

1
Find R1 + R2 per metre — from OSG Table I1: 1.5 mm² line with 1.0 mm² CPC → 30.20 mΩ/m.
2
Calculate R1 + R2:
30.20 × 10⁻³ × 25 = 0.755 Ω
3
Temperature correction (Table I3, factor 1.20):
0.755 × 1.20 = 0.906 Ω
4
Calculate Zs:
Zs = 0.8 + 0.906 = 1.706 Ω
5
Check — from OSG Table B6, max Zs for a 6 A Type B MCB = 7.28 Ω.
✓ PASS — 1.706 Ω is well within 7.28 Ω.
Smaller MCB ratings have much higher permitted Zs values (a 6 A MCB allows 7.28 Ω, but a 45 A MCB allows only 0.88 Ω). This is because a small MCB needs less fault current to trip, so it can tolerate higher loop impedance. Larger circuits with bigger MCBs are the ones most likely to fail the shock protection check.

Calculating Prospective Fault Current (PFC) from Zs

PFC = V / Zs

V = nominal voltage (230 V). Zs = total earth fault loop impedance.

Example: if Zs = 1.052 Ω → PFC = 230 ÷ 1.052 = 218.6 A.

The highest PFC at the origin of the installation should be recorded and compared against the breaking capacity of every protective device. For a typical 230 V single-phase domestic supply, the worst-case PFC is normally up to 16 kA.

🌡️ Step 9 — Thermal Constraint (Adiabatic Equation)

What Is Thermal Constraint?

When an earth fault occurs, a very large current flows through the CPC for a short time — until the protective device disconnects. During that time, the CPC heats up rapidly. Thermal constraint is a calculation to check whether the CPC is big enough to survive this heating without its insulation being damaged.

Why Is This Particularly Important?

In PVC flat-profile (twin & earth) cable, the CPC is smaller than the line conductors. For example:

• 1.0 mm² cable → 1.0 mm² CPC
• 1.5 mm² cable → 1.0 mm² CPC
• 2.5 mm² cable → 1.5 mm² CPC
• 4.0 mm² cable → 1.5 mm² CPC
• 6.0 mm² cable → 2.5 mm² CPC
• 10.0 mm² cable → 4.0 mm² CPC
• 16.0 mm² cable → 6.0 mm² CPC

These CPC sizes are from OSG Table 7.1(i). Because the CPC is the thinnest conductor in the cable, it heats up the fastest under fault conditions — so we must check it can cope.

The Adiabatic Equation

S = √(I² × t) / k

S = minimum cross-sectional area of the CPC (mm²) needed to survive the fault.

I = the fault current in Amperes (the prospective fault current, or the current that will flow through the CPC under fault conditions).

t = the operating time of the protective device in seconds (how long the fault current flows before the device trips).

k = a material constant for the type of conductor and insulation.

Where Does Each Value Come From?

ValueSourceHow to Find It
I (fault current)Calculated from Zs, or given in the questionI = V ÷ Zs (e.g. 230 ÷ 1.15 = 200 A). Or the PFC may be measured on site or stated in the question.
t (disconnection time)Reg 411.3.2.2 or the device's time-current curveFor a final circuit ≤ 32 A, use 0.4 s. For circuits > 32 A on a TN system, use 5 s. For specific fault currents, the time can be read from the device's time-current characteristic curve in Appendix 3 of BS 7671 (e.g. Fig. 3A4 for MCBs, Fig. 3A2 for BS 3036 fuses).
k (material constant)BS 7671 Table 54.3For a copper conductor with 70 °C PVC insulation (the standard domestic cable), k = 115. For 90 °C thermosetting insulation, k = 143.
Actual CPC sizeOSG Table 7.1(i)Look up the cable size under the "Cable size (mm²)" column to find the CPC size. For example, 2.5 mm² flat cable has a 1.5 mm² CPC.

The Decision

Calculate S using the adiabatic equation. Then compare S with the actual CPC size in the cable:

• If the actual CPC ≥ S → thermal constraint is MET ✓

• If the actual CPC < S → thermal constraint is NOT MET ✗ — a larger cable or a separate, larger CPC is needed.

Worked Examples — Thermal Constraint

📐 Example 1: 2.5 mm² Ring Circuit with BS 3036 Fuse

Scenario: A 230 V ring main is wired in 2.5 mm² PVC cable with a separate 1.5 mm² CPC in plastic conduit. The circuit is protected by a 30 A semi-enclosed fuse (BS 3036). An earth fault loop impedance test gives Zs = 1.15 Ω.

1
Calculate the fault current (I):
I = V / Zs = 230 / 1.15 = 200 A
2
Find the disconnection time (t) — this is a final circuit ≤ 32 A, so the maximum disconnection time is 0.4 s (Reg 411.3.2.2). From the BS 3036 time-current curve (Fig. 3A2(a) of BS 7671 Appendix 3), a 30 A fuse clears 200 A in approximately 0.4 s.
3
Identify k — copper conductor with 70 °C PVC insulation → k = 115 (from BS 7671 Table 54.3).
4
Calculate the minimum CPC size (S):
S = √(200² × 0.4) / 115
S = √(40000 × 0.4) / 115
S = √16000 / 115
S = 126.49 / 115 = 1.10 mm²
5
Compare — actual CPC = 1.5 mm². Is 1.5 ≥ 1.10?
✓ PASS — 1.5 mm² CPC is greater than the minimum 1.10 mm² required. The CPC can safely carry the fault current.
📐 Example 2: 2.5 mm² Immersion Heater Circuit with BS 3036 Fuse

Scenario: A TN supply feeds a domestic immersion heater wired in 2.5 mm² PVC cable incorporating a 1.5 mm² CPC. The circuit is protected by a 16 A semi-enclosed fuse (BS 3036).

1
Find the fault current (I) — from the BS 3036 time-current curve (Fig. 3A2(a)), a current of about 90 A will trip a 16 A fuse in 0.4 s.
2
Values: I = 90 A, t = 0.4 s, k = 115.
3
Calculate S:
S = √(90² × 0.4) / 115
S = √(8100 × 0.4) / 115
S = √3240 / 115
S = 56.92 / 115 = 0.49 mm²
4
Compare — actual CPC = 1.5 mm². Is 1.5 ≥ 0.49?
✓ PASS — 1.5 mm² CPC comfortably exceeds the 0.49 mm² minimum.
With a smaller fuse (16 A vs 30 A), the fault current needed to trip the device is much lower (90 A vs 200 A), so the CPC needs to be much smaller. Thermal constraint is easier to pass when the protective device trips at a lower current.
📐 Example 3: 6 mm² Cable with High PFC — Thermal Constraint FAILS

Scenario: A 6 mm² PVC flat-profile cable has a 2.5 mm² CPC (from OSG Table 7.1(i)). The PFC of the circuit is 1.7 kA (1700 A) and the circuit must disconnect in 0.4 s. k = 115.

1
Values: I = 1700 A, t = 0.4 s, k = 115.
2
Calculate S:
S = √(1700² × 0.4) / 115
S = √(2,890,000 × 0.4) / 115
S = √1,156,000 / 115
S = 1075.17 / 115 = 9.35 mm²
3
Compare — actual CPC = 2.5 mm². Is 2.5 ≥ 9.35?
✗ FAIL — The CPC needs to be at least 9.35 mm², but the cable only has 2.5 mm². The CPC would overheat and its insulation could be damaged or catch fire.
Why does this fail? The fault current is very high (1700 A) and it flows for 0.4 s — that is a lot of energy (I²t) forced through a thin 2.5 mm² conductor. The designer would need to either use a cable with a much larger CPC (e.g. a 16 mm² cable with a 6 mm² CPC), install a separate larger CPC alongside the cable, or reduce the fault current by increasing Zs. This example comes directly from the textbook and demonstrates why thermal constraint is a critical final check — the cable may carry the load current and pass voltage drop, but the CPC cannot survive a fault.

Summary — The Complete 9-Step Design Check

All 9 steps must pass before the design is accepted:
  1. Ib — Design current calculated ✓
  2. In — Protective device selected (In ≥ Ib) ✓
  3. Installation method identified ✓
  4. Correction factors applied, It calculated ✓
  5. Cable size selected (tabulated rating ≥ It) ✓
  6. Voltage drop calculated ✓
  7. Voltage drop within limits (3% lighting / 5% power) ✓
  8. Shock protection — Zs ≤ tabulated maximum ✓
  9. Thermal constraint — actual CPC ≥ calculated S ✓

Failing any one step means the design must be revised. The most common fixes are: increasing the cable size (helps steps 5–7), shortening the cable route (helps steps 7–8), or installing an RCD (helps step 8 on TT systems).

🧪 Voltage Drop — Multiple Choice Quiz

Test your knowledge on voltage drop and cable sizing. Select an answer then check your results.

0/20
Correct Answers
📊 Maximum Demand & Diversity (OSG Appendix A)

What is Diversity?

Diversity recognises that not all appliances in a building will operate at full power at the same time. For example, in a typical home, it is highly unlikely that every light, the oven, the shower, and all plug-in heaters will all be switched on simultaneously.

OSG Appendix A provides diversity factors based on years of industry experience that allow us to estimate a realistic maximum demand rather than an oversized theoretical total.

Why Use Diversity?

① Prevents over-engineering — avoids unnecessarily large, expensive cables and switchgear.
② Reflects realistic usage patterns — load is spread over time.
③ Balances safety and economy — the installation is safe but not wastefully oversized.

Key Diversity Rules (Domestic)

Circuit TypeDiversity Rule
Cooking ApplianceFirst 10 A + 30% of remainder + 5 A if socket on cooker unit
Lighting66% of total connected load
Socket Outlets100% of largest circuit + 40% of each additional circuit
Space Heating100% of largest + 75% of remainder
📐 Example 1: Cooking Appliance (with socket)

Scenario: A 32 A rated electric cooker with a socket on the control unit.

1
Take the first 10 A.
2
Remaining current: 32 − 10 = 22 A
3
30% of remainder: 22 × 0.3 = 6.6 A
4
Add 5 A for the socket on the cooker unit.
Assumed Demand: 10 + 6.6 + 5 = 21.6 A
Even though the cooker could draw 32 A, we assume only about 21.6 A because thermostats will cycle the heating elements on and off.
📐 Example 2: Lighting Circuits

Scenario: A house has 20 LED downlights, each rated at 10 W. Total load = 200 W.

1
Find the current: I = P / V = 200 / 230 = 0.87 A
2
Apply 66% diversity: 0.87 × 0.66 = 0.57 A
Assumed Demand: 0.57 A
📐 Example 3: Socket Outlets (Multiple Circuits)

Scenario: A house with three 32 A ring final circuits (Kitchen, Downstairs, Upstairs).

1
Circuit 1 (Kitchen — largest): 100% = 32 A
2
Circuit 2 (Downstairs): 40% × 32 = 12.8 A
3
Circuit 3 (Upstairs): 40% × 32 = 12.8 A
Assumed Demand: 32 + 12.8 + 12.8 = 57.6 A
Without diversity: 32 + 32 + 32 = 96 A. Diversity acknowledges you won't be boiling three kettles and running three vacuums in different rooms at the same second!

Practice Questions

❓ Practice 1: 45 A Cooker (No Socket)

Task: Using the standard cooker diversity rule, calculate the assumed demand for a 45 A rated electric range cooker. The cooker control unit does not have a socket outlet.

Reveal Answer
1
Take the first 10 A.
2
Remainder: 45 − 10 = 35 A
3
30% of remainder: 35 × 0.3 = 10.5 A
4
No socket, so nothing extra to add.
Assumed Demand: 10 + 10.5 = 20.5 A
❓ Practice 2: Small Studio Flat — Total Demand

Task: Calculate the total assumed demand for a flat with: one 32 A Ring (sockets), one 32 A Radial (additional sockets), and one 6 A lighting circuit (actual load 4 A).

Reveal Answer
1
Socket Circuit 1 (largest): 100% = 32 A
2
Socket Circuit 2: 40% × 32 = 12.8 A
3
Lighting (66% of actual load, not breaker size): 4 × 0.66 = 2.64 A
Total Assumed Demand: 32 + 12.8 + 2.64 = 47.44 A
Without diversity the total would be 32 + 32 + 6 = 70 A. Many older UK flats have a 60 A main fuse — without diversity it looks inadequate, but with diversity the 47.44 A load is perfectly safe.
🛡️ Overcurrent Protection & Protective Devices

Types of Faults

Overload Fault
Circuit is electrically sound, but too many appliances draw more current than the circuit is designed for.
Short Circuit Fault
An overcurrent resulting from a fault of small resistance between two live conductors at different potentials under normal conditions.
Earth Fault
Current flows between an exposed or extraneous conductive part and earth via the earth path to the protective device.
In-Rush Current
A brief, high surge of current when equipment first starts — common with motors, transformers, and discharge lighting.

Tap / click each card to reveal the definition

Key Definitions

Fuse: Provides local protection of the circuit from overload. Used in domestic and similar premises.

Basic Protection: A physical barrier between a person / livestock and a live part.
Fault Protection: Current flowing between exposed/extraneous conductive parts and earth — the earth path allows this current to reach the protective device.

Overcurrent Selection Rule — The Golden Triangle

Ib ≤ In ≤ Iz

Ib (design current) ≤ In (device rating) ≤ Iz (cable's current-carrying capacity under installed conditions).

In plain English: the device must be big enough to carry the load without nuisance tripping, but small enough to protect the cable from overheating. This is Step 2 of the 9-step design process — and the It calculation (Step 4) is how you verify that Iz is adequate.

Prospective Fault Current (PFC)

Ipf = U₀ / Ze
TNS System — Ze = 0.8 Ω
PFC = 230 / 0.8 = 287.5 A ≈ 290 A
TNC-S System — Ze = 0.35 Ω
PFC = 230 / 0.35 = 657 A ≈ 660 A

Protective Devices Comparison

DeviceAdvantagesDisadvantages
Re-wireable Fuse
(BS 3036)		 No moving parts · Cheap · Low replacement cost Incorrect element size can be fitted · Slow to repair · Poor breaking capacity
Cartridge Fuse
(BS 1361)		 No moving parts · Small physical size · Accurate current rating Incorrect cartridge can be fitted · Not suitable for high fault current · Can be shorted out
MCB
(BS EN 60898)		 Tamper proof · Supply quickly restored · Pre-set tripping characteristics Expensive · Regular testing required · Has moving parts

MCB Types

TypeTrip RangeBest Suited For
Type B3–5 × InDomestic ring mains, lighting, sockets
Type C5–10 × InDischarge lighting in large premises (e.g. supermarket)
Type D10–20 × InWelding and X-ray machines (large in-rush currents)

RCD & RCBO

RCD
Protects against earth faults only. Measures incoming vs outgoing current and trips if there is a difference.
RCBO
Combines overload protection AND earth fault protection in one device.
An RCD does not protect against overload or short-circuit current — it only protects against earth fault current. That's why RCBOs are increasingly preferred.

Discrimination

For effective discrimination, protective devices should be sized from smallest at the load to largest at the supply: e.g. 13 A at the socket → 32 A at the distribution board → 100 A at the main switch.

🔌 Standard Ring & Radial Final Circuits

Ring vs. Radial — Key Difference

Ring Final Circuit (A1)
Phase, neutral and CPC all loop out from the CU, through each socket, and return to the CU.
Radial Circuit (A2 / A3)
Phase, neutral and CPC loop out from the CU, through each socket, and do not return to the CU.

Circuit Specifications (BS 7671)

SpecA1 — RingA2 — Radial (20 A)A3 — Radial (32 A)
MCB Size32 A20 A32 A
Live Cable2.5 mm²2.5 mm²4 mm²
Max Floor Area100 m²50 m²75 m²
Max SocketsNo limit——

Spurs & Additional Rules

RuleDetail
Non-fused spurMay only feed one single or twin socket outlet
Fused spursUnlimited number allowed from a ring
Fused spur fuse (BS 1362)Must not exceed 13 A
Permanently connected equipmentProtected by breaker not exceeding 16 A
Cooker control unit distanceWithin 2 m of the cooker
Cooker control unit positionMust not be located directly above the unit
Water heater (> 15 litres)Must be on its own separate circuit
Socket/switch heights (new dwellings)450 mm to 1200 mm from floor
RCD protection for sockets < 20 A30 mA RCD required

Recommended Socket Outlets per Room

RoomSmall (<12 m²)Medium (12–25 m²)Large (>25 m²)
Living Room468
Dining Room345
Single Bedroom234
Double Bedroom345
Study456
Kitchen6810
Bathroom000
📏 Trunking Size Calculation

The Rule

Sum of Cable Factors ≤ Trunking Factor

Look up each cable's factor from the tables (e.g. Table 3.4 / OSG Table E3), multiply by the quantity, sum them, then pick a trunking size whose factor meets or exceeds that sum (Table 3.5 / OSG Table E5).

📐 Example 1: Small Load

Requirement: 10 × 1.5 mm² solid PVC + 5 × 2.5 mm² solid PVC.

1
Find factors: 1.5 mm² solid = 8.0 · 2.5 mm² solid = 11.9
2
Calculate sum:
(10 × 8.0) + (5 × 11.9) = 80 + 59.5 = 139.5
3
Select Trunking: We need a factor ≥ 139.5.
Minimum Size: 50 × 38 mm (Factor 767)
📐 Example 2: Medium Load

Requirement: 30 × 4.0 mm² stranded PVC + 20 × 6.0 mm² stranded PVC.

1
Find factors: 4.0 mm² stranded = 16.6 · 6.0 mm² stranded = 21.2
2
Calculate sum:
(30 × 16.6) + (20 × 21.2) = 498 + 424 = 922
3
Select Trunking: We need a factor ≥ 922.
Minimum Size: 100 × 25 mm (Factor 993)
📐 Example 3: Larger Load

Requirement: 50 × 2.5 mm² stranded + 40 × 4.0 mm² stranded + 20 × 10 mm² stranded PVC.

1
Find factors: 2.5 mm² stranded = 12.6 · 4.0 mm² stranded = 16.6 · 10 mm² stranded = 35.3
2
Calculate sum:
(50 × 12.6) + (40 × 16.6) + (20 × 35.3)
= 630 + 664 + 706 = 2000
3
Select Trunking: We need a factor ≥ 2000.
Minimum Size: 100 × 50 mm (Factor 2091)
📚 OSG Appendix Revision — Quick-Fire Quiz

Test yourself across all OSG Appendices (A–F). Select your answer then check your results.

0/20
Correct Answers

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For educational use only. Content aligned to BS 7671 18th Edition & On-Site Guide.