12.2 Voltage Drop Principles & Calculations

12.2 Voltage Drop Principles & Calculations

When electric current flows through an installation cable, it encounters the internal electrical resistance of the metallic conductors. By Ohm's law ($V = I \times R$), this resistance produces an internal drop in electrical potential along the length of both the outgoing line conductor and the returning neutral conductor. As a result, the terminal voltage available at the point of utilization (such as a luminaire, electric motor, or appliance socket) is lower than the nominal supply voltage present at the origin of the consumer unit or distribution board.

Controlling voltage drop within strict regulatory limits is essential for electrical safety, energy efficiency, and equipment longevity.

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Detrimental Effects of Excessive Voltage Drop

Allowing cables to operate with excessive voltage drop impairs the operational efficiency and lifespan of connected loads:

• Lighting Luminaires: In incandescent and halogen lamps, a small voltage drop causes a substantial decrease in luminous output (lumen output falls by approximately 3% for every 1% drop in voltage). In modern LED and fluorescent luminaires, under-voltage causes high-frequency driver flicker, stroboscopic effects, or complete failure to strike.
• Electric Motors: The starting and running torque developed by an induction motor is proportional to the square of the applied voltage ($T \propto V^2$). Excessive voltage drop causes prolonged motor starting periods, increased stator winding currents, severe overheating, and premature tripping of thermal overloads.
• Electronic Control Gear & Computers: Microprocessor equipment, programmable logic controllers (PLCs), contactor coils, and sensitive instrumentation may malfunction, suffer data corruption, or shut down when supply potential drops below minimum tolerance thresholds.
• Heating Appliances: Thermal power output decreases as the square of the voltage ($P = \frac{V^2}{R}$). An immersion heater or electric shower operating under severe volt drop will produce substantially less hot water, increasing energy run times.

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Permitted Voltage Drop Limits under BS 7671

BS 7671 Regulation 525 and Appendix 4 set the maximum voltage drop, between the origin of the installation and any load point, that is deemed to satisfy the regulations.

For low-voltage installations supplied directly from a public distribution network (with nominal voltages of $230\text{ V}$ single-phase and $400\text{ V}$ three-phase), the maximum allowable limits are:

For private high-voltage supplies (where the consumer owns the supply substation and HV/LV step-down transformer), the limits are relaxed to 6.0% for lighting ($13.8\text{ V}$) and 8.0% for other uses ($18.4\text{ V}$). Most domestic, commercial and small industrial installations are supplied from the public network, so the 3% and 5% values are the ones you will design to.

                  ┌──────────────────────────────────────────────────┐
                  │      BS 7671 Maximum Permitted Voltage Drop      │
                  │           (Public Low-Voltage Supply)            │
                  └─────────────────────────┬────────────────────────┘
                                            │
               ┌────────────────────────────┴────────────────────────────┐
               ▼                                                         ▼
     Lighting Circuits (3.0%)                                   Power Circuits (5.0%)
     • 230 V single-phase: 6.9 V                                • 230 V single-phase: 11.5 V
     • 400 V three-phase: 12.0 V                                • 400 V three-phase: 20.0 V

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Tabulated Millivolt per Ampere per Meter ($mV/A/m$)

To make voltage drop calculations straightforward on site, BS 7671 and the On-Site Guide publish pre-calculated millivolt drop figures known as $(mV/A/m)$ values.

What Does $(mV/A/m)$ Mean?

The $(mV/A/m)$ rating defines the potential drop in millivolts that will occur across both line and neutral conductors when an electric current of 1 Ampere travels through a cable run length of 1 Meter.

Because electrical resistance is inversely proportional to conductor cross-sectional area ($R = \frac{\rho L}{A}$), larger conductors have lower resistance, resulting in smaller $(mV/A/m)$ ratings:

Values from BS 7671 Table 4D5 (reproduced in the On-Site Guide) for single-phase circuits at normal conductor operating temperature.

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The Voltage Drop Calculation Formula

The actual voltage drop ($V_d$) across any cable run is calculated using the standard formula:

$V_d = \frac{(mV/A/m) \times I_b \times L}{1000}$

Where:

• $V_d$ = Actual voltage drop along the circuit run in Volts (V).
• $(mV/A/m)$ = Tabulated millivolt drop figure from BS 7671 / On-Site Guide tables for the specific cable conductor size.
• $I_b$ = Circuit design current in Amperes (A).
• $L$ = One-way route length of the circuit run in meters (m) (measured from the consumer unit to the furthest load terminal).
• $1000$ = Mathematical divisor converting millivolts (mV) to Volts (V).

Percentage Voltage Drop

To express the calculated volt drop as a percentage of the nominal supply voltage ($V_0 = 230\text{ V}$):

$\% V_d = \frac{V_d}{V_0} \times 100\% = \frac{V_d}{230} \times 100\%$

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Verification & Upsizing Methodology

Once $V_d$ is calculated, it must be evaluated against the permitted threshold:

1. If $V_d \le V_{d(\text{permitted})}$, the cable conductor cross-sectional area satisfies BS 7671 voltage drop criteria.
2. If $V_d > V_{d(\text{permitted})}$, the cable is non-compliant. The installer must upsize the conductor cross-sectional area to the next standard manufactured size with a lower $(mV/A/m)$ value and re-run the calculation until $V_d$ falls within the permitted limit.

Maximum Permissible Circuit Length ($L_{\text{max}}$)

By transposing the voltage drop formula, an electrician can determine the maximum distance a circuit can extend before exceeding the voltage drop limit:

$L_{\text{max}} = \frac{V_{d(\text{perm})} \times 1000}{(mV/A/m) \times I_b}$

For a 230 V single-phase power circuit with a maximum volt drop of $11.5\text{ V}$:

$L_{\text{max}} = \frac{11500}{(mV/A/m) \times I_b}$

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Worked Calculation Examples

Example 1: Workshop Radial Power Circuit

Scenario: A dedicated 230 V single-phase radial socket-outlet circuit is being installed to feed machinery in a detached garage workshop. The circuit design current is $I_b = 20\text{ A}$, protected by a 20 A Type B MCB. The physical route length from the house consumer unit to the workshop socket is $L = 36\text{ m}$. The installer initially selects $2.5\text{ mm}^2$ flat twin and earth cable ($mV/A/m = 18$).

1. Determine Permitted Voltage Drop: For a power circuit supplied at 230 V, permitted drop is $5\%$: $V_{d(\text{perm})} = 230\text{ V} \times 0.05 = 11.5\text{ V}$

2. Calculate Voltage Drop for $2.5\text{ mm}^2$ Conductor: $V_d = \frac{18 \times 20\text{ A} \times 36\text{ m}}{1000} = \frac{12960}{1000} = 12.96\text{ V}$

3. Evaluate Compliance: $12.96\text{ V} > 11.5\text{ V} \quad \text{--- FAILS BS 7671!}$ The $2.5\text{ mm}^2$ cable produces a $5.63\%$ voltage drop, exceeding the $5.0\%$ limit.

4. Upsize Conductor to $4.0\text{ mm}^2$: From Table 4D5, a $4.0\text{ mm}^2$ cable has $(mV/A/m) = 11$. $V_d = \frac{11 \times 20\text{ A} \times 36\text{ m}}{1000} = \frac{7920}{1000} = 7.92\text{ V}$

5. Re-evaluate Compliance: $7.92\text{ V} \le 11.5\text{ V} \quad \text{--- PASSES!}$ $\% V_d = \frac{7.92}{230} \times 100\% = 3.44\%$ The $4.0\text{ mm}^2$ cable complies fully with BS 7671 Regulation 525.

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Example 2: Commercial Office Lighting Circuit

Scenario: A 230 V commercial office LED lighting installation has a total connected design load of $1800\text{ W}$ operating at unity power factor ($\cos\phi = 1.0$). The one-way run length from the distribution board to the furthest luminaire is $L = 32\text{ m}$. The specification calls for $1.5\text{ mm}^2$ cable ($mV/A/m = 29$). Determine if this cable meets BS 7671 requirements.

1. Calculate Design Current ($I_b$): $I_b = \frac{P}{V} = \frac{1800\text{ W}}{230\text{ V}} = 7.83\text{ A}$

2. Determine Permitted Voltage Drop: For a lighting circuit on a public supply, permitted drop is $3\%$: $V_{d(\text{perm})} = 230\text{ V} \times 0.03 = 6.90\text{ V}$

3. Calculate Voltage Drop for $1.5\text{ mm}^2$ Conductor: $V_d = \frac{29 \times 7.83\text{ A} \times 32\text{ m}}{1000} = \frac{7266.24}{1000} = 7.27\text{ V}$

4. Evaluate Compliance: $7.27\text{ V} > 6.90\text{ V} \quad \text{--- FAILS BS 7671!}$ Although it exceeds the limit by only $0.37\text{ V}$ (giving $3.16\%$), the design does not meet the limit and must be changed.

5. Upsize Conductor to $2.5\text{ mm}^2$: From Table 4D5, a $2.5\text{ mm}^2$ cable has $(mV/A/m) = 18$. $V_d = \frac{18 \times 7.83\text{ A} \times 32\text{ m}}{1000} = \frac{4510.08}{1000} = 4.51\text{ V}$

6. Re-evaluate Compliance: $4.51\text{ V} \le 6.90\text{ V} \quad \text{--- PASSES!}$ $\% V_d = \frac{4.51}{230} \times 100\% = 1.96\%$ The $2.5\text{ mm}^2$ cable is compliant, ensuring no luminaire flicker or premature driver failure.