Voltage drop
Voltage drop
This is an excerpt from the sizing and protection of conductors chapter in The Electrical Installation Guide. You can download this exerpt which looks at the determination of voltage drop by simply downloading the sizing and protection of conductors chapter by clicking here.
Determination of voltage drop
The impedance of circuit conductors is low but not negligible: when carrying load current there is a voltage drop between the origin of the circuit and the load terminals. The correct operation of a load (a motor, lighting circuit, etc.) depends on the voltage at its terminals being maintained at a value close to its rated value. It is necessary therefore to determine the circuit conductors such that at full-load current, the load terminal voltage is maintained within the limits required for correct performance.
This section deals with methods of determining voltage drops, in order to check that:
- They comply with the particular standards and regulations in force
- They can be tolerated by the load
- They satisfy the essential operational requirements
Maximum voltage drop
Maximum allowable voltage-drop vary from one country to another. Typical values for
LV installations are given below in Figure G25.
These voltage-drop limits refer to normal steady-state operating conditions and do
not apply at times of motor starting, simultaneous switching (by chance) of several
loads, etc. as mentioned in Chapter A Sub-clause 4.3 (factor of simultaneity, etc.).
When voltage drops exceed the values shown in Figure G25, larger cables (wires)
must be used to correct the condition.
The value of 8%, while permitted, can lead to problems for motor loads; for example:
- In general, satisfactory motor performance requires a voltage within ± 5% of its rated nominal value in steady-state operation,
- Starting current of a motor can be 5 to 7 times its full-load value (or even higher). If an 8% voltage drop occurs at full-load current, then a drop of 40% or more will occur during start-up. In such conditions the motor will either:
- Stall (i.e. remain stationary due to insufficient torque to overcome the load torque) with consequent over-heating and eventual trip-out
- Or accelerate very slowly, so that the heavy current loading (with possibly undesirable low-voltage effects on other equipment) will continue beyond the normal start-up period
- Finally an 8% voltage drop represents a continuous power loss, which, for continuous loads will be a significant waste of (metered) energy. For these reasons it is recommended that the maximum value of 8% in steady operating conditions should not be reached on circuits which are sensitive to under-voltage problems (see Fig. G26).
Calculation of voltage drop in steady load conditions
Use of formulae
Figure G27 below gives formulae commonly used to calculate voltage drop in a given circuit per kilometre of length.
If:
- IB: The full load current in amps
- L: Length of the cable in kilometres
- R: Resistance of the cable conductor in Ω/km
Note: R is negligible above a c.s.a. of 500 mm2
- X: inductive reactance of a conductor in Ω/km
Note: X is negligible for conductors of c.s.a. less than 50 mm2. In the absence of any other information, take X as being equal to 0.08 Ω/km. -
: phase angle between voltage and current in the circuit considered, generally:
- Incandescent lighting: cos = 1
- Motor power:
- At start-up: cos = 0.35
- In normal service: cos = 0.8
- Un: phase-to-phase voltage
- Vn: phase-to-neutral voltage
For prefabricated pre-wired ducts and bustrunking, resistance and inductive reactance values are given by the manufacturer.
Simplified table
Calculations may be avoided by using Figure G28 next page, which gives, with an adequate approximation, the phase-to-phase voltage drop per km of cable per ampere, in terms of:
- Kinds of circuit use: motor circuits with cos close to 0.8, or lighting with a cos close to 1.
- Type of cable; single-phase or 3-phase
K x IB x L
K is given by the table,
IB is the full-load current in amps,
L is the length of cable in km.
The column motor power “cos = 0.35” of Figure G28 may be used to compute the
voltage drop occurring during the start-up period of a motor (see example no. 1 after
the Figure G28).
*You might also find this voltage calculator from NoOutage useful.
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