Before we learn how to drop voltage calculations, we must understand what voltage drop is.
As current flows through a conductor or wire, its flow is inherently resisted by the wire itself, this is called
impedance. The impedance caused by the wire is determined by the physical properties of the wire; these physical properties include material type, cross-sectional area, and length. The increase in impedance causes or voltage drop.
How did we jump from impedance to voltage dropping?
How are these two things related? Recall that impedance is measured in ohms
(a measurement of resistance) and also recall that voltage, resistance, and current are related (ohm's law).
Let's think about the formula V = I x R. If I increase R and have to keep I the same, what must happen to V in
order for the equation to remain true (left side of the equal sign the same as the right side)?
V must decrease.
Now that voltage drop makes a list more theoretical sense, let's examine one of its practical implications.
Looking at the diagram alone, assuming it is to scale, what can we assume about the voltage drop to be
calculated at A and B? The voltage drop is bigger at B, why? Because it is further away from the 120V source.
Recall that the NEC (National Electrical Code) limits the amount of voltage drop allowed at the furthest point of a circuit.
Calculating Voltage Drop
Before attempting to solve the problem, make sure to have the following information:
- Voltage (at the source of the circuit)
- Current in Amperes
- Length of wire
- Resistance factors for 1,000 feet of wire at 167 degrees F (75 degrees C) aluminum 21.2, copper 12.9
- Kemil of wire used (see table 8, chapter 9 in NEC)
In single-phase circuits use
Voltage Drop = 2 x Length x Current x Resistance factor / Kemil
In three-phase circuits use
Voltage Drop = 1.732 x Length x Current x Resistance factor / Kemil
Notice that if you are asked to find the maximum length of wire for a given voltage drop you just need to solve for length on the equations alone.
If asked to find the maximum current, solve for it in the equations alone.
*If single-phase, use 2. If three-phase, use 1.73
*If single-phase, use 2. If three-phase, use 1.73.
How are the resistance factor for aluminum and copper calculated?
You were given a resistance factor (per 1000 ft of wire) of 21.2 for aluminum and 12.9 for copper
We did so by
- Identifying the resistances (KFT) for 1000 -Kemil
Copper = 0.0129 Ω per 1000'
= 0.0000129 Ω per 1'
Aluminum = 0.0212 Ω per 1000'
= 0.0000212 Ω per 1'
See table 8, Chapter 9 in the NEC
- Recalling that 1 kemil is equal to 1000 circular mills this means there are 1 million circular mills in 1000 kemil
- Multiplying the resistance of the conductor per foot times the circular mills of the conductor.
Copper = 0.0000129 x 1 million Aluminum = 0.0000212 x 1 million
Copper = 12.9 Aluminum = 21.2