Applied Math - Occupations - Electrician

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Introduction

Disclaimer: The author of this article is not a licensed electrician. No content in this article should be put into practice unless researched and confirmed with the local appropriate electrical codes.

Electricians need to know some mathematics in order to make decisions about the size of wiring, the power capacity of a system, and in other applications.

Voltage is often described as the pressure in an electrical system. High voltage means there is significant pressure on the electrons to move in a system.

Voltage alone cannot move electrons to produce power. Current is a measure of the flow of electrons through an electrical circuit. If the voltage is high but the current is low, the power of the system might be low because the electrons cannot move as well as they could in a high current system.

The power of a system is the voltage multiplied by the current.

\[P = V \cdot I\]

In the equation above, \(P\) is power, \(V\) is voltage, and \(I\) is current. Power is typically measured in the unit of "watts", but sometimes British Thermal Units (BTUs) are used in heating and air conditioning. Voltage is measured in volts. Current is typically measured in Ampere units, or A for short.

If you substitute \(V = 100\) volts and \(I = 2\) amperes in the above equation, then the power is \(P = 200\) volt-amperes. Note the units are volt-amperes and not volts per amperes, as we are multiplying volts by amperes.

As electricity flows through a wire, cable, or some other component, the voltage drops over the distance. If voltage is low, machinery can overheat, devices can shut down, etc.

The formula for voltage drop, called ohm’s law, is

\[V = R \cdot I\]

where voltage is \(V\), \(I\) is current, and \(R\) is the resistance, measured in ohms. A more expanded version of this formula is

\[V = \frac{2 \cdot L \cdot I \cdot R}{C}\]

where \(L\) is the length if the circuit (which includes the wiring), and \(C\) is, roughly speaking, the thickness of the wire.

Example:

1. A 120-volt, 12 kVA (kilovolt amps) generator can be loaded on to a maximum of how many amperes per line?

If we substitute these two numbers into the power equation, including the units, we will have

\[P = V \cdot I\]

\[12 \textbf{ kilovolt-amperes} = 120 \textbf{ volts} \cdot I\]

Note that the voltage units are close but are not the same. Before we perform any math, we should convert kilovolts to volts. "kilo" means one thousand, as 12 kilovolt-amperes is 12,000 volt-amperes.

\[12,000 \textbf{ volt-amperes} = 120 \textbf{ volts} \cdot I\]

To solve for \(I\), we do the opposite of what 120 is doing and divide 120 from both sides.

\[\frac{12,000 \textbf{ volt-amperes}}{120 \textbf{ volts}} = \frac{120 \textbf{ volts}}{120 \textbf{ volts}} \cdot I\]

\[100 \textbf{ amperes} = I\]

The maximum current is 100 amperes, according to this formula.

2. The National Electrical Code (NEC) recommends that electrical systems be chosen so that the branch circuit (the electrical wiring that delivers power to outlets or devices in a room) and feeders (the wiring that supplies powers to branch circuits) have a combined maximum voltage drop of no more than 5%. What is the minimum recommended voltage for a 240 volt load?

According to the NEC, we should plan for a 5% voltage drop. 5% of 240 volts is

\[240 \cdot 0.05 = 12\]

so in order to avoid low voltage problems, the minimum supplied voltage should be \(240\) volts - \(12\)volts dropped = \(228\) volts.

3. A 15 amperes breaker is loaded with a 10 amperes load. What percentage is the breaker loaded?

10 out of 15 amperes are "loaded". To calculate a percentage, divide the amount you’re interested in (in this case, the loaded amperes) by the starting amount (15 amperes) and multiply by 100.

\[\% = \frac{10}{15} \cdot 100\]

4. A 60-ampere, 240-volt, single-phase load is 250 feet from a panel board and is supplied with a copper wire. The copper wire has a resistance of 12.9 ohms. The copper wire has a circular mil area of 52,620. What is the voltage drop?

Using the voltage drop equation mentioned above,

\[V = \frac{2 \cdot L \cdot I \cdot R}{C}\]

We substitute \(L = 250\), \(I = 60\), \(R = 12.9\), and \(C = 52,620\) into the formula.

\[V = \frac{2 \cdot 250 \cdot 60 \cdot 12.9}{52,620}\]

\[V = \frac{387000}{52,620}\]

\[V = 7.355\]

The voltage drop should be expected to be a little more than 7 volts.

5. It is said that a breaker under continuous load should not be loaded over 80% of its circuit rating.

There are several large, industrial lights at a job site. They require 200 watts each. The voltage being supplied is 120 volts. The circuit connecting the lights is 20 amps. How many of these lights can be connected to the circuit without loading the breaker over 80%?

80% of the total power supply is

\[P = V \cdot I\]

\[P = 120 \textbf{ volts} \cdot 20 \textbf{ amperes} \cdot 0.80\]

\[P = 1,920\] watts

Each light requires 200 watts. That means we can have

\[1,920 \div 200 = 9.6\]

but we round down for safety - 9 lights are possible.

Practice Problems

1. A 240-volt, 12 kVA (kilovolt amps) generator can be loaded on to a maximum of how many amperes per line?

2. The National Electrical Code (NEC) recommends that electrical systems be chosen so that the branch circuit (the electrical wiring that delivers power to outlets or devices in a room) and feeders (the wiring that supplies powers to branch circuits) have a combined maximum voltage drop of no more than 5%. What is the minimum recommended voltage for a 120 volt load?

3. A 20 amperes breaker is loaded with a 15 amperes load. What percentage is the breaker loaded?

4. An 40-ampere, 120-volt, single-phase load is 300 feet from a panel board and is supplied with a copper wire. The copper wire has a resistance of 12.9 ohms. The copper wire has a circular mil area of 52,620. What is the voltage drop?

5. It is said that a breaker under continuous load should not be loaded over 80% of its circuit rating.

There are several large, industrial lights at a job site. They require 400 watts each. The voltage being supplied is 120 volts. The circuit connecting the lights is 30 amps. How many of these lights can be connected to the circuit without loading the breaker over 80%?

Theory Questions

1. What is the relationship between power and current? Is it linear, quadratic, etc.?

2. If the power stays constant, what is the relationship between voltage and current? In other words, if the power of a system is 100 watts and the voltage increases by 2 volts, what happens to the current?

3. If the voltage stays constant, what is the relationship between power and current? In other words, if the voltage is 12 volts and the power has increased, what must have happened to the current?