Understanding Ohm’s Law
To fully understand the principles behind Ohm’s Law the electrician must learn the relationships between current (I), resistance (R), and voltage (E). The three basic formula for ohm’s law is I = E ÷ R, R = E ÷ I, and E = I x R. However, it will not do you any good if you do not know what these letters mean, so let us examine them in detail.
Voltage is expressed as (E), sometimes also as (V) on various exams. The “voltage” expressed as E is the pressure that is required to force (1) ampere through a resistance value of (1) ohm. To get “current to flow”, which is the movement of electrons in a conductive material, such as copper or aluminum, you need a force to move those electrons. By applying voltage (E) the force is applied creating something called Electromotive Force, otherwise known as EMF. Remember, the voltage, expressed as (E) in our formula is simply the force that drives the movement of electrons.
Ampere is expressed as (I), sometimes also as (A) on various exams. (I) is electrical current, expressed in amperes, that will flow through (1) ohm under the voltage (V) pressure of (1) volt. The (I) also can be referenced as “Intensity”, but more often it is simply expressed as Amps or Amperes. The amount of current (intensity) that flows through a conductive material, such as a copper or aluminum conductor, is measured in amperes. We can measure this intensity using an Amprobe measurement device.
Resistance is expressed as (R), referred to as Resistance, and sometimes followed by the Greek Omega symbol Ω. Resistance represents opposition to current (I) flow. The resistance or opposition to current (I) flow can be attributed to many factors. The length, conductive material, temperature, as well as the overall cross-sectional area of the conductive material can have an overall negative contribution to current (I) flow.
There is one additional value that we need to discuss. While it is not part of a traditional ohm’s law calculation it is often found in ohm’s law wheels and is quite valuable for the electrician to be aware of. In all electrical circuits, you have various loads that consume POWER so naturally, we need a value for power, which is expressed in watts, and thus you will see the symbol (W) or (P) used to express this value.
Power is expressed as (W) or (P), referred to as Watts. Power is equating to the electrical energy that is delivered and ultimately consumed, the formula utilizing the (W) or (P) value helps express that power value and overall consumption.
When using the National Electrical Code®, electricians express power in volt-amperes or VA for short.
However, as you know there are other expressions of power you will encounter, such as kW (kilowatts), kVA (kilo-volt-amperes). In most situations, and for exams it holds true, the electrician would convert kW into VA for the overall calculation. For example, 8kW would be 8000 watts, since the k stands for 1,000, which is then converted to 8000 VA (volt-amperes) or 8kVA depending on what the electrical exam question is looking for. For most service and feeder calculations it is best to always convert the values down to the volt-ampere (VA) value.
In your career, you will hear other electricians refer to a term called “proportionality”. Knowing that that word means in relationship to using ohm’s law is beneficial to the electrician and likewise on electrical exams.
When the current (I) flows on a conductor, such as a copper conductor, it is directly proportional to the voltage that is applied to the conductor. The current (I), expressed in amperes, is inversely proportional to the conductor’s resistance of the given circuit.
Principals of Proportionality Examined.
When a voltage (E) is increased, and resistance (R) remains the same, the current (I) will increase proportionately.
When resistance (R) is increased, and voltage (E) remains the same, the current will decrease proportionately.
For example, If the voltage is 120 V, and the current in amperes is 10 amps, and the resistance of 12 ohms.
If the voltage is increased to 240 V, the resistance remains unchanged, the current will jump to 20 amperes.
Now, if you increase the resistance in the same example, say to 24 ohms, and the voltage remains the same at 120 V, the current, in amperes will be decreased to 5 amperes.
Let look at what we call an Ohm’s Law Wheel.
Note: unless expressed otherwise, assume that all power factors are at 100% Unity. Just remember that Power Factor is only an issue in Alternating Current (A/C) systems and not to be considered for Direct Current (D/C) systems. Do not worry we will talk about Power Factors later in another article.
Using an ohm’s where (Formula Pie) is rather simple. You will need no less than (2) values to solve an equation. Once you have two values, the electrician can utilize them to resolve the final value and answer your question. As with all exams make sure you read the question multiple times to make sure you are answering the question in a manner that is being desired on the exam.
As you can see by the ohm’s law wheel, we have a lot of formulas to work with. Just remember we need at least two values to solve a question. So, let us do a few examples and see how it breaks down.
A 6 AWG THWN conductor is acceptable for how many kW at a value of 120 volts.
Step 1 – The question desires an answer in kW. Do not make the mistake of answering in watts when the question is asking for the answer in kW. Now, we need to determine what formula from the ohm’s law wheel you are going to use. In this case, we are solving for (P) watts we would use W = E x I.
Step 2 – Since we need no less than two values, we have the voltage (120 V) but no (I) Current, in amperes. To find this current value the electrician goes to Table 310.16, under the 75°C Column and selects the ampacity for the 6 AWG Copper, always assume copper unless your exam states otherwise, is 65 amperes.
Step 3 – Resolve using the formula. 120 (E) x 65 (I) = 7,800 watts. However, be careful on your exam since the question wanted the answer in kW. To convert this, take 7,800 ÷ 1000 = 7.8 kW.
There are many ways to use the Ohm’s Law Wheel, the choice of the formula to use is directly relevant to the information that is being provided in the electrical exam question. Let us look at another example.
How much current (I) is being drawn from a specific load of 75 watts at a 115-volt source?
Step 1 – In this example, we are given (2) values to work with. You would select your formula based on what you have been presented. You would select (I) = W ÷ E.
Step 2 – Input your known factors and solve – 75 (W) ÷ 115 (E) = 0.6521 Amperes (I)
How about this one, the same 75 watts (W) of load rated at 115 volts (E) but now has a source voltage of 120 volts (E). Now, the wattage of the load is 75 watts but because the source voltage (120 V) differs from the load's rated voltage (115 V) the actual watts will be different at 120 volts.
Step 1 – The electrician needs to find the resistance value of the actual 75-watt load. Selection of the proper formula would be using R = E²÷ W. Why this formula? It happens to be the only one we can use since we have no current (I) values shown and we are solving for resistance (R).
115 x 115 ÷ 75 =176.33 ohms
Step 2 – Now that you have the resistance of the load, you can find the current draw the question is asking for.
E ÷ R = I
120 (E) ÷ 176.33 (R) = 0.6805 (I) in amperes
Note: The principles of proportionality taking place, the amperes at 115 volts were 0.6521 in the previous example but at 120 volts it increased to 0.6805. The resistance did not change while the other factors, in this case, voltage changed.
Now, assume the exam question then asked you to express the power being consumed in the above example.
Step 1 – Pick the right formula, in this case, W = E² ÷ R would be the correct choice.
Step 2 – 120 x 120 (E²) ÷ 176.33 (R) = 81.6650 watts (P) of actual consumption.
Again, it is important to state the obvious, be sure to read the question on your exam closely so that you can answer the question appropriately.
Looking for a quicker way to solve equations without remembering all the values of the Ohm’s Law Wheel? Here are some smaller, easier to remember “PIE” and “EIR” diagrams that can be used as well.
Simply cover up the unknown value with your thumb and either multiple or divide, as necessary.
E = I x R
I = E ÷ R
R = E ÷ I
P = I x E
I = P ÷ E
E = P ÷ I
Here are a few additional examples to practice on, use the formulas to solve.
Question: If the current is 10 amperes and the resistance is 12 ohms, what is the voltage?
Step 1 – Using the formulas in the examples above, cover the unknown value, which is (E). That would result in I x R = E.
Step 2 – 10 (I) x 12 (R) = 120 volts is the answer.
Since each smaller “PIE” has three (3) formulas and “EIR” has three (3) formulas, you can combine them depending on the values given to answer any ohm’s law question. Let us try one.
Question: What is the resistance of an electric radiant heater rated at 1400 watts at 120 volts?
Step 1 – Since the question is looking for resistance but you are provided with watts, use the “PIE” to solve for (I) first. Select the right formula, which is Watts (P) ÷ Volts (E) = (I) Amperes.
Step 2 – 1400 Watts (P) ÷ 120 Volts (E) = 11.66 rounded to 11.67 Amperes (I).
Step 3 – Now since you have the ampere value you can now switch to the “EIR” to solve for resistance. The formula to use is Resistance (R) = Volts (E) ÷ Amperes (I).
Step 4 – 120 Volts (E) ÷ 11.67 Amperes (I) = 10.28 Ohms (R), which is 1400 watts (P).
The goal of this lesson is to help the electrical student better understand the principles of Ohms Law and how they apply to real-world situations. If you desire to learn more about the National Electrical Code please consider signing up for our Fast Trax™ NEC Learning Program. Check out www.masterthenec.com/fasttraxprogram
Article By: Paul W Abernathy, CEO, and Founder of Electrical Code Academy, Inc.
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