Steve Humphreys, Technical Commercial Manager at NAPIT, stresses the importance of understanding and applying electrical formula.
Electrical formulae and its application can be an area that apprentices, trainees and electricians can find both baffling and frustrating. However, it is essential in everything we do, from selecting the correctly sized protective devices, to designing and installing complex electrical installations.
I think back to my teaching days as an electrical lecturer and remember the look of dread on the learners’ faces as we covered science and principles involving maths subjects, such as algebra and trigonometry.
The availability of software applications for things such as circuit design have certainly made life easier, but it is extremely important to understand the basic building blocks of electrical formula to help us learn and become better electricians and designers.
Back to basics
Probably the most common and well-known formula is Ohm’s Law. In 1826, Georg Simon Ohm published details of an experiment in which he investigated the relationship between current, voltage and resistance. The formula is easily remembered by using the Ohms Law triangle as displayed in Fig 1.
Ohm’s Law states that if you have two of the values you can always find the third. In order to see this, we can use the following example:
Example:
“We have a circuit where the current is 32 A and we have a resistance of 0.5 Ω. What is the voltage?”
V = I x R, which gives us:
V = 32 x 0.5 = 16 V
Power triangle
The same method can be applied when we need to calculate the power in a circuit. We use a slightly different triangle that contains the values power in watts (W), current in amps (I) and voltage (V) as shown in Fig 2.
Example:
“We have a circuit where the power is 3000 W and the current is 13 A. What is the voltage?”
V = P ÷ I, which gives:
V = 3000 ÷ 13 = 230 V
The power triangle is particularly useful for finding out the current of a particular appliance and then sizing a fuse accordingly.
We can also calculate the values of voltage, current, resistance and power using the handy formula wheel shown in Fig 3.
Whilst handy triangles and wheels can help us with formulae, it’s always good to be able to do it ourselves using transposition. There are many ways of transposing a formula but the simplest way is to use the opposite rule.
Rule 1: The opposite of addition (+) is subtraction (-)
Rule 2: The opposite of multiplication (×) is division (÷)
Rule 3: The opposite of squaring (x2) is square root (√)
Rule 4: The most important rule to remember is whatever you do to one side you need to do to the other side i.e. either side of the equals sign.
We can now apply this rule to find any value in a formula as shown below in the following examples.
Example 1:
“As we saw earlier Ohm’s Law states that V = I x R. However, what if we need to find the current (I)?”
Step 1: Firstly, we need to get I by itself, so we simply put I first in the rearranged formula: I=
Step 2: Next, we can move the original value that was before the equals sign (or on its own) to the other side of the equals sign, in this case it was V.
I=V
Step 3: Lastly, we take the remaining value, which is R, and apply opposite rule 2. This means that we move it from the top (multiply) to the bottom (divide) and underneath V:
I=VorI=V÷R R
Example 2:
A harder example would be to find b from the formula: a2 +b2 =c
Step 1: Firstly, we need to get b² by itself, so we simply put it first in the rearranged formula. b² =
Step 2: Next, we can move the original value that was before the equals sign (or on its own) to the other side of the equals sign, in this case it was c: b2 = c
Step 3: Next, we take the remaining value, which is a², and apply opposite rule 1. This means that the + (addition) now becomes – (minus): b² =c-a²
There is one more step to do because we need to find b not b².
Step 4: Therefore, using the opposite rule 3 to remove the square from b² we apply square root. However, whatever we do to one side, we need to do to the other side to give: b = √c – a²
So far, we haven’t really applied these formulae practically to help us in our day-to-day work. Let’s look at a worked example using the adiabatic equation.
The adiabatic equation is very useful for finding a protective conductor size required for a circuit if we do not use the tabulated values in BS 7671.
It is important to establish the correct size of our protective conductor to ensure it has a large enough cross-sectional area (CSA). This is because it needs to carry any fault current to disconnect our protective device before the cable is damaged.
Adiabatic equation
S is the size or CSA of the conductor in mm² (see Fig 4)
t is the time in seconds
I is the fault current in amps
k is a factor for temperature and resistivity etc. for different types of cable
However, for this example, let’s say that we already know the size of the conductor but we want to know the time it takes for the conductor to reach its limiting temperature (normally 70 ̊C for polyvinyl chloride (PVC) thermoplastic cable).
Firstly, we have to transpose or rearrange the original formula and get t on its own.
Example 3:
Step 1: We need to get t by itself. To make life easier it would be better to get rid of the k from the bottom right-hand side first. We do this by multiplying k on the right-hand side, which cancels out, but remember that we need to also do it on the left-hand side: S x k = √I²t
Step 2: Next, we can get rid of the square root on the right-hand side by squaring it by itself and then do the same on the left-hand side: S² xk² =I²t
Step 3: Next, we can get rid of I² on the right-hand side by dividing it by itself and then do the same on the left-hand side, so we end up with:
Let’s now apply some real-life values for a final circuit on a TN system with a maximum disconnection time of 0.4 seconds as follows:
S = 4 mm² (CSA of the circuit protective conductor (CPC))
k = factor is 115
I = fault current is 650 amps
t= 
t= 
t = 0.5 seconds
So, we can see that our CPC will carry the fault current for 0.5 seconds before it is damaged. As our protective device will operate in less than that, within 0.4 seconds, the CPC size is satisfactory.
Conclusion
Hopefully, you can now see the value of formulae within electrical installations and circuit design. The maths side of working out circuit design doesn’t need to be something to shy away from.
By using a logical and methodical approach, electrical formulae and calculations can be a very useful tool to meet the requirements of BS 7671 and ensure that our electrical installations are safe and compliant.
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