The transformer transforms the energy from one electrical circuit to another. This occurs by using electromagnetic induction. It is known as an efficient voltage converter, and it can reduce the high voltage to low voltage and vice versa. A goodcondition transformer consists of two windings, which primary winding and secondary are winding. There are two types of the transformer which are step up and step down transformers. In this article, we will discuss the transformer formula with examples. Let us learn the concept!
Transformer Formula
What is a Transformer?
The transformer is an electrical device that allows us to increase or decrease the voltage in an alternating current electrical circuit, maintaining power. The power that enters the equipment, in the case of an ideal transformer, is equal to that obtained at the output.
Real machines have a small percentage of losses. It is a device that converts the alternating electrical energy of a certain voltage level into alternating energy of another voltage level, based on the phenomenon of electromagnetic induction.
It is made up of two coils of conductive material, wound on a closed nucleus of ferromagnetic material, but electrically isolated from each other. The only connection between the coils is the common magnetic flux established in the core. The coils are called primary and secondary according to the input or output of the system in question, respectively.
Transformer Formula
The value of the power for an electric circuit is the value of the voltage by the value of the current intensity. As in the case of a transformer, the value of the power in the primary is the same value for the power in the secondary we have:
(input voltage on the primary coil)Â Ã— (input current on the primary coil ) = (output voltage on the secondary coil )Â Ã— (output current on the secondary coil)
This can be written as an equation:
\(V_p\) Ã— \(I_p\) = \(V_s\) Ã— \(I_s\)
We can also work out the transformer output voltage if we know the input voltage and the number of turns on the primary and secondary coils.
\(\frac{input voltage on the primary coil}{ output voltage on the secondary coil}\) = \(\frac {number of turns of wire on the primary coil} { number of turns of wire on the secondary coil}\)
This can be written as an equation:
\(\frac {V_p}{V_s}\) = \(\frac {n_p}{n_s}\)
Where,
\(V_p\)  input voltage on the primary coil. 
\(V_s\)  input voltage on the secondary coil. 
\(I_p\)  input current on the primary coil.

\(I_s\)  input current on the secondary coil. 
\(n_p\)  the number of turns of wire on the primary coil. 
\(n_s\)  the number of turns of wire on the secondary coil. 
Solved Examples
Q. 1: We have a transformer with a current in the primary coil of 10 A and input voltage in the primary coil of 120 V, if the voltage in the output of the secondary coil is 50 V, calculate the current in the output of the secondary coil.
Solution: As we have to determine the output current in the secondary coil, we will the first equation i.e.
\(V_p \)Ã— \(I_p\) = \(V_s\) Ã— \(I_s\)
As given,
\(V_p\) = 120 V
\(V_s\)= 50 V
\(I_p\) = 10 A
Putting values given in the question:
\(I_s\) = \(\frac{V_p}{V_s}\)Â Ã— \(I_p\)
\(I_s\) =\(\frac{120}{50}\) Ã— 10
= 24 A
Therefore, current in the output of the secondary coil is 24 A.
Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s
I have realy intrested to to this topic
m=f/a correct this
Interesting studies
It is already correct f= ma by second newton formula…