Physics Lesson 16.17.3 - How does a Transformer Work?

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Welcome to our Physics lesson on How does a Transformer Work?, this is the third lesson of our suite of physics lessons covering the topic of Power in an Alternating Circuit. Transformers, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

How does a Transformer Work?

An ideal transformer contains a rectangular core (empty at middle) with two coils wound around the opposing arms. The core is made of insulating material to prevent the flow of undesired currents (Foucault currents). The coil connected with the input is called the primary coil while the one connected with the output is called secondary coil.

Physics Tutorials: This image provides visual information for the physics tutorial Power in an Alternating Circuit. Transformers

As seen from the figures, step-up transformers have more turn in the secondary coil while step-down transformers have more turns in the primary coil. The alternating current flowing through the primary coil produces a variable magnetic flux in the core. Since the number of turns is different, the flux produced in the secondary coil changes, generating an induced emf in the secondary coil (in the primary coil the emf is due to the input source). However, the frequency of current (and voltage) doesn't change. This is an indicator of the relationship between currents and voltages in each coil of transformer.

The symbol of transformers in electric circuits is Physics Tutorials: This image provides visual information for the physics tutorial Power in an Alternating Circuit. Transformers

An ideal transformer has no any power loss during the induction from primary to secondary coil. However, this is not possible. Some of the energy (power) is lost during this process despite the coils are made of soft iron to reduce this loss. Hence, we can write

P1 ≥ P2

where 1 stands for input (primary coil) and 2 for output (secondary coil). The equality sign stands for ideal transformers. Thus, we obtain for the efficiency of transformer:

e = P2/P1 × 100%

Obviously, an ideal transformer is 100% efficient as no energy is lost during the induction.

We will explain the rest of transformers theory considering only ideal transformers. Thus, based on the power formula shown above, the relationship between currents and voltages in an ideal transformer is

i1 ∙ ∆V1 = i2 ∙ ∆V2

The relationship between the emf's and the number of turns in each coil is

ε1/ε2 = N1/N2

where N1 and N2 are the numbers of turns in the primary and secondary coil respectively. The ratio N1/N2 is called the turns factor.

For ideal transformers, the emf's in each coil are equal to the corresponding voltages at their ends. Hence, we can write

i1/i2 = ∆V2/∆V1

Combining all the above equations, we obtain for an ideal transformer

i1/i2 = ε2/ε1 = ∆V2/∆V1 = N2/N1

This means the current is inversely proportional to electromotive force, voltage and number of turns in the transformer.

The equation

∆V2/∆V1 = N2/N1

is known as the equation of transformer.

Example 2

The voltage in the primary coil of a transformer is 220 V and the number of turns in this coil is 1000. If the number of turns in the secondary coil is 50, find:

  1. What type of transformer is this?
  2. The voltage in the secondary coil is the transformer is ideal
  3. The voltage in the secondary coil if the transformer is 80% efficient
  4. The current in the secondary coil for both situations above if the input current is 4.4 A.

Solution 2

  1. This is a step-down transformer as the number of turns in the secondary coil is smaller than in the primary coil (N1 = 1000, N2 = 50).
  2. If the transformer is ideal, we have
    ∆V2/∆V1 = N2/N1
    Substituting the known values, we obtain
    ∆V2/220 V = 50/1000
    Thus,
    ∆V2 = 50 ∙ 220 V/1000
    = 11 V
  3. If the transformer is 80% efficient, we obtain for the output voltage
    ∆V1 = 80% ∙ ∆V2(ideal) = 0.80 ∙ 11V = 8.8 V
  4. For ideal transformer we have
    i1 ∙ ∆V1 = i2 ∙ ∆V2
    Since i1 = 4.4 A, we have
    (4.4 A) ∙ (220 V) = i2 ∙ (11 V)
    i2 = (4.4 A) ∙ (220 V)/(11 V)
    = 88 A
    Since we have considered the efficiency when finding the output voltage, it is obvious that the current remains the same even in the real transformer. Hence, we have
    i2 = 88 A

You have reached the end of Physics lesson 16.17.3 How does a Transformer Work?. There are 3 lessons in this physics tutorial covering Power in an Alternating Circuit. Transformers, you can access all the lessons from this tutorial below.

More Power in an Alternating Circuit. Transformers Lessons and Learning Resources

Magnetism Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
16.17Power in an Alternating Circuit. Transformers
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
16.17.1Electric Power as a Rate of Electric Energy Transfer
16.17.2Transformers
16.17.3How does a Transformer Work?

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