IGCSEPhysicsEdexcelPhysics (Modular)Unit 2Revision Notes7. Magnetism & ElectromagnetismElectromagnetic InductionTransformer Equations

Transformer Equations (Edexcel IGCSE Physics (Modular))

Revision Note

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Ashika

Written by: Ashika

Reviewed by: Caroline Carroll

The transformer equation

  • The output potential difference (voltage) of a transformer depends on:

    • The number of turns on the primary and secondary coils

    • The input potential difference (voltage)

  • It can be calculated using the transformer equation below:

fraction numerator potential space difference space across space primary space coil over denominator potential space difference space across space secondary space coil end fraction space equals space fraction numerator number space of space turns space on space primary space coil over denominator number space of space turns space on space secondary space coil end fraction

  • This equation for transformers can be written using symbols as follows:

V subscript p over V subscript s space equals space n subscript p over n subscript s

  • Where

    • Vp = potential difference (voltage) across the primary coil in volts (V)

    • Vs = potential difference (voltage) across the secondary coil in volts (V)

    • np = number of turns on primary coil

    • ns = number of turns on secondary coil

  • The transformer equation above can be flipped upside down to give:

V subscript s over V subscript p space equals space n subscript s over n subscript p

  • The equations above show that:

    • The ratio of the potential differences across the primary and secondary coils of a transformer is equal to the ratio of the number of turns on each coil

Step-up transformer

  • A step-up transformer increases the potential difference of a power source

  • A step-up transformer has more turns on the secondary coil than on the primary coil (Ns > Np)

Step-down transformer

  • A step-down transformer decreases the potential difference of a power source

  • A step-down transformer has fewer turns on the secondary coil than on the primary coil (Ns < Np)

Worked Example

A transformer has 20 turns on the primary coil and 800 turns on the secondary coil. The input potential difference across the primary coil is 500 V.

a) Calculate the output potential difference

b) State what type of transformer it is

Part (a)

Step 1: List the known quantities

  • Number of turns in primary coil, Np = 20 

  • Number of turns in secondary coil, Ns = 800

  • Voltage in primary coil, Vp = 500 V

Step 2: Write out the transformer equation

  • Use the version with the secondary coil quantities on the top to minimise the amount of rearranging

N subscript s over N subscript p space equals space V subscript s over V subscript p

Step 3: Rearrange for Vs

V subscript s space equals space N subscript s over N subscript p space cross times space V subscript p

Step 4: Substitute values into the equation

V subscript s space equals space 800 over 20 space cross times space 500 space equals space 20 space 000 space straight V

Part (b)

  • The transformer is a step-up transformer

  • This is because the transformer has:

    • More secondary coils

    • A greater secondary voltage

Examiner Tips and Tricks

When you are using the transformer equation make sure you have used the same letter (p or s) in the numerators (top line) of the fraction and the same letter (p or s) in the denominators (bottom line) of the fraction.There will be less rearranging to do in a calculation if the variable which you are trying to find is on the numerator (top line) of the fraction.The individual loops of wire going around each side of the transformer should be referred to as turns and not coils.

The ideal transformer equation

  • An ideal transformer would be 100% efficient

    • Although transformers can increase the voltage of a power source, due to the law of conservation of energy, they cannot increase the power output

  • If a transformer is 100% efficient:

Input space power space equals space Output space power

  • The equation to calculate electrical power is:

P space equals space V space cross times space I

  • Where:

    • P = power in Watts (W)

    • V = potential difference in volts (V)

    • I = current in amps (A)

  • Therefore, if a transformer is 100% efficient then:

V subscript p space cross times space I subscript p space equals space V subscript s space cross times space I subscript s

  • Where:

    • Vp = potential difference across primary coil in volts (V)

    • Ip = current through primary coil in Amps (A)

    • Vs = potential difference across secondary coil in volts (V)

    • Is = current through secondary coil in Amps (A)

  • The equation above could also be written as:

P subscript s space equals space V subscript p space cross times space I subscript p

  • Where:

    • Ps = output power (power produced in the secondary coil) in Watts (W)

Worked Example

A transformer in a travel adapter steps up a 115 V ac mains electricity supply to the 230 V needed for a hair dryer. A current of 5 A flows through the hairdryer. Assuming that the transformer is 100% efficient, calculate the current drawn from the mains supply. 

Step 1: List the known quantities

  • Voltage in primary coil, Vp = 115 V

  • Voltage in secondary coil, Vs = 230 V

  • Current in secondary coil, Is = 5 A

Step 2: Write the equation linking the known values to the current drawn from the supply, Ip

Vp × Ip = Vs × Is

Step 3: Substitute in the known values

115 × Ip = 230 × 5

Step 4: Rearrange the equation to find Ip

I subscript p space equals space fraction numerator 230 space cross times space 5 over denominator 115 end fraction

Step 5: Calculate a value for Ip and include the correct unit

Ip = 10 A

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    Author
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    Ashika

    Author: Ashika

    Expertise: Physics Project Lead

    Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.