Revision NotesExam QuestionsPast Papers
O LevelPhysicsCambridge (CIE)Revision Notes4. Electricity & Magnetism4.5 Electromagnetic EffectsTransformer Calculations

Transformer Calculations (Cambridge (CIE) O Level Physics): Revision Note

Leander Oates

Last updated

Transformer Calculations

  • 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 equation:

fraction numerator straight P. straight D. space across space primary space coil over denominator straight P. straight D. space across space secondary space coil end fraction space equals fraction numerator space 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 can be written using symbols as follows:

V subscript p over V subscript s equals 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 equation above can be flipped upside down to give:

V subscript s over V subscript p equals 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

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 this is

Answer

Part (a)

Step 1: List the known quantities

  • Number of turns in primary coil, N subscript P = 20

  • Number of turns in secondary coil, N subscript S = 800

  • Voltage in primary coil, V subscript P = 500 V

Step 2: Write the equation linking the output potential difference (bold italic V subscript bold S) to the known quantities

  • There will be less rearranging to do if V subscript S is on the top of the fraction

N subscript S over N subscript P space equals fraction numerator space V subscript S over denominator V subscript P end fraction

Step 3: Rearrange the equation to make bold italic V subscript bold S the subject

V subscript S space equals fraction numerator space N subscript S V subscript P over denominator N subscript P end fraction

Step 4: Substitute the known values into the equation

V subscript S space equals fraction numerator space 800 space cross times space 500 over denominator 20 end fraction space equals space 20 space 000 space straight V

Part (b)

The secondary voltage is larger than the primary, therefore this is a step-up transformer

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.

High-Voltage Transmission

  • Transformers have a number of roles:

    • They are used to increase the potential difference of electricity before it is transmitted across the national grid

    • They are used to lower the high voltage electricity used in power lines to the lower voltages used in houses

    • They are used in adapters to lower mains voltage to the lower voltages used by many electronic devices

Advantages of High Voltage Transmission

  • When electricity is transmitted over large distances, the current in the wires heats them, resulting in energy loss

  • To transmit the same amount of power as the input power the potential difference at which the electricity is transmitted should be increased

    • This will result in a smaller current being transmitted through the power lines

    • This is because P = IV, so if V increases, I must decrease to transmit the same power

  • A smaller current flowing through the power lines results in less heat being produced in the wire

    • This will reduce the energy loss in the power lines

Model of High-Voltage Transmission

power-lines, IGCSE & GCSE Physics revision notes

Electricity is transmitted at high voltage, reducing the current and hence power loss in the cables

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:TransformersNext:Uses of an Oscilloscope
    Leander Oates

    Author: Leander Oates

    Expertise: Physics

    Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.