Transformer Calculations (CIE IGCSE Physics: Co-ordinated Sciences (Double Award))

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Katie M

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Katie M

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Transformer calculations

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  • The voltages across the primary and secondary coils of a transformer can be calculated using the transformer equation, which states

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

  • It can be expressed by the equation:

fraction numerator primary space voltage over denominator secondary space voltage end fraction space equals space fraction numerator number space of space turns space on space primary over denominator number space of space turns space on space secondary end fraction

  • It can be expressed in symbols as follows:

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

  • Where
    • V subscript p = voltage across the primary coil, in volts (V)
    • V subscript s = voltage across the secondary coil, in volts (V)
    • N subscript p = number of turns on the primary coil
    • N subscript s = number of turns on the secondary coil
  • The transformer equation can be flipped upside down to give:

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

  • Rearranging for the secondary voltage:

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

  • This equation shows that the output (secondary) voltage of a transformer depends on:
    • the number of turns on the primary and secondary coils
    • the input (primary) voltage
  • In a step-up transformer, V subscript s space greater than space V subscript p and N subscript s space greater than space N subscript p
  • In a step-down transformer, V subscript s space less than space V subscript p and N subscript s space less than space N subscript p

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Worked example

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

a)
Calculate the output voltage of the secondary coil.
b)
State whether this is a step-up or step-down transformer.
 

Answer

Part (a)

Step 1: List the known quantities

  • Number of turns on the primary coil, N subscript p = 20
  • Number of turns on the secondary coil, N subscript s = 800
  • Voltage across the primary coil, V subscript P = 500 V

Step 2: Write down the transformer equation

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

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

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

V subscript s space equals space V subscript p cross times fraction numerator space N subscript s over denominator N subscript p end fraction

Step 4: Substitute the known values into the equation

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

Part (b)

  • The secondary voltage is larger than the primary, V subscript s space greater than space V subscript p
  • There are more turns on the secondary coil than on the primary, N subscript s space greater than space N subscript p
  • Therefore, this is a step-up transformer

Exam Tip

When carrying out transformer calculations, 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.

Ideal transformer equation

Extended tier only

  • A transformer which is 100% efficient is called an ideal transformer
  • 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, then the input power in the primary coil is equal to the output power of the secondary coil:

power space in space primary space equals space power space in space secondary

  • The equation to calculate electrical power is:

P space equals space V space cross times space I

  • Where:
    • P = power, in watts (W)
    • V = voltage, in volts (V)
    • I = current, in amps (A)
  • Therefore, the equation for an ideal transformer is:

I subscript p V subscript p space equals space I subscript s V subscript s

  • Where:
    • I subscript p = primary current, in amps (A)
    • V subscript p = primary voltage, in volts (V)
    • I subscript s = secondary current, in amps (A)
    • V subscript s = secondary voltage, in volts (V)
  • The equation above could also be written as a ratio:

I subscript s over I subscript p space equals space V subscript p over V subscript s

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Worked example

A transformer in a travel adapter steps up a 115 V a.c. 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. 

Answer:

Step 1: List the known quantities

  • Voltage in primary coil, V subscript p = 115 V
  • Voltage in secondary coil, V subscript s = 230 V
  • Current in secondary coil, I subscript s = 5 A

Step 2: Write down the equation for an ideal transformer

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

Step 3: Substitute in the known values

115 space cross times space I subscript p space equals space 230 space cross times space 5

Step 4: Rearrange the equation to find the primary current

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

I subscript p space equals space 10 space straight A

High-voltage transmission

Extended tier only

  • Electricity is transmitted through power cables at a low current to prevent dissipation of energy
    • When current flows in a wire, there is heating in the wire due to resistance
    • Therefore, energy is dissipated to the surroundings, this energy is wasted
    • The lower the current, the more efficient the energy transfer
  • Electrical power is equal to voltage × current, or P space equals space V I
  • This means that a low current can be achieved by increasing the voltage, so electricity must be transmitted at a high voltage
    • A smaller current flowing through the power lines results in less heat being produced in the wire
    • This reduces the energy loss in the power lines

Power Loss National Grid, downloadable AS & A Level Physics revision notes

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

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Calculating power losses

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  • The power dissipated in the wire due to resistance is given by:

P space equals space I squared R

  • Where:
    • P = power, in watts (W)
    • I = current, in amps (A)
    • R = resistance, in ohms (Ω)
  • A step-up transformer is used to increase the voltage and decrease the current of electricity before transmission
    • A high-voltage transmission ensures the same power transfer with a smaller current
    • A smaller current means less thermal energy will be lost due to the resistance in the wire
  • A step-down transformer is used to decrease the voltage and increase the current of electricity after transmission
    • High-voltage electricity is dangerous for use in homes, so it must be lowered before the current reaches consumers

Exam Tip

If you forget the equation P = I2R just remember 'Twinkle twinkle little star, power equals I squared R''.

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Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.