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

Transformer calculations

Extended tier only

• 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:

$\frac{\mathrm{primary}\mathrm{voltage}}{\mathrm{secondary}\mathrm{voltage}}=\frac{\mathrm{number}\mathrm{of}\mathrm{turns}\mathrm{on}\mathrm{primary}}{\mathrm{number}\mathrm{of}\mathrm{turns}\mathrm{on}\mathrm{secondary}}$

• It can be expressed in symbols as follows:

$\frac{V_p}{V_s}=\frac{N_p}{N_s}$

• Where

  • $V_p$ = voltage across the primary coil, in volts (V)

  • $V_s$ = voltage across the secondary coil, in volts (V)

  • $N_p$ = number of turns on the primary coil

  • $N_s$ = number of turns on the secondary coil

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

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$

• Rearranging for the secondary voltage:

$V_s=V_p\times \frac{N_s}{N_p}$

• 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, $V_s>V_p$ and $N_s>N_p$

• In a step-down transformer, $V_s<V_p$ and $N_s<N_p$

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:

$\mathrm{power}\mathrm{in}\mathrm{primary}=\mathrm{power}\mathrm{in}\mathrm{secondary}$

• The equation to calculate electrical power is:

$P=V\times I$

• Where:

  • P = power, in watts (W)

  • V = voltage, in volts (V)

  • I = current, in amps (A)

• Therefore, the equation for an ideal transformer is:

$I_p{V}_p=I_s{V}_s$

• Where:

  • $I_p$ = primary current, in amps (A)

  • $V_p$ = primary voltage, in volts (V)

  • $I_s$ = secondary current, in amps (A)

  • $V_s$ = secondary voltage, in volts (V)

• The equation above could also be written as a ratio:

$\frac{I_s}{I_p}=\frac{V_p}{V_s}$

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 $P=VI$

• 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

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

Calculating power losses

Extended tier only

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

$P=I^{2}R$

• 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