Transformer Calculations (CIE IGCSE Physics)

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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:

  • 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

Exam Tip

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.

Transformer Efficiency

EXTENDED

  • 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 power = Output power

  • The equation to calculate electrical power is:

P = V × 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:

Vp × Ip = Vs × Is

  • 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:

Ps = Vp × Ip

  • Where:
    • Ps = output power (power produced in 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

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

Ip = 10 A

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

power-lines, IGCSE & GCSE Physics revision notes

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

Calculating Power Losses

EXTENDED

  • When a current passes through a wire, the current creates a heating effect which means the wires warm up
  • This means they lose electrical energy as heat which reduced the efficiency of the transformer
    • This is due to electrical resistance which is present in all wires

  • The power (energy per second) lost in the wire is given by the following equation

P = I2R

  • Where:
    • P = power in watts (W)
    • I = current in amps (A)
    • R = resistance in ohms (Ω)

  • Since the power is the energy lost per second, the total energy lost in a time t will be:

E = P × t

  • Where:
    • E = energy in joules (J)
    • = time in seconds (s)

  • A step-up transformer may be used to increase the voltage of a power supply from the power station to the transmission wires
  • The number of turns and voltage for the transformer is related by the following equation:

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

  •  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 the primary coil
    • ns = number of turns on the secondary coil

  • A step-up transformer has more turns on the secondary coil, Ns, than on the primary coil, Np
  • Since a transformer cannot output more power than is put into it, increasing the voltage must result in the current being lowered

IpVp = IsVs

  • Where:
    • Ip = current in the primary coil in amps (A)
    • Is = current in the secondary coil in amps (A)

  • Lower current results in less power and energy loss in the cables
    • This makes the transfer of electrical energy through the wires more efficient

Exam Tip

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

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Ashika

Author: Ashika

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.