AC & High Voltage Transmission (OCR Gateway GCSE Physics)

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AC & High Voltage Transmission (HT only)

  • Remember, in a transformer, an alternating current is supplied to the primary coil
  • The current is continually changing direction
    • This means it will produce a changing magnetic field around the primary coil
    • A changing magnetic field leads to the generation of an alternating potential difference
    • If connected in a complete circuit then an electric current will flow

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

Structure of a Transformer

  • A basic transformer consists of:
    • A primary coil
    • A secondary coil
    • An iron core

  • Iron is used because it is easily magnetised

Transformer Diagram, downloadable AS & A Level Physics revision notes

Structure of a transformer

How a Transformer Works

  • An alternating current is supplied to the primary coil
  • The current is continually changing direction
    • This means it will produce a changing magnetic field around the primary coil

  • The iron core is easily magnetised, so the changing magnetic field passes through it
  • As a result, there is now a changing magnetic field inside the secondary coil
    • This changing field cuts through the secondary coil and induces a potential difference

  • As the magnetic field is continually changing the potential difference induced will be alternating
    • The alternating potential difference will have the same frequency as the alternating current supplied to the primary coil

  • If the secondary coil is part of a complete circuit it will cause an alternating current to flow

The Ideal Transformer

  • 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 = VI

  • 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

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

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

Ip = 10 A

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Joanna

Author: Joanna

Joanna obtained her undergraduate degree in Natural Sciences from Cambridge University and completed her MSc in Education at Loughborough University. After a decade of teaching and leading the physics department in a high-performing academic school, Joanna now mentors new teachers and is currently studying part-time for her PhD at Leicester University. Her passions are helping students and learning about cool physics, so creating brilliant resources to help with exam preparation is her dream job!