AC & High Voltage Transmission (OCR GCSE Physics A (Gateway)): Revision Note

Exam code: J249

Joanna

Written by: Joanna

Reviewed by: Caroline Carroll

Updated on

AC & High Voltage Transmission

Higher Tier 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.

Answer:

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

Ip=230×5115

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

Ip = 10 A

Unlock more, it's free!

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

the (exam) results speak for themselves:

Build on this topic

Joanna

Author: Joanna

Expertise: Physics Content Creator

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!

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.