Syllabus Edition

First teaching 2023

First exams 2025

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Electric Resistance (SL IB Physics)

Revision Note

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Electric Resistance

  • As electrons move through a conductor within a circuit (or any other component), they collide with the metal ions and transfer some of their electrical potential energy to the positive ions of the metal

Electrons and resistance, downloadable AS & A Level Physics revision notes

Free electrons collide with metal ions which resist their flow

  • This transfer of energy results in an increase in the kinetic energy of the atoms in the lattice
    • This raises the overall internal energy of the metal
  • The macroscopic result of this transfer is the heating up of the wire which causes resistance
  • Some metals heat up more than others
    • The greater the heating effect, the higher the resistance
    • Copper has a low electrical resistance, making it an ideal material to make wires from

  • All electrical components have resistance to different degrees, including the wires and batteries
  • Voltmeters and ammeters are said to be ideal when
    • An ideal voltmeter has infinite resistance, such that no current passes through it
    • An ideal ammeter has zero resistance, such that all the current passes through it

Exam Tip

Unless otherwise stated, voltmeters and ammeters will always be ideal in exam questions. This simply means you do not need to consider the resistance of an ammeter or voltmeter as part of the total resistance of the circuit

When non-ideal meters are used, the resistance will always be constant

Calculating Resistance

  • The resistance R of a component is defined as:

The ratio of the potential difference across the component to the current flowing through it 

  • It is calculated as follows:

R space equals space fraction numerator space V over denominator I end fraction

  • Where:
    • V = potential difference (V)
    • I = electric current (A)
    • R = resistance (Ω)

  • The units for resistance is ohms represented by the greek letter 'omega', Ω
  • The higher the resistance of a component, the lower the current flowing through it and vice versa
  • In terms of SI base units: 1 Ω = 1 kg m2 s–3 A–2

Worked example

A charge of 5.0 C passes through a resistor at a constant rate in 30 s. The potential difference across the resistor is 2.0 V.

Calculate the resistance R of the resistor.

Answer:

Step 1: Write down the known quantities 

  • Charge, Δq = 5.0 C
  • Time, Δt = 30 s
  • Potential difference, V = 2.0 V

Step 2: Write down the equation for the resistance R 

R space equals fraction numerator space V over denominator I end fraction

Step 3: Calculate the current I from the charge and time 

I space equals space fraction numerator increment q space over denominator increment t end fraction

Step 4: Substitute the numbers into the above equation 

I space equals fraction numerator space 5.0 over denominator 30 end fraction space equals space 0.17 space straight A

Step 5: Substitute this value of the current into the equation for the resistance given in Step 2

R space equals space fraction numerator 2.0 over denominator 0.17 end fraction space space equals space 12 space straight capital omega

Exam Tip

It is common for current to be given in mA or potential difference in kV. For resistance to be in Ω, you must convert mA into A and kV into V!

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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.