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4.3 Potential Difference and Resistance

GCSE Physics · AQA · Chapter 4: Electric Circuits

A free GCSE Physics lesson for AQA (8463) from OBA School: using ammeters and voltmeters, V = E/Q, resistance V = I R, Ohm’s law and I–V graphs. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Connect an ammeter and a voltmeter correctly in a circuit
  • Define potential difference and use E = Q V
  • Explain what resistance is and use V = I R, including rearranging and converting units
  • State Ohm’s law and describe the I–V graph for an ohmic conductor, including reversing the p.d.

Key words

  • Potential difference: The energy transferred per unit charge (per coulomb) passing between two points. Unit: volt (V). Also called p.d. or voltage.
  • Volt: One volt is one joule per coulomb (1 V = 1 J/C).
  • Resistance: How much a component opposes the current through it. R = V ÷ I. Unit: ohm (Ω).
  • Ohm’s law: The current through a conductor at constant temperature is directly proportional to the potential difference across it.
  • Ohmic conductor: A conductor whose resistance stays constant as the current changes (at constant temperature).
  • Voltmeter: A meter connected in parallel across a component to measure the p.d. across it.

Measuring current and potential difference

  • An ammeter is connected in series, so the same current passes through it and the component.
  • A voltmeter is connected in parallel (across) the component, with one lead on each side of it.
  • Small currents are often measured in milliamperes: 1 mA = 0.001 A.

Potential difference

As charge passes through a component, energy is transferred to the component (for example, to heat a filament). The potential difference across the component is the energy transferred for each coulomb of charge.

energy transferred = charge flow × potential difference E = Q V

E in joules (J) Q in coulombs (C) V in volts (V)

Rearranged: V = E ÷ Q Q = E ÷ V

Worked example: energy transferred by a heater

1500 C of charge flows through a heater connected to a 230 V supply.
E = Q V = 1500 × 230

Answer: E = 345 000 J (= 345 kJ)

Worked example: finding a p.d.

A motor transfers 4.8 kJ when 400 C flows through it.
E = 4.8 × 1000 = 4800 J; V = E ÷ Q = 4800 ÷ 400

Answer: V = 12 V

Resistance

Electrons moving through a metal keep colliding with the vibrating metal ions. These collisions oppose the flow of charge – this is resistance. For a fixed p.d., the bigger the resistance, the smaller the current.

potential difference = current × resistance V = I R

V in volts (V) I in amperes (A) R in ohms (Ω)

Rearranged: R = V ÷ I I = V ÷ R

Worked example: resistance of a lamp

A lamp has a p.d. of 6.0 V across it and a current of 0.25 A through it.
R = V ÷ I = 6.0 ÷ 0.25

Answer: R = 24 Ω

Worked example: current in milliamps

A 470 Ω resistor has a p.d. of 9.4 V across it. Find the current in mA.
I = V ÷ R = 9.4 ÷ 470 = 0.020 A

Answer: I = 20 mA

Ohm’s law and current–potential difference graphs

If you measure the current through a resistor (or a metal wire) for different p.d.s and plot current (y-axis) against p.d. (x-axis), you get a straight line through the origin. The current is directly proportional to the p.d., so the resistance is constant. This is Ohm’s law, and it only holds if the temperature stays constant.

  • A component that follows Ohm’s law is called an ohmic conductor.
  • Reversing the p.d. reverses the direction of the current. The line continues straight into the bottom-left quarter, so the resistance is the same in both directions.
  • With current on the y-axis, the steeper line belongs to the smaller resistance (more current for the same p.d.).

Exam tip: To find resistance from an I–V graph, read one point off the line and work out R = V ÷ I. For the 20 Ω line: 4.0 V ÷ 0.20 A = 20 Ω. Do not use “gradient = resistance”.

Common mistakes

  • Connecting the voltmeter in series. It must go across the component, in parallel.
  • Saying p.d. “flows” through a component. Current flows through; p.d. is across.
  • Forgetting to change kJ to J or mA to A before using E = Q V or V = I R.
  • Thinking a steeper I–V line means a bigger resistance. It means a smaller resistance.
  • Forgetting the condition in Ohm’s law: the temperature must be constant.

Revision questions

Try these, then check your answers below each one.

Question 1. When 40 C of charge passes through a heater, 9200 J of energy is transferred. Calculate the potential difference across the heater.

Answer: 230 V

Working: V = E ÷ Q = 9200 ÷ 40 = 230 V.

Question 2. A 12 Ω resistor carries a current of 0.75 A. Calculate the potential difference across it.

Answer: 9 V

Working: V = I R = 0.75 × 12 = 9.0 V.

Question 3. A potential difference of 6.0 V across a resistor gives a current of 30 mA. Calculate its resistance.

Answer: 200 Ω

Working: I = 30 ÷ 1000 = 0.030 A. R = V ÷ I = 6.0 ÷ 0.030 = 200 Ω.

Related lessons

Continue in the full course

This lesson is section 4.3 (AGCPH0403) of our free GCSE Physics course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Physics course.

 

 

 

 

 

 

 

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