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How are current, voltage and resistance defined, and how does Ohm's law relate them?

Circuit concepts: charge, current, voltage (potential difference) and resistance, their units, and Ohm's law relating voltage, current and resistance.

An Eduqas GCSE Electronics answer on the core circuit concepts: charge and current, voltage as energy per coulomb, resistance, their units, and applying Ohm's law to find voltage, current or resistance in a circuit.

Generated by Claude Opus 4.813 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

Eduqas wants you to define charge, current, voltage (potential difference) and resistance, give their units, and use Ohm's law to find any one of voltage, current and resistance from the other two. These four quantities and this one equation are the tools every later topic in the course assumes, so they must be automatic.

The answer

Charge and current

Voltage (potential difference)

Resistance

Ohm's law

Examples in context

Current, voltage, resistance and Ohm's law are used in every circuit you will meet. The current-limiting resistor for an LED, the potential-divider output that feeds a comparator, the base resistor that sets a transistor's switching current and the timing resistor of a 555 are all found with Ohm's law. Getting fluent with the three rearrangements and the unit conversions now saves marks throughout both written papers.

Try this

Q1. State Ohm's law as an equation. [1 mark]

  • Cue. V=IRV = IR (current proportional to voltage for a fixed resistor at constant temperature).

Q2. A 470 Ω470\ \Omega resistor has 9.0 V9.0\ \text{V} across it. Find the current. [2 marks]

  • Cue. I=VR=9.0470=0.019 A=19 mAI = \frac{V}{R} = \frac{9.0}{470} = 0.019\ \text{A} = 19\ \text{mA}.

Q3. A current of 0.25 A0.25\ \text{A} flows through a resistor when 5.0 V5.0\ \text{V} is across it. Find its resistance. [2 marks]

  • Cue. R=VI=5.00.25=20 ΩR = \frac{V}{I} = \frac{5.0}{0.25} = 20\ \Omega.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20193 marksA resistor in a circuit has a current of 20 mA20\ \text{mA} flowing through it when the potential difference across it is 6.0 V6.0\ \text{V}. Calculate its resistance.
Show worked answer →

Use Ohm's law rearranged for resistance: R=VIR = \dfrac{V}{I}.

Convert the current to amperes: 20 mA=20×10−3 A=0.020 A20\ \text{mA} = 20 \times 10^{-3}\ \text{A} = 0.020\ \text{A}.

Substitute: R=6.00.020=300 ΩR = \dfrac{6.0}{0.020} = 300\ \Omega.

Markers reward the correct rearrangement, the conversion of milliamps to amps, and the answer 300 Ω300\ \Omega. The most common error is leaving the current in milliamps, which gives 0.3 Ω0.3\ \Omega (a factor of 1000 out).

Eduqas 20222 marksDefine potential difference and state its unit.
Show worked answer →

Definition (1 mark): potential difference (voltage) is the energy transferred per unit charge as charge moves between two points in a circuit, V=WQV = \dfrac{W}{Q} where WW is the energy transferred and QQ is the charge.

Unit (1 mark): the volt (V); one volt is one joule per coulomb.

Markers reward the energy-per-charge idea (not just "the push" or "the voltage") and the correct unit. A statement such as "the energy given to each coulomb of charge" earns the definition mark.

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