Many pupils do not recognise that the potential difference across each resistor in a simple parallel circuit is the potential difference across the battery

Electricity and Magnetism


Many pupils approach questions about simple parallel circuits by working out the effective resistance of the load, rather than by recognising that the potential difference across each resistor is the potential difference across the battery.

Diagnostic Resources

The following worksheets may help to identify whether students hold this particular misconception.

For more information, see the University of York EPSE website.

Resources to Address This

  • Voltage in parallel Circuits (11-14)

    Source - SPT / El03 PN12

    As the second bulb is added, there is a current in both loops. The power in both bulbs is equal and set by the current in and the voltage across each bulb. Both bulbs are as bright as they would be in a simple loop (one battery, one bulb).

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  • Predicting and measuring voltages in a parallel circuit (11-14)

    Source - SPT/ El03 TA06

    The approach taken is to encourage the pupils to make predictions of voltage values before they make each measurement.

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  • Loops in parallel circuits (14-16)

    Source - SPT/ Ee01 PN11

    Circuits with only one parallel connection are best analysed by thinking of the two separate loops that make up the circuit. The introduction to parallel connections in circuits was made in this way in the SPT: Electric circuits topic. The loops have nothing in common, other than the battery, so you can treat them as two separate loops.

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  • Lamps in parallel (11-16)

    Source - Practical physics / Electric circuits and fields/ Potential difference/ ...

    The current flowing in a circuit increases as more lamps are added in parallel with each other - but the pd across each lamp remains the same.

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  • Episode 114: Components in series and parallel (16-19)

    Source - TAP/ Electricity/ Series and parallel circuits/ ...

    Ask students to recall the equation which defines resistance. (R = V / I where V is the pd across a component and I is the current through it.) Connect 1, 2 and then 3 lamps in series across a supply (constant pd) showing them the reduction in current at the same pd ....


    Repeat the experiment but this time add lamps in parallel. The current increases and the effective (load) resistance decreases. If you calculate the resistance values you should get 1:1/2:1/3 as you add the lamps.

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The following studies have documented this misconception:

  • Millar, R. and Beh, K. L. () Students’ understanding of voltage in simple parallel electric circuits. International Journal of Science Education, 15 (4),


    Review sheet

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