For components connected in series: there is the same current through each…

The idea

In a series circuit there is only one path and charge is not used up, so the current is the same at every point: an ammeter reads the same wherever you place it. Each coulomb passes through every component in turn and gives up some energy at each one, so the potential differences across the components add up to the potential difference of the supply. Because the charge has to push through both resistances one after the other, the total resistance is the sum, R = R₁ + R₂. Put a 4 Ω and a 2 Ω resistor across a 6 V supply and the total resistance is 6 Ω, so the current is 6 ÷ 6 = 1 A everywhere, and the potential differences are 4 V and 2 V, which add back to 6 V. The larger resistor always takes the larger share of the supply potential difference.

What earns the marks

• The current being the same through every component in the series loop.
• Adding the resistances, R = R₁ + R₂, to find the total resistance.
• Using the total resistance in I = V ÷ R to find the single circuit current.
• A potential difference found from V = I × R, with the individual values adding to the supply value.

Common mistakes and how to put them right

• Sharing the current between series components in the way that potential difference is shared.

• Giving individual potential differences that add to more than the supply potential difference.

• Using the full supply potential difference across one resistor when calculating the current.

• Assuming the component nearest the cell receives more current because the charge reaches it first.

Exam tips

• Work in the order total resistance, then current, then each individual potential difference; that route collects the marks.
• Check your answer by adding the individual potential differences, which must total the supply value.
• If two potential differences are given and a third is asked for, subtract from the supply rather than starting a calculation.