Series and parallel circuits: calculate the currents, potential differences and resistances in dc series circuits

The idea

In a dc series circuit there is a single loop, so the current is the same at every point: I₁ = I₂ = I₃. The supply potential difference is shared between components, so V_total = V₁ + V₂ + V₃, and the total resistance is the sum of the individual resistances, R_total = R₁ + R₂ + R₃. To calculate an unknown quantity, first identify what is known, then choose the relationship linking the known and unknown values. For example, a 6.0 V supply with a 2.0 Ω and a 4.0 Ω resistor in series gives R_total = 6.0 Ω, so I = V ÷ R = 6.0 V ÷ 6.0 Ω = 1.0 A; the potential differences are V = I × R = 2.0 V and 4.0 V, which sum to the supply value. Always include units and check that the shared potential differences add to the supply value.

What earns the marks

• States that current is the same at all points in a series circuit because there is only one path for charge to flow.
• Uses V = I × R correctly, substituting values with consistent units and rearranging when the unknown is current or resistance.
• Applies V_total = V₁ + V₂ + V₃ to find a missing potential difference or to check that the individual values sum to the supply value.
• Applies R_total = R₁ + R₂ + R₃ for resistors in series and uses it with the supply potential difference to find the current.
• Calculates a missing resistance from R = V ÷ I using the potential difference across that component and the series current.
• Checks the final answer by confirming that the separate potential differences add to the supply value and that the current is unchanged throughout.

Common mistakes and how to put them right

• Adding the currents at different points in a series circuit.

  Put it right: Current is the same everywhere in a single loop, so no addition is needed.

• Using the supply potential difference with a single component's resistance to find the current.

  Put it right: Either use the supply potential difference with the total resistance, or a component's potential difference with its own resistance.

• Forgetting to convert milliamperes or kilohms before substituting.

  Put it right: Convert to amperes and ohms first, for example 250 mA = 0.25 A and 2.2 kΩ = 2200 Ω.

• Treating the supply potential difference as the value across each component.

  Put it right: The supply value is shared, so each component has a smaller potential difference than the supply, even if the resistances are equal.

Exam tips

• Write down the series rules before substituting numbers so the method is visible to the examiner.
• Show the rearranged equation with values substituted, then give the answer with its unit.
• Use the check that the component potential differences sum to the supply value to catch arithmetic slips.
• Keep full calculator precision until the final line, then round to a sensible number of significant figures.