Series and Parallel Circuits

In a series circuit the current is the same everywhere and the p.d.s add up; in a parallel circuit the current splits between branches and each branch gets the full supply p.d. Examiners test these rules every session because one swapped rule wrecks a whole multi-part question. Learn the two current rules and two voltage rules below and most circuit questions become routine.

What are the rules for series and parallel circuits?

In a series circuit there is one loop. The current is the same at every point. The p.d.s across the components add up to the supply p.d. Total resistance is the sum: $R = R_1 + R_2$.

In a parallel circuit the current splits between branches. The current from the source equals the sum of the branch currents. The p.d. across each branch equals the supply p.d. The combined resistance of two resistors in parallel is less than either resistor alone. Extended candidates calculate it with: $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$.

Rule	Series	Parallel
Current	Same everywhere	Splits; branch currents add to supply current
p.d.	Shares the supply p.d.	Same across every branch
Resistance	$R = R_1 + R_2$ (adds)	$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ (less than smallest)
One component breaks	Whole circuit stops	Other branches keep working

The breakage row explains real wiring. Houses use parallel circuits so each appliance gets the full mains p.d. and works independently. Decorative lights wired in series all die when one bulb fails.

Why is parallel resistance smaller, not larger?

Adding a parallel branch opens an extra path for charge. More paths mean more total current for the same supply p.d., and $R = \dfrac{V}{I}$ then gives a smaller resistance. State it that way in explanations. A quick check for two resistors: the parallel answer must be smaller than the smaller resistor. Two equal resistors in parallel give exactly half of one of them.

Worked Exam Question

A 6.0 V battery connects to a 4.0 Ω resistor in parallel with a 12 Ω resistor. (a) Calculate the combined resistance. (2 marks) (b) Calculate the current supplied by the battery. (2 marks)

Solution. (a) Equation: $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$. Substitute: $\dfrac{1}{R} = \dfrac{1}{4.0} + \dfrac{1}{12} = \dfrac{3}{12} + \dfrac{1}{12} = \dfrac{4}{12}$. Rearrange: $R = \dfrac{12}{4}$. Answer: $R = 3.0\ \Omega$. Sense check: 3.0 Ω is less than 4.0 Ω. ✓ (b) Equation: $V = IR$, so $I = \dfrac{V}{R}$. Substitute: $I = \dfrac{6.0}{3.0}$. Answer: $I = 2.0\ \text{A}$.

Mark scheme:

• M1: $\dfrac{1}{R} = \dfrac{1}{4.0} + \dfrac{1}{12}$ with correct substitution.
• A1: 3.0 Ω with unit.
• M1: $I = \dfrac{V}{R}$ using their combined resistance (error carried forward allowed).
• A1: 2.0 A with unit.

Common Mistakes

• Forgetting the final reciprocal. Fix: $\dfrac{1}{R} = \dfrac{4}{12}$ is not the answer. Flip it: $R = 3.0\ \Omega$. Students lose this mark constantly.
• Adding parallel resistances directly. Fix: 4 Ω and 12 Ω in parallel is 3 Ω, never 16 Ω. Run the “smaller than smallest” check.
• Applying the series current rule to parallel branches. Fix: write S or P next to the diagram before touching numbers.
• Assuming equal current in unequal branches. Fix: more current flows through the smaller resistance, in inverse proportion.
• Voltmeter readings that exceed the supply. Fix: series p.d.s must sum to the e.m.f., so total your answers and check.

Exam Technique Tip

Before any circuit calculation, annotate the diagram. Label every component with known V, I and R values, mark series sections and parallel blocks, then reduce the circuit one block at a time. Examiners award method marks for a correct partial reduction, so written stages beat mental arithmetic. This habit alone recovers 2-3 marks on a typical Paper 4 circuits question.

How This Is Examined

A CS subtopic with a clear tier split. Core candidates (Papers 1 and 3) state the qualitative rules, add series resistances, and know parallel resistance is smaller, without the reciprocal formula. Extended candidates (Papers 2 and 4) calculate combined parallel resistance for two resistors and handle circuits mixing series and parallel sections. Paper 4 usually embeds these rules inside a longer circuit-calculations question worth 6-8 marks. Practical papers (P5/P6) test the rules through building circuits and explaining unexpected ammeter readings. Most Malaysian school mocks copy this Paper 4 style, so practise the full mixed-circuit reduction, not just isolated rules.