Series and Parallel Circuits

This lesson explains Series and Parallel Circuits for Cambridge IGCSE Physics 0625. It separates the Core requirements from the additional Supplement work for Extended candidates. Use the diagrams, method steps and error checks rather than memorising a paragraph. Work through the example before testing the same skill without notes.

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 example

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: $I = 2.0\ \text{A}$.

Original marking points:

• 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.

These marking points belong to this original example. They are not an official Cambridge mark scheme.

Common errors and how to correct them

• Forgetting the final reciprocal. Fix: $\dfrac{1}{R} = \dfrac{4}{12}$ is not the answer. Flip it: $R = 3.0\ \Omega$. This error can cost the mark.
• 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.

How to apply this in an exam

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. Mark schemes can award method marks for a correct partial reduction, so written stages beat mental arithmetic. This method makes partial working visible and easier to check.

Where this skill matters

This page contains both Core and Supplement requirements. 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 can embed these rules inside a longer circuit calculation. Practical papers (P5/P6) test the rules through building circuits and explaining unexpected ammeter readings.