Circuit Calculations

Circuit calculations are multi-step problems worth 6-8 marks on Paper 4, combining $V=IR$ with the series and parallel rules. Each step depends on the one before, which is how examiners separate grades A and A* from B. A fixed method turns them into routine marks.

How do you solve any 0625 circuit calculation?

Use the same five steps every time. One: redraw or annotate the circuit, labelling all known values. Two: mark which sections are series and which are parallel. Three: reduce resistances to a single total. Four: find the supply current with $I = \dfrac{V}{R}$. Five: work back outwards to find individual currents and p.d.s.

The core equations:

Relationship	Words	Symbols
Ohm’s law form	p.d. = current × resistance	$V = IR$
Series resistance	total = sum of resistances	$R = R_1 + R_2$
Parallel resistance	reciprocals add	$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$
Charge	charge = current × time	$Q = It$

Two component behaviours feed these questions. A thermistor’s resistance falls as temperature rises. A light-dependent resistor’s (LDR) resistance falls as light intensity rises. Examiners drop them into circuits and ask how readings change.

What is a potential divider and why does it keep appearing?

A potential divider is two resistors in series across a supply. The supply p.d. divides between them in proportion to their resistances. The larger resistance takes the larger share. For Extended candidates the share across $R_2$ is $V_2 = V \times \dfrac{R_2}{R_1 + R_2}$. Pair a fixed resistor with a thermistor and the output p.d. changes with temperature. That is the standard sensing-circuit question. When the thermistor warms, its resistance falls, so its share of the p.d. falls and the fixed resistor’s share rises.

Worked Exam Question

A 12 V battery connects in series with a 2.0 Ω resistor and a parallel pair: 6.0 Ω and 3.0 Ω. (a) Calculate the total circuit resistance. (3 marks) (b) Calculate the battery current. (1 mark) (c) Calculate the p.d. across the parallel pair. (2 marks)

Solution. (a) Parallel pair first: $\dfrac{1}{R} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{6} + \dfrac{2}{6} = \dfrac{3}{6}$, so $R = 2.0\ \Omega$. Then series: total $= 2.0 + 2.0 = 4.0\ \Omega$. (b) $I = \dfrac{V}{R} = \dfrac{12}{4.0} = 3.0\ \text{A}$. (c) p.d. across the 2.0 Ω series resistor: $V = IR = 3.0 \times 2.0 = 6.0\ \text{V}$. Parallel pair takes the rest: $12 - 6.0 = 6.0\ \text{V}$. (Check: $V = 3.0 \times 2.0 = 6.0\ \text{V}$ directly across the pair. ✓)

Mark scheme:

• M1: reciprocal method for the parallel pair.
• A1: parallel pair = 2.0 Ω.
• A1: total = 4.0 Ω.
• A1: $I = 3.0\ \text{A}$.
• M1: $V = IR$ using their current, or subtraction from 12 V.
• A1: 6.0 V with unit.

Common Mistakes

• Reducing the circuit in the wrong order. Fix: always collapse parallel blocks first, then add the series chain.
• Using the supply p.d. across a single component. Fix: 12 V is across the whole circuit; each component only gets its share.
• Dropping the reciprocal flip mid-question. Fix: write "$1/R =$" and "$R =$" on separate lines so the flip is visible.
• Thermistor/LDR direction reversed. Fix: heat ↓ resistance for thermistors; light ↓ resistance for LDRs. Then trace the effect step by step.
• No sense checks. Fix: branch currents must sum to the supply current, and component p.d.s must sum to the e.m.f.

Exam Technique Tip

Multi-step questions carry “error carried forward”. If part (a) is wrong but part (b) uses your (a) value correctly, part (b) still scores full marks. So never abandon a question after a doubtful step. Write the method clearly and keep going. Showing $I = \dfrac{V}{R}$ with your substituted numbers is what triggers the forward credit.

How This Is Examined

This CS subtopic is Paper 4’s favourite long question. Extended candidates face full mixed circuits with potential dividers, thermistors and LDRs. Core candidates (Papers 1 and 3) get shorter versions: series totals, single $V = IR$ steps and qualitative parallel statements. Papers 2 MCQs often hide a two-step calculation behind four plausible answers, so estimate before choosing. Practical papers use these calculations to process measured data. In our 1.5-hour classes we time students on one full Paper 4 circuit question per week; the five-step method above usually halves their solving time within a month.