IGCSE Physics revision · Electricity & magnetism
Electric circuits
One of the most heavily examined areas of the course, and a common source of lost marks. Everything rests on knowing exactly what current, voltage and resistance each measure — and on the series/parallel rules being automatic.
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What the syllabus demands
- —Define current, potential difference, e.m.f. and resistance
- —Use I = Q/t, V = IR (Ohm's law), and the resistance rules for series and parallel circuits
- —Recall current and voltage rules in series and parallel circuits
- —Describe how resistance varies with length and cross-sectional area of a wire
- —Calculate electrical power (P = IV) and energy (E = IVt); understand fuses and earthing
Definitions that earn marks
Clear definitions to practise — check your course mark scheme
- Electric current
- The charge passing a point per unit time: I = Q ÷ t. Measured in amperes.
- Potential difference (voltage)
- The work done (energy transferred) per unit charge passing between two points: V = W ÷ Q.
- e.m.f.
- The electromotive force of a source: the energy given to each unit of charge by the source.
- Resistance
- The opposition to current: R = V ÷ I, measured in ohms (Ω).
The equations
More equations to practise: the IGCSE formula sheet.
Where the marks die
Common mistakes to check
- 01
Thinking current is used up around a circuit. Current is the same at every point in a series circuit — what is 'used' is energy, measured by the potential difference across each component.
- 02
Reversing the meter connections: ammeters go in series, voltmeters in parallel across the component. Drawn wrong, the circuit mark is gone.
- 03
Forgetting the parallel resistance result must be smaller than the smallest branch. If your answer is bigger, you forgot to take the reciprocal at the end of 1/R = 1/R₁ + 1/R₂.
- 04
Wire resistance rules half-remembered: resistance increases with length (double length, double resistance) and decreases with cross-sectional area (double area, half resistance).
- 05
Fuse explanations without melting: too large a current melts the fuse wire, breaking the circuit and protecting the cable. The fuse rating must be just above the appliance's normal current.
One worked example, done properly
Question
A 6.0 Ω and a 3.0 Ω resistor are connected in parallel to a 6.0 V supply. Calculate the total resistance and the current from the supply.
Method
- 1.1/R = 1/6.0 + 1/3.0 = 1/6 + 2/6 = 3/6, so R = 2.0 Ω.
- 2.Check: 2.0 Ω is smaller than the smallest branch (3.0 Ω) — as parallel must be.
- 3.I = V ÷ R = 6.0 ÷ 2.0.
R = 2.0 Ω, I = 3.0 A
Test yourself
10 questions · instant marking
Question 1 of 10
A 12 V supply drives a current of 3 A through a resistor. Its resistance is:
Question 2 of 10
In a series circuit, the current:
Question 3 of 10
An ammeter is connected:
Question 4 of 10
Two 10 Ω resistors are connected in parallel. Their combined resistance is:
Question 5 of 10
Doubling the length of a wire (same material and thickness):
Question 6 of 10
A charge of 60 C passes a point in 20 s. The current is:
Question 7 of 10
A 230 V kettle draws 10 A. Its power is:
Question 8 of 10
The purpose of a fuse is to:
Question 9 of 10
In a parallel circuit, the potential difference across each branch is:
Question 10 of 10
As a filament lamp heats up, its resistance: