A circuit diagram is a map, not a picture. Each standard symbol stands for a component; straight lines stand for connecting wire assumed to have no resistance. A circuit only works when the loop from one terminal of the supply to the other is complete.
Simple definition
A circuit diagram uses standard symbols and straight lines to show how components are connected into loops.
Example
A cell, a switch, and a lamp drawn in one rectangle of lines means: close the switch and charge can flow around one complete loop, lighting the lamp.
Complete loopsupply → components → supply
Charge can only flow when there is an unbroken path from one terminal of the supply, through the components, and back to the other terminal.
Current needs an unbroken conducting path
Ammeter placementin series
An ammeter counts the charge flowing through a point, so it sits in the path of the current — in series with the component.
An ideal ammeter has zero resistance
Voltmeter placementin parallel
A voltmeter compares the energy per charge between two points, so it connects across a component — in parallel with it.
An ideal voltmeter has infinite resistance
01
Sources
A cell is one unit; a battery is two or more cells. The longer line marks the positive terminal. A power supply symbol stands for mains-driven sources.
02
Control and protection
Switches break the loop; fuses melt to break it automatically when current is too large; variable resistors adjust current smoothly.
03
Measurement and output
Ammeters read current in series, voltmeters read potential difference in parallel, and lamps, motors, buzzers, and resistors transfer energy out of the circuit.
04
Sensing and one-way components
A thermistor is a resistor rectangle with a diagonal bar across it. The kind used at this level is NTC: its resistance falls sharply as it gets hotter. A light-dependent resistor is the same rectangle with two arrows pointing in at it, and its resistance falls as the light on it grows brighter. Both are input components — they turn a temperature or a light level into a resistance the rest of the loop can respond to. A diode is a triangle pointing into a bar: it passes conventional current only in the direction the triangle points, and blocks it the other way, which is what makes it useful for turning alternating current into direct. A light-emitting diode is the same symbol with two arrows pointing away from it.
Interactive circuit graph
Read the topology before the values.
Use this connectivity map alongside 13.1 Circuit Symbols & Diagrams. Change the network, then select a node or branch to see what the graph preserves.
Diagram mode
Nodes
Points where electrical connections meet.
Branches
Lines or general impedances connecting two nodes.
Simplification
Component details, such as specific resistor values, are hidden to focus purely on connectivity.
Series loopEach junction leads into the next branch, so the components share one continuous route. It has 3 nodes and 3 branches. Component names and example values are shown.
Series loop: 3 nodes and 3 branches. Select a node or branch to investigate it.
02
Change one variable at a time
Make the relationship visible.
Complete the loop
The ammeter sits in the loop (series); the voltmeter reaches across the lamp (parallel).
Loopbroken — no current
Ammeter reads0 A
Voltmeter readsthe full supply e.m.f.
03
Catch the common trap
Explain before calculating.
How must an ammeter and a voltmeter be connected to measure a lamp's current and potential difference?
Choose an answer to test the model.
04
Worked examples
State the rule, substitute, then check units.
EasyName the two meters in a test circuit and how each must be connected to a lamp.
The ammeter reads current: in series, in the lamp's loop.
The voltmeter reads p.d.: in parallel, across the lamp.
AnswerAmmeter in series, voltmeter in parallel
MediumDraw a circuit to measure the resistance of a lamp using a cell, a switch, an ammeter, and a voltmeter.
Draw one loop: cell → switch → ammeter → lamp → back to the cell.
Connect the voltmeter in parallel across the lamp only.
Close the switch, read I and V, then use R = V/I.
AnswerSeries loop of cell, switch, ammeter, and lamp, with the voltmeter across the lamp
HardTwo wiring faults, each made on its own. (a) The ammeter is connected in parallel with the lamp. (b) In a separate attempt, the voltmeter is connected in series in the loop. Predict the meter readings and the lamp in each case, treating the faults separately.
Fault (a) alone: the ammeter's near-zero resistance short-circuits the lamp, so almost all the current bypasses it. The lamp goes out and the ammeter carries a dangerously large current, limited only by the cell and the leads.
Fault (b) alone: the voltmeter's near-infinite resistance now sits in the only loop, so the current falls to almost nothing. The lamp stays dark and the voltmeter reads close to the full e.m.f., because nearly all of it is dropped across the voltmeter itself.
Judge each fault on its own. If both were made at the same time the series voltmeter would still choke the loop, so the parallel ammeter would not draw a large current — the danger in (a) exists only while the rest of the circuit is wired correctly.
Answer(a) lamp out, ammeter passes a dangerously large current; (b) current ≈ 0, lamp dark, voltmeter reads ≈ the e.m.f.
ChallengingDesign a circuit to measure a lamp's resistance at three different brightnesses using one cell, one variable resistor, an ammeter, and a voltmeter. Explain each component's role.
Series loop: cell → variable resistor → ammeter → lamp → cell; voltmeter across the lamp only.
The variable resistor sets the loop current, giving three operating points; the meters record (I, V) pairs at each.
R = V/I at each point — expect R to rise with brightness as the filament heats.
AnswerRheostat in series to set current; R = V/I from each (I, V) pair, rising as the lamp brightens
Exam diagrams for this topic2 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.Open subsectionHide subsection
See it. Read it. Work it.
These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.
Fig. 3.1A circuit diagram of two resistors connected side by side between the same pair of points. A wire arrives from the left and reaches a junction, marked with a dot, where it divides into two branches. The upper branch contains a resistor labelled 6.0 ohm and the lower branch a resistor labelled 3.0 ohm. The two branches rejoin at a second junction dot on the right, from which a single wire continues away to the right.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. Both resistors lie between the same two junction dots, so they have the same p.d. across them; it is the current, not the voltage, that divides at the left-hand dot.
aState State which of the two resistors in Fig. 3.1 carries the larger current, and give a reason.
recall2 marks
Check answer · 2 marks
the 3.0 Ω resistor
both branches lie between the same two junctions and so have the same p.d. across them, and I = V/R, so the smaller resistance carries the larger current
bDetermine The potential difference between the two junction dots in Fig. 3.1 is 12 V. Determine the current in each resistor and the current in the single wire arriving from the left.
routine3 marks
Check answer · 3 marks
current in the 6.0 Ω resistor = 12/6.0 = 2.0 A
current in the 3.0 Ω resistor = 12/3.0 = 4.0 A
current in the wire from the left = 2.0 + 4.0 = 6.0 A
cDetermine Each resistor in Fig. 3.1 is rated at a maximum power of 12 W. Determine the largest potential difference that may safely be applied between the two junction dots, and the current drawn at that potential difference.
demanding4 marks
Check answer · 4 marks
for the 3.0 Ω resistor, V = √(PR) = √(12 × 3.0) = 6.0 V
for the 6.0 Ω resistor, V = √(12 × 6.0) = 8.5 V, so the 3.0 Ω resistor is the one that limits the pair
largest safe p.d. = 6.0 V
current drawn = 6.0/6.0 + 6.0/3.0 = 3.0 A
dExplain A student claims that connecting a third resistor between the same two junction dots in Fig. 3.1 must raise the total resistance, because there is then more resistance present. Explain why the student is wrong, and state the largest total resistance the arrangement could ever have once a third resistor is added there.
top of the paper4 marks
Check answer · 4 marks
a resistor connected between the same two dots is in parallel with the others and gives the current an extra path
for the same p.d. the total current is therefore larger, and R = V/I, so the total resistance falls
the pair alone gives 1/R = 1/6.0 + 1/3.0, so R = 2.0 Ω
adding a third branch can only bring the total below 2.0 Ω, so 2.0 Ω is a value the arrangement approaches but never reaches
Transfer challenge
Three identical lamps, each of resistance 240 Ω, are connected in parallel across a 240 V supply. Calculate the total current drawn, and state the effect on the other two lamps if one filament breaks.
Check answer · 3 marks
current in each lamp = 240/240 = 1.0 A
total current = 3 × 1.0 = 3.0 A
the other two lamps still have the full 240 V across them and are unaffected, and the total current falls to 2.0 A
02Fig. 7.1Electrical quantities · Electric circuitsIGCSEOpen diagramClose diagram
Figure comment
Fig. 7.1Circuit diagram. From the positive terminal of the 12 V battery, drawn at the foot of the circuit with its long plate on the left, the wire runs round to a resistor labelled 40 ohm. Beyond that resistor the circuit reaches a junction and divides into two parallel branches, the upper one containing a resistor labelled 30 ohm and the lower one a resistor labelled 60 ohm. The branches rejoin at a second junction, and a single wire returns from there to the negative terminal of the battery.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. The 40 Ω sits before the first junction dot, so it carries the whole battery current; only past that dot is the current shared between the 30 Ω and the 60 Ω.
aDetermine The current drawn from the battery in Fig. 7.1 is 0.20 A. Determine the potential difference across the 40 Ω resistor.
recall2 marks
Check answer · 2 marks
the 40 Ω resistor lies before the junction, so it carries the whole 0.20 A
V = IR = 0.20 × 40 = 8.0 V
bDetermine Determine the power dissipated in the 40 Ω resistor in Fig. 7.1, and the total power supplied by the battery.
routine2 marks
Check answer · 2 marks
power in the 40 Ω resistor = I²R = 0.20² × 40 = 1.6 W
total power from the battery = VI = 12 × 0.20 = 2.4 W
cDetermine The 40 Ω resistor in Fig. 7.1 is replaced by a variable resistor. Determine the resistance it must be set to for the potential difference across the parallel pair to be 6.0 V.
demanding3 marks
Check answer · 3 marks
the parallel pair has a combined resistance of (30 × 60)/90 = 20 Ω, so the current is 6.0/20 = 0.30 A
p.d. across the variable resistor = 12 − 6.0 = 6.0 V
R = V/I = 6.0/0.30 = 20 Ω
dExplain The 30 Ω resistor in Fig. 7.1 is replaced by a filament lamp whose resistance rises as it warms up. Explain what happens to the potential difference across the 40 Ω resistor as the lamp warms, and state whether the current in the 60 Ω resistor rises or falls.
top of the paper4 marks
Check answer · 4 marks
the resistance of the lamp rises, so the combined resistance of the parallel pair rises
the total resistance of the circuit rises, so the current drawn from the battery falls
the p.d. across the 40 Ω resistor is that current × 40, so it falls
the parallel pair therefore takes a larger share of the 12 V, so the current in the 60 Ω resistor, equal to that p.d. divided by 60, rises
Transfer challenge
A 9.0 V battery is connected to a 200 Ω resistor in series with a thermistor whose resistance is 400 Ω at room temperature. Calculate the potential difference across the 200 Ω resistor, and state how it changes when the thermistor is warmed.
Check answer · 3 marks
total resistance = 200 + 400 = 600 Ω, so I = 9.0/600 = 0.015 A
p.d. across the 200 Ω resistor = 0.015 × 200 = 3.0 V
warming lowers the resistance of the thermistor, so the current rises and the p.d. across the 200 Ω resistor rises