What Is Electric Circuits? A Visual Guide for IGCSE Physics Students

Get the circuit rules right or lose marks

Electric circuits are tested on every single IGCSE Physics paper. Not most papers. Every paper. You need to know how current flows, how voltage splits, how resistance adds up, and how to use V = IR without hesitation. This guide gives you exactly that: the rules, the calculations, and the common traps students fall into.

No fluff. Just what you need to score marks.

Key facts: electric circuits at a glance

• Current (I) is the rate of flow of charge, measured in amperes (A). Formula: I = Q / t.
• Potential difference (V) is the energy transferred per unit charge, measured in volts (V). Formula: V = W / Q.
• Resistance (R) measures how much a component opposes current, measured in ohms (Ω). Formula: R = V / I.
• Ohm's law: V = IR applies to ohmic conductors at constant temperature.
• Series circuit: one path. Current is the same everywhere. Voltages add up.
• Parallel circuit: multiple paths. Voltage is the same across each branch. Currents add up.

Circuit symbols you must recognise

Examiners expect clean, correct circuit diagrams. Draw a wonky symbol or mislabel a component, and you've already lost marks before the calculation even starts.

Series circuits: one path, one set of rules

A series circuit has a single path for current. Every component sits on the same loop. Break one component and the whole circuit goes dead.

The rules are straightforward:

• Current is the same through every component.
• Voltage is shared between components: V_total = V₁ + V₂ + ...
• Total resistance adds up: R_total = R₁ + R₂ + ...

Think of it as a single road. Every car (unit of charge) passes through every toll booth (component). The flow rate (current) stays constant, but each booth takes its share of the fare (voltage).

Worked example: series circuit

A 9 V battery is connected in series with a 3 Ω resistor and a 6 Ω resistor. Find the current and the p.d. across each resistor.

1. Total resistance: R = 3 + 6 = 9 Ω
2. Current: I = V / R = 9 / 9 = 1.0 A
3. P.d. across 3 Ω: V = IR = 1.0 × 3 = 3.0 V
4. P.d. across 6 Ω: V = IR = 1.0 × 6 = 6.0 V
5. Check: 3.0 + 6.0 = 9.0 V. Matches the battery. Done.

That check at the end? Do it every time. If your voltages don't add up to the supply, you've made an error somewhere.

Parallel circuits: multiple paths, shared voltage

A parallel circuit splits into branches. Current divides at junctions and recombines afterwards. Each branch gets the full supply voltage.

The rules flip compared to series:

• Voltage is the same across every branch.
• Current is shared: I_total = I₁ + I₂ + ...
• Total resistance decreases: 1/R_total = 1/R₁ + 1/R₂ + ...

That last point trips students up constantly. Adding a resistor in parallel always makes the total resistance smaller, not bigger. More paths means less overall opposition to current.

Worked example: parallel circuit

A 12 V supply is connected across two resistors in parallel: 4 Ω and 12 Ω. Find the current through each and the total current.

1. P.d. across each = 12 V (same in parallel).
2. Current through 4 Ω: I = 12 / 4 = 3.0 A
3. Current through 12 Ω: I = 12 / 12 = 1.0 A
4. Total current: 3.0 + 1.0 = 4.0 A
5. Check with combined resistance: 1/R = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3, so R = 3 Ω. Then I = 12 / 3 = 4.0 A. Confirmed.

V = IR: the equation that runs the show

Almost every circuit calculation comes back to V = IR. Rearrange it three ways and know when to use each one:

Don't just memorise these. Practise rearranging them until it's automatic. In the exam, you won't have time to sit there thinking about which version to use. You should see the values, grab the right formula, and go.

Worked example: finding resistance from a graph

An I-V graph for an ohmic conductor shows a straight line through the origin. At V = 6.0 V, the current reads 0.30 A. What's the resistance?

R = V / I = 6.0 / 0.30 = 20 Ω

For a filament lamp, the graph curves. The resistance isn't constant because the filament heats up as current flows through it. If the exam asks for resistance at a specific point on a curved I-V graph, read off V and I at that exact point and use R = V / I. Don't try to use the gradient. The gradient of an I-V graph gives 1/R, and on a curve that value changes at every point.

Resistance in combined circuits

Real exam circuits mix series and parallel sections. The trick is to break them into chunks and solve one chunk at a time.

Worked example: combination circuit

Two 6 Ω resistors are connected in parallel with each other. This parallel combination is then connected in series with a 4 Ω resistor. The whole circuit is powered by a 10 V battery. Find the total resistance and the current from the battery.

1. Parallel section first: 1/R_p = 1/6 + 1/6 = 2/6, so R_p = 3 Ω
2. Now add the series resistor: R_total = 3 + 4 = 7 Ω
3. Total current: I = V / R = 10 / 7 = 1.43 A (3 s.f.)

Always solve the parallel section first, then combine with the series section. Trying to do everything in one step is how students make errors that cascade through the whole question.

Potential dividers (Extended syllabus)

A potential divider is just two resistors in series. The output voltage is taken from the junction between them. The formula:

V_out = V_in × R₂ / (R₁ + R₂)

Nothing complicated here. The bigger resistor takes a bigger share of the voltage. That's all it does.

Where it gets interesting is when you swap one resistor for a thermistor or LDR. Suddenly V_out changes with temperature or light intensity. That's a sensor circuit, and Cambridge examiners test it regularly.

Worked example: potential divider

A 12 V supply connects to a 2 kΩ resistor (R₁) in series with a 4 kΩ resistor (R₂). Find V_out across R₂.

V_out = 12 × 4000 / (2000 + 4000) = 12 × 4000 / 6000 = 8.0 V

R₂ is twice the size of R₁, so it takes two-thirds of the voltage. Quick sense check: 8 out of 12 is indeed two-thirds. If your answer doesn't pass a rough proportion check like this, go back and look for a mistake.

Sensor circuit: thermistor as R₁

Replace R₁ with a thermistor. As temperature rises, the thermistor's resistance drops. R₁ gets smaller, so R₂ now takes a larger proportion of V_in. V_out across R₂ goes up. If you need the opposite effect (V_out dropping with temperature), put the thermistor as R₂ instead.

The same logic works for an LDR. More light means lower LDR resistance. Work out which position gives you the response you want.

Practical skills: ammeter and voltmeter placement

This catches students out badly on Papers 5 and 6. The rules are non-negotiable:

• Ammeter: always in series with the component you're measuring current through. It has very low resistance so it barely affects the circuit.
• Voltmeter: always in parallel across the component you're measuring p.d. across. It has very high resistance so almost no current flows through it.

Swapping them is a disaster. An ammeter in parallel creates a near-short circuit because of its low resistance across the supply. A voltmeter in series blocks almost all current because of its high resistance in the path. Both give readings that are either useless or dangerous.

When drawing circuit diagrams in the exam, place the ammeter first (in the loop, in series), then add the voltmeter last (connected across the component of interest, in parallel). This order makes it harder to get them mixed up.

Do and Don't: circuit question mistakes

Test yourself

Work through each problem on paper before looking at the answers. Doing them in your head doesn't build exam technique.

1. A 6 V battery connects in series with a 2 Ω and a 4 Ω resistor. What is the current? What is the p.d. across each resistor?
2. Two resistors, 10 Ω and 15 Ω, are connected in parallel across a 30 V supply. Find the current through each resistor and the total current.
3. A combination circuit has a 5 Ω resistor in series with a parallel pair of 8 Ω and 8 Ω. The supply is 15 V. Find the total resistance and the current from the supply.
4. In a potential divider, R₁ = 3 kΩ and R₂ = 6 kΩ. The supply is 9 V. What is V_out across R₂?
5. A thermistor replaces R₁ in a potential divider. What happens to V_out across R₂ when temperature increases? Explain your reasoning.

Circuit questions: how to pick up every mark

Cambridge IGCSE circuit questions follow predictable patterns. Here's how to handle them efficiently:

• Read the circuit diagram first. Trace the path from positive terminal to negative. Count how many paths exist.
• Write down what you know: supply voltage, component values, any given currents or voltages.
• Identify what you need to find. Pick the right formula before touching your calculator.
• Show every step of your working. Method marks are your safety net if you make an arithmetic slip.
• Include units on every single answer. No unit means no mark on a calculation question.
• Verify: do your voltages add up to the supply in a series section? Do your branch currents add up to the total in a parallel section?

Circuit problems account for a hefty chunk of your IGCSE Physics exam. Get these rules drilled, practise the calculations until they're second nature, and you'll pick up marks that other students leave on the table.