What is an electric circuit and why is it important in Physics?

An electric circuit is a closed loop that allows electric current to flow through it, typically consisting of a power source, conductors, and a load. It is crucial in Physics, especially in the Cambridge IGCSE curriculum, as it helps students understand how electrical energy is transferred and used in various devices, forming the foundation for more advanced topics in Electricity and Magnetism.

How do series and parallel circuits differ in terms of current and voltage?

In a series circuit, the current is the same through all components, but the voltage is divided among them. In contrast, a parallel circuit has the same voltage across all components, but the current is divided. Understanding these differences is essential for Cambridge IGCSE Physics students to analyze circuit behavior and solve related problems.

What role do resistors play in an electric circuit?

Resistors are components that limit the flow of electric current in a circuit. They are used to control voltage and current levels, protect sensitive components, and divide voltages. In the Cambridge IGCSE Physics syllabus, students learn how to calculate resistance using Ohm's Law and understand its impact on circuit functionality.

How is Ohm's Law applied in electric circuits?

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points, with the resistance being the constant of proportionality. It is expressed as V = IR, where V is voltage, I is current, and R is resistance. This fundamental principle is vital for Cambridge IGCSE students to solve circuit problems and understand the relationship between voltage, current, and resistance.

What is the significance of a circuit diagram in understanding electric circuits?

A circuit diagram is a graphical representation of an electric circuit, using symbols to depict components like resistors, batteries, and switches. It is significant because it provides a clear and simplified way to visualize and analyze the arrangement and connections within a circuit. Cambridge IGCSE Physics students use circuit diagrams to predict circuit behavior and troubleshoot issues effectively.

Why is "Forgetting to invert at the end of the parallel calculation" a common mistake in Electric Circuits?

Why it happens: Stopping at 1/R_T. How to avoid it: After summing reciprocals, INVERT to get R_T. Source: 0625/42 — every series

Why is "Getting parallel resistance larger than each individual" a common mistake in Electric Circuits?

Why it happens: Adding instead of using the reciprocal rule. How to avoid it: Parallel total must be SMALLER than the smallest of the resistors. If not, recheck.

Why is "Saying current is the same in every branch of a parallel combination" a common mistake in Electric Circuits?

Why it happens: Mixing series and parallel rules. How to avoid it: Series → same CURRENT. Parallel → same PD. Switch your reasoning to match the topology.

Why is "Inverting the potential divider fraction" a common mistake in Electric Circuits?

Why it happens: Memory slip. How to avoid it: V across one resistor = V_T × that resistorsum. The bigger the resistor, the bigger its share of pd.

Cambridge IGCSE Physics (0625)

Electric Circuits

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Short Notes - Electric Circuits

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Detailed Study Notes

Detailed notes on Electricity and Magnetism for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Electric Circuits — Cambridge IGCSE 0625 Physics Extended (2026)

Series and parallel circuits, the rules for current and voltage, combined resistance, plus thermistors, LDRs and the potential divider. The skill of reading a circuit diagram.

At a glance

Series: same current everywhere; voltages add up.
Parallel: same voltage across each branch; currents add up.
Series resistors: $R_{\text{tot}} = R_{1} + R_{2} + \ldots$
Parallel resistors: $\dfrac{1}{R_{\text{tot}}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \ldots$
Thermistor: resistance DROPS when heated.
LDR (light-dependent resistor): resistance DROPS in bright light.
Potential divider: two resistors in series share the supply voltage in ratio.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

4.3 — State the current and voltage rules for series and parallel circuits.
4.3 — Calculate combined resistance for series and parallel arrangements.
4.3 — Describe action of thermistor and LDR.
4.3 — Use the potential divider rule.

Series circuits

Same $I$ through every component; $V$ is shared, sum equals supply.

Series circuit. Components connected end-to-end in a single loop. Only one path for current.

Rules.

Current: SAME at every point in the circuit.
Voltage: total supply voltage = sum of voltages across each component.
Resistance: $R_{\text{tot}} = R_{1} + R_{2} + R_{3} + \ldots$

Why current is the same. Charge has nowhere else to go — what flows in must flow out at every point.

One path: current is identical throughout; the resistor voltages add to the 9 V supply.

Worked. A $9\,\text{V}$ battery in series with $R_{1} = 2\,\Omega$ and $R_{2} = 4\,\Omega$ .

$R_{\text{tot}} = 2 + 4 = 6\,\Omega$ .
$I = V/R_{\text{tot}} = 9/6 = 1.5\,\text{A}$ .
$V_{1} = IR_{1} = 1.5 \times 2 = 3\,\text{V}$ .
$V_{2} = IR_{2} = 1.5 \times 4 = 6\,\text{V}$ . ( $3 + 6 = 9\,\text{V}$ ✓.)

Disadvantage. If one bulb fails (open), the WHOLE circuit breaks — like Christmas lights in older designs.

Series: same $I$ everywhere.
Voltages add up to supply.
$R_{\text{tot}} = R_{1} + R_{2} + \ldots$
One bulb fails → all fail.

Parallel circuits

Same $V$ across each branch; total $I$ = sum of branch currents. Total $R$ less than smallest.

Parallel circuit. Components on separate branches between the same two points. Multiple paths for current.

Rules.

Voltage: SAME across each parallel branch (= supply voltage if branches connect directly to the cell).
Current: total current from supply = sum of currents in each branch.
Resistance: $\dfrac{1}{R_{\text{tot}}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \ldots$ The total resistance is LESS than the smallest individual resistor.

Worked. Two resistors in parallel: $R_{1} = 4\,\Omega$ , $R_{2} = 6\,\Omega$ .

$\dfrac{1}{R_{\text{tot}}} = \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}$ .
$R_{\text{tot}} = 12/5 = 2.4\,\Omega$ .
Confirm: less than $4$ ✓.

Each branch sees the full supply voltage; branch currents add up — total R is below the smallest resistor.

Two-resistor shortcut. $R_{\text{tot}} = \dfrac{R_{1}R_{2}}{R_{1} + R_{2}}$ .

Worked. Same as above: $R_{\text{tot}} = (4 \times 6)/(4 + 6) = 24/10 = 2.4\,\Omega$ ✓.

Advantage. If one bulb fails, the others stay on (each on its own branch). Used for household wiring.

Parallel: same $V$ across branches.
Branch currents add to total.
$1/R_{\text{tot}} = \sum 1/R_{i}$ .
Two-resistor: $R_{1}R_{2}/(R_{1}+R_{2})$ .
Total $R$ < smallest individual.

Thermistors and LDRs

Thermistor: $R$ drops when hot. LDR: $R$ drops in light.

Thermistor (NTC type).

Resistance DECREASES as temperature increases.
Used in: temperature-sensing circuits, fire alarms, oven controllers.
Rule of thumb: at room temp $\sim$ kΩ, at boiling point $\sim$ tens of Ω.

LDR (Light-Dependent Resistor).

Resistance DECREASES as light intensity increases.
Used in: street lights (turn on at dusk), camera light meters, automatic doors (sometimes).

Both are non-linear, non-ohmic components. A graph of $R$ vs temperature (thermistor) or $R$ vs light intensity (LDR) is a curve, not a straight line.

Cambridge tip. When asked "describe the action of a thermistor" — say something like "as temperature rises, the resistance of the thermistor falls". Cambridge mark schemes look for this exact phrasing.

Thermistor: $R$ DROPS when HOTTER.
LDR: $R$ DROPS in BRIGHTER light.
Both non-ohmic.
Used in sensing circuits.

Potential divider

Two resistors in series across a supply: voltage splits in ratio of the resistors.

Potential divider. Two resistors $R_{1}$ and $R_{2}$ in series across a supply $V_{\text{in}}$ . The output voltage taken across $R_{2}$ is: $V_{\text{out}} = V_{\text{in}} \times \dfrac{R_{2}}{R_{1} + R_{2}}.$

So the supply voltage divides between the two resistors in proportion to their values.

The larger resistor takes the larger share: Vout = 12 × 400 ÷ (200 + 400) = 8 V.

Worked. $V_{\text{in}} = 12\,\text{V}$ , $R_{1} = 200\,\Omega$ , $R_{2} = 400\,\Omega$ . Find $V_{\text{out}}$ across $R_{2}$ .

$V_{\text{out}} = 12 \times 400/(200+400) = 12 \times 400/600 = 8\,\text{V}$ .

Use with sensors. Replace one of the resistors with a thermistor or LDR. Then $V_{\text{out}}$ varies with temperature or light — useful for triggering alarms or switches.

Worked qualitative. Light-sensitive circuit: replace $R_{2}$ with an LDR. In darkness LDR has high $R$ → $V_{\text{out}}$ across it is large. In bright light LDR has low $R$ → $V_{\text{out}}$ small. Feed $V_{\text{out}}$ to a transistor that switches a streetlight: light goes ON in the dark.

$V_{\text{out}} = V_{\text{in}} \times R_{2}/(R_{1}+R_{2})$ .
Larger resistor takes larger voltage.
Replace one resistor with LDR/thermistor for a sensor circuit.

Quick recap

Series: same $I$ , voltages add, $R_{\text{tot}} = \sum R_{i}$ .
Parallel: same $V$ , currents add, $1/R_{\text{tot}} = \sum 1/R_{i}$ .
Thermistor: $R \downarrow$ when $T \uparrow$ .
LDR: $R \downarrow$ when light $\uparrow$ .
Potential divider: $V_{\text{out}} = V_{\text{in}} \times R_{2}/(R_{1}+R_{2})$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Series circuit — components in a single loop; same current, voltages add.
Parallel circuit — branches between common points; same voltage, currents add.
Thermistor — resistor whose resistance varies with temperature.
LDR — Light-Dependent Resistor; resistance falls with brightness.
Potential divider — series resistor pair that splits a voltage.

How it’s examined

Circuits appear every Paper 4 (8-12 marks) — combined resistance, current and voltage rules, often combined with a sensor and potential divider. Examiner reports flag forgetting to take the reciprocal in parallel resistance, and mixing up series and parallel rules in mixed circuits.

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (4.3); 0625/42 Oct/Nov 2024 — Q17 (parallel circuit, potential divider); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — Electric Circuits

Step-by-step solutions to past-paper-style questions on electric circuits, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Electric Circuits

Almost every electric circuits exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

One resistor combination to evaluate — find the total resistance, find the current in this branch or find the pd across this resistor — with a single rule to apply.

How to approach it

Identify the topology first: series adds directly, parallel uses $\frac{1}{R_T}=\sum\frac{1}{R_n}$ , the divider gives $V_n=V_T\times\frac{R_n}{R_T}$ . Substitute and finish the calculation.

Common trap

Stopping at $1/R_T$ without inverting. Examiner reports flag this every series — and a parallel total must always be smaller than the smallest resistor, so check.

Multi-step problem

Recognise it by

A mixed series-and-parallel network: you must combine resistors, then find a current, then a pd or an unknown resistor — several stages chained together.

How to approach it

Reduce the circuit one block at a time: collapse the parallel section to a single resistor, add the series parts, then apply $V=IR$ working outwards from the supply.

Common trap

Applying the unloaded divider formula when a load draws current. Examiner reports note that once a load is connected, the bottom resistor and load must be recombined in parallel first.

Show that / prove

Recognise it by

Explain or state qualitatively what happens — a reasoning answer about how an output changes, e.g. a thermistor or LDR divider as temperature or light changes.

How to approach it

State how the sensor's resistance changes, substitute that change into the divider rule, and trace the effect on the fraction to predict whether $V_{out}$ rises or falls.

Common trap

Forgetting which component the output is taken across. Examiner reports flag that the resistor the output sits across must be the numerator of the divider fraction.

1Resistors in series
CoreDirect calculation• series
Question
Three resistors $4\,\Omega$ , $6\,\Omega$ , $10\,\Omega$ in series. Find the total resistance.
Step-by-step solution
1. Step 1
  Add directly.
  $R_{T} = 4 + 6 + 10 = 20\,\Omega$
Answer
$20\,\Omega$
2Resistors in parallel
ExtendedDirect calculation• Adapted from 0625/42 May/Jun 2024 Q17• parallel
Question
Two resistors $6\,\Omega$ and $3\,\Omega$ in parallel. Find the total resistance.
Step-by-step solution
1. Step 1
  $\dfrac{1}{R_{T}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}$ .
  $\dfrac{1}{R_{T}} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{6} + \dfrac{2}{6} = \dfrac{3}{6}$
2. Step 2
  Invert.
  $R_{T} = 2\,\Omega$
Answer
$2\,\Omega$
Examiner tip
Parallel resistance is ALWAYS less than the smallest individual resistor.
3Current in a parallel branch
ExtendedDirect calculation• current
Question
A $12\,\text{V}$ supply drives the parallel pair above. Find the current through the $6\,\Omega$ resistor.
Step-by-step solution
1. Step 1
  PD across each branch in parallel = supply pd = $12\,\text{V}$ .
2. Step 2
  $I = V/R$ .
  $I = \dfrac{12}{6} = 2\,\text{A}$
Answer
$2\,\text{A}$
4Potential divider
ExtendedDirect calculation• divider
Question
A $9\,\text{V}$ battery is connected across $4\,\Omega$ and $5\,\Omega$ in series. Find the pd across the $5\,\Omega$ resistor.
Step-by-step solution
1. Step 1
  $V = V_{\text{total}} \times \dfrac{R_{\text{branch}}}{R_{\text{total}}}$ .
  $V_{5\Omega} = 9 \times \dfrac{5}{4 + 5} = 5\,\text{V}$
Answer
$5\,\text{V}$
5Ammeter and voltmeter readings — mixed circuit
ExtendedMulti-step problem• Adapted from 0625/42 Oct/Nov 2023 Q9• ammeter, voltmeter, mixed
Question
A $12\,\text{V}$ battery is connected to a $4\,\Omega$ resistor in SERIES with a parallel combination of $6\,\Omega$ and $3\,\Omega$ . An ammeter reads the total current; a voltmeter reads the pd across the parallel block. Find both readings.
Step-by-step solution
1. Step 1
  Parallel block: $\dfrac{1}{R_{p}} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}$ , so $R_{p} = 2\,\Omega$ .
2. Step 2
  Total circuit resistance: $R_{T} = 4 + 2 = 6\,\Omega$ .
3. Step 3
  Total current (ammeter): $I = V/R_{T}$ .
  $I = 12/6 = 2.0\,\text{A}$
4. Step 4
  Voltmeter reads pd across parallel block: $V_{p} = I R_{p}$ .
  $V_{p} = 2.0 \times 2 = 4.0\,\text{V}$
Answer
Ammeter: $2.0\,\text{A}$ . Voltmeter: $4.0\,\text{V}$ .
6Find a missing resistor in parallel
ExtendedDirect calculation• parallel, missing R
Question
Two resistors are connected in parallel. One is $12\,\Omega$ and the total resistance is $4\,\Omega$ . Find the second resistor.
Step-by-step solution
1. Step 1
  Use the parallel formula and rearrange for the unknown.
  $\dfrac{1}{R_{T}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} \Rightarrow \dfrac{1}{R_{2}} = \dfrac{1}{R_{T}} - \dfrac{1}{R_{1}}$
2. Step 2
  Substitute.
  $\dfrac{1}{R_{2}} = \dfrac{1}{4} - \dfrac{1}{12} = \dfrac{3 - 1}{12} = \dfrac{2}{12}$
3. Step 3
  Invert.
  $R_{2} = 6\,\Omega$
Answer
$6\,\Omega$
7Currents at a junction
CoreDirect calculation• Kirchhoff, junction
Question
At a junction, currents of $0.6\,\text{A}$ and $0.9\,\text{A}$ flow in along two wires. A third wire carries the rest away from the junction. Find this current.
Step-by-step solution
1. Step 1
  Kirchhoff's first law: total current in = total current out (conservation of charge at a junction).
2. Step 2
  Solve.
  $I_{\text{out}} = 0.6 + 0.9 = 1.5\,\text{A}$
Answer
$1.5\,\text{A}$
8LDR in a potential divider
ExtendedDirect calculation• LDR, divider
Question
A $5.0\,\text{V}$ supply is connected to a $2\,\text{k}\Omega$ fixed resistor in series with an LDR. The output is taken across the LDR. In bright light the LDR has resistance $500\,\Omega$ ; in darkness $20\,\text{k}\Omega$ . Find $V_{\text{out}}$ for each case.
Step-by-step solution
1. Step 1
  $V_{\text{out}} = V_{\text{in}} \times \dfrac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}}$ .
2. Step 2
  Bright (LDR $= 500\,\Omega$ ).
  $V_{\text{out}} = 5.0 \times \dfrac{500}{2000 + 500} = 5.0 \times \dfrac{500}{2500} = 1.0\,\text{V}$
3. Step 3
  Dark (LDR $= 20\,\text{k}\Omega = 20000\,\Omega$ ).
  $V_{\text{out}} = 5.0 \times \dfrac{20000}{2000 + 20000} = 5.0 \times \dfrac{20000}{22000} \approx 4.5\,\text{V}$
Answer
Bright: $\approx 1.0\,\text{V}$ . Dark: $\approx 4.5\,\text{V}$ .
Examiner tip
The examiner report flags candidates often forgetting that 'output across the LDR' means the LDR's resistance appears in the NUMERATOR of the divider fraction.
9Thermistor as a temperature switch
ExtendedShow that / prove• thermistor, divider
Question
A $9.0\,\text{V}$ supply, a thermistor and a $3\,\text{k}\Omega$ fixed resistor are connected in series. The output is taken across the FIXED resistor. As temperature RISES, state qualitatively what happens to $V_{\text{out}}$ and explain.
Step-by-step solution
1. Step 1
  Thermistor resistance DECREASES as temperature rises (negative temperature coefficient — standard 0625 thermistor).
2. Step 2
  Divider rule: $V_{\text{out}} = V_{\text{in}} \times \dfrac{R_{\text{fixed}}}{R_{\text{thermistor}} + R_{\text{fixed}}}$ .
3. Step 3
  When the thermistor's $R$ falls, the denominator falls, so the fraction → $1$ and $V_{\text{out}}$ rises towards $V_{\text{in}} = 9.0\,\text{V}$ .
Answer
$V_{\text{out}}$ RISES as temperature increases (approaches $9\,\text{V}$ when the thermistor's resistance becomes very small compared with $3\,\text{k}\Omega$ ).
10Current splitting between two parallel branches
ExtendedMulti-step problem• parallel, current
Question
A $9.0\,\text{V}$ supply is connected across two resistors in parallel: $R_{1} = 6\,\Omega$ and $R_{2} = 3\,\Omega$ . Find the current in each branch and the total supply current.
Step-by-step solution
1. Step 1
  In parallel, each branch has the same pd as the supply: $9.0\,\text{V}$ .
2. Step 2
  Branch currents: $I = V/R$ .
  $I_{1} = 9.0/6 = 1.5\,\text{A};\quad I_{2} = 9.0/3 = 3.0\,\text{A}$
3. Step 3
  Total supply current (Kirchhoff at the junction): $I = I_{1} + I_{2}$ .
  $I = 1.5 + 3.0 = 4.5\,\text{A}$
Answer
$I_{1} = 1.5\,\text{A}$ , $I_{2} = 3.0\,\text{A}$ , total $4.5\,\text{A}$ .
Examiner tip
The examiner report flags candidates often calculating one branch current then assuming the OTHER branch carries the rest of an assumed total — start from the supply pd, not from a guess at total current.
11Combined series-and-parallel — find unknown resistor
ChallengeMulti-step problem• multi-loop, mixed
Question
A $12\,\text{V}$ battery is connected to a circuit consisting of a $2\,\Omega$ resistor in SERIES with a parallel combination of an unknown $R$ and a $6\,\Omega$ resistor. The total current from the battery is $3.0\,\text{A}$ . Find $R$ .
Step-by-step solution
1. Step 1
  Total resistance: $R_{T} = V/I = 12/3.0 = 4\,\Omega$ .
2. Step 2
  Series chain: $R_{T} = 2 + R_{p}$ , so $R_{p} = 2\,\Omega$ .
3. Step 3
  Parallel of $R$ and $6\,\Omega$ equals $2\,\Omega$ :
  $\dfrac{1}{2} = \dfrac{1}{R} + \dfrac{1}{6} \Rightarrow \dfrac{1}{R} = \dfrac{1}{2} - \dfrac{1}{6} = \dfrac{3-1}{6} = \dfrac{1}{3}$
4. Step 4
  Invert.
  $R = 3\,\Omega$
Answer
$R = 3\,\Omega$
12Potential divider with a load — synoptic
ChallengeMulti-step problem• Adapted from 0625/42 May/Jun 2022 Q10• divider, Kirchhoff, synoptic
Question
A $12\,\text{V}$ supply is connected across a series chain of two $6\,\Omega$ resistors ( $R_{1}$ on top, $R_{2}$ on bottom). The output is taken across $R_{2}$ . Now an external load of $6\,\Omega$ is connected in parallel with $R_{2}$ . Find the new output voltage and the current drawn from the supply.
Step-by-step solution
1. Step 1
  With the load attached, the bottom half is $R_{2}$ ( $6\,\Omega$ ) in parallel with the load ( $6\,\Omega$ ).
  $R_{\text{bottom}} = \dfrac{6 \times 6}{6 + 6} = 3\,\Omega$
2. Step 2
  Total resistance: $R_{T} = R_{1} + R_{\text{bottom}} = 6 + 3 = 9\,\Omega$ .
3. Step 3
  Supply current: $I = V/R_{T}$ .
  $I = 12/9 \approx 1.33\,\text{A}$
4. Step 4
  Output pd (across the parallel block): $V_{\text{out}} = I \times R_{\text{bottom}}$ .
  $V_{\text{out}} = 1.33 \times 3 = 4.0\,\text{V}$
5. Step 5
  Compare with the unloaded value of $V_{\text{out}} = 6.0\,\text{V}$ — connecting a load PULLS DOWN the divider's output.
Answer
$V_{\text{out}} \approx 4.0\,\text{V}$ , supply current $\approx 1.33\,\text{A}$ . The load loads the divider, dropping the output from $6\,\text{V}$ to $4\,\text{V}$ .
Examiner tip
The examiner report flags candidates often applying the unloaded divider formula even when a load is connected — once a load draws current, the bottom resistor is in parallel with the load and must be recombined first.

Model Answers — Electric Circuits

High-scoring sample answers for electric circuits on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
State one difference between how current behaves in a series circuit and in a parallel circuit.
Model answer
In a series circuit the current is the same at every point (one single path). In a parallel circuit the current splits between the branches, so different branches can carry different currents (and they add up to the supply current).
Why this scores
One mark for any correct distinction — same current everywhere in series vs current divides in parallel. The matching rule is: series shares the voltage, parallel shares the current.
Question 2
Paper 2/4 style2 marks
A $4.0\,\Omega$ resistor and a $12\,\Omega$ resistor are connected (a) in series and (b) in parallel. Calculate the total resistance in each case.
Model answer
(a) Series: $R_{T} = 4.0 + 12 = \mathbf{16\,\Omega}$ .

(b) Parallel: $\dfrac{1}{R_{T}} = \dfrac{1}{4.0} + \dfrac{1}{12} = \dfrac{3}{12} + \dfrac{1}{12} = \dfrac{4}{12}$ , so $R_{T} = \mathbf{3.0\,\Omega}$ .
Why this scores
One mark for the series total, one for the parallel total. Remember to invert at the end of the parallel calculation — and the parallel total ( $3.0\,\Omega$ ) must be smaller than the smallest resistor.
Question 3
Paper 4 structured style3 marks
A $12\,\text{V}$ battery is connected across two resistors in parallel: $6.0\,\Omega$ and $3.0\,\Omega$ . Calculate (a) the current in each resistor and (b) the total current drawn from the battery.
Model answer
(a) In parallel, each resistor has the full supply pd of $12\,\text{V}$ across it. So $I_{6} = \dfrac{12}{6.0} = \mathbf{2.0\,\text{A}}$ and $I_{3} = \dfrac{12}{3.0} = \mathbf{4.0\,\text{A}}$ .

(b) The total current is the sum of the branch currents (Kirchhoff's first law): $I = 2.0 + 4.0 = \mathbf{6.0\,\text{A}}$ .
Why this scores
Three marks: each branch has the full pd (1); both branch currents (1); total = sum of branches (1). The smaller resistor carries the larger current.
Question 4
Paper 4 structured style4 marks
A $12\,\text{V}$ supply is connected across a $4.0\,\Omega$ resistor and an $8.0\,\Omega$ resistor in series. (a) Calculate the potential difference across the $8.0\,\Omega$ resistor. (b) State which resistor has the larger pd across it and why.
Model answer
(a) The same current flows through both. Using the potential-divider relationship:

$V_{8} = V_{\text{total}} \times \dfrac{R_{8}}{R_{4} + R_{8}} = 12 \times \dfrac{8.0}{4.0 + 8.0} = 12 \times \dfrac{8.0}{12} = \mathbf{8.0\,\text{V}}.$

(b) The $8.0\,\Omega$ resistor has the larger pd. Because the current is the same through both, the larger resistance has the larger pd across it ( $V = IR$ ) — it takes the bigger share of the supply voltage.
Why this scores
Four marks: same current in series (1); correct divider substitution (1); $V_{8} = 8.0\,\text{V}$ (1); larger resistor takes the larger share of pd (1). A useful check: the two pds ( $4.0\,\text{V} + 8.0\,\text{V}$ ) add to the supply $12\,\text{V}$ .
Question 5
Paper 4 application style5 marks
A thermistor and a fixed resistor are connected in series across a $9.0\,\text{V}$ supply, with the output voltage taken across the fixed resistor. (a) State how the resistance of a thermistor changes as it gets hotter. (b) Explain what happens to the output voltage as the temperature rises, and how this circuit could switch on a cooling fan.
Model answer
(a) As the thermistor gets hotter, its resistance decreases.

(b) The output is taken across the fixed resistor, so $V_{\text{out}} = V_{\text{in}} \times \dfrac{R_{\text{fixed}}}{R_{\text{thermistor}} + R_{\text{fixed}}}$ . As the temperature rises, the thermistor's resistance falls, so the denominator falls and the fraction increases — the output voltage rises (towards $9.0\,\text{V}$ ). This rising voltage can be fed to a switching circuit so that, once it reaches a set value, it switches on a cooling fan when the temperature gets too high.
Why this scores
Five marks: thermistor resistance falls as temperature rises (1); output across the fixed resistor (1); thermistor resistance drop raises the share of pd on the fixed resistor (1); so output voltage rises (1); high output triggers the fan (1). Identifying which component the output is across is essential to get the direction right.
Question 6
Paper 4 multi-part structured style6 marks
A $12\,\text{V}$ battery is connected to a $4.0\,\Omega$ resistor in series with a parallel combination of a $6.0\,\Omega$ and a $3.0\,\Omega$ resistor. Calculate (a) the total resistance, (b) the current from the battery, (c) the potential difference across the parallel combination, and (d) the current in the $6.0\,\Omega$ resistor.
Model answer
(a) Parallel block: $\dfrac{1}{R_{p}} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{2}$ , so $R_{p} = 2.0\,\Omega$ . Total: $R_{T} = 4.0 + 2.0 = \mathbf{6.0\,\Omega}$ .

(b) $I = \dfrac{V}{R_{T}} = \dfrac{12}{6.0} = \mathbf{2.0\,\text{A}}$ .

(c) pd across the parallel block: $V_{p} = I R_{p} = 2.0 \times 2.0 = \mathbf{4.0\,\text{V}}$ .

(d) Current in the $6.0\,\Omega$ resistor: $I_{6} = \dfrac{V_{p}}{6.0} = \dfrac{4.0}{6.0} = \mathbf{0.67\,\text{A}}$ .
Why this scores
Six marks: parallel block $2.0\,\Omega$ (1); total $6.0\,\Omega$ (1); supply current $2.0\,\text{A}$ (1); pd across the parallel block $4.0\,\text{V}$ (1); use $V_p$ across the $6.0\,\Omega$ branch (1); $0.67\,\text{A}$ (1). Reduce the circuit one block at a time, then work outward from the supply.

Key Formulae — Electric Circuits

The formulae you need to memorise for electric circuits on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Resistors in series
$R_{T} = R_{1} + R_{2} + \ldots$
When to use
All current flows through every resistor in turn.
Resistors in parallel
$\dfrac{1}{R_{T}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \ldots$
When to use
Current splits between separate paths.
Potential divider
$V_{n} = V_{T} \times \dfrac{R_{n}}{R_{T}}$
When to use
Read off pd across one resistor in a series chain.
Series vs parallel rules
$\text{Series: same current.}\ \text{Parallel: same pd.}$
When to use
Quickly checking your answer.

Key Definitions and Keywords — Electric Circuits

Definitions to memorise and the exact keywords mark schemes credit for electric circuits answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Series
Examiner keyword
Components connected end-to-end so the same current flows through each.
Parallel
Examiner keyword
Components connected across the same two nodes — same pd; current splits between them.
Thermistor
Examiner keyword
A resistor whose resistance falls when temperature rises. Used in temperature sensors.
Light-dependent resistor (LDR)
Examiner keyword
Resistance falls in brighter light. Used in light-level sensors.
Diode
Examiner keyword
Allows current in one direction only (low resistance forward; very high reverse).

Common Mistakes and Misconceptions — Electric Circuits

The traps other students keep falling into on electric circuits questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to invert at the end of the parallel calculation
0625/42 — every series
Why it happens
Stopping at $1/R_{T}$ .
How to avoid it
After summing reciprocals, INVERT to get $R_{T}$ .
✕Getting parallel resistance larger than each individual
Why it happens
Adding instead of using the reciprocal rule.
How to avoid it
Parallel total must be SMALLER than the smallest of the resistors. If not, recheck.
✕Saying current is the same in every branch of a parallel combination
Why it happens
Mixing series and parallel rules.
How to avoid it
Series → same CURRENT. Parallel → same PD. Switch your reasoning to match the topology.
✕Inverting the potential divider fraction
Why it happens
Memory slip.
How to avoid it
$V$ across one resistor = $V_{T} \times \dfrac{\text{that resistor}}{\text{sum}}$ . The bigger the resistor, the bigger its share of pd.

Electric Circuits — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Electric Circuits

Short Notes - Electric Circuits

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Electric Circuits

1Resistors in series

2Resistors in parallel

3Current in a parallel branch

4Potential divider

5Ammeter and voltmeter readings — mixed circuit

6Find a missing resistor in parallel

7Currents at a junction

8LDR in a potential divider

9Thermistor as a temperature switch

10Current splitting between two parallel branches

11Combined series-and-parallel — find unknown resistor

12Potential divider with a load — synoptic

Resistors in series

Resistors in parallel

Potential divider

Series vs parallel rules

Series

Parallel

Thermistor

Light-dependent resistor (LDR)

Diode

✕Forgetting to invert at the end of the parallel calculation

✕Getting parallel resistance larger than each individual

✕Saying current is the same in every branch of a parallel combination

✕Inverting the potential divider fraction

Electric Circuits — frequently asked questions