The key idea
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.
Understand it · 01
Keep current and voltage conceptually separate
Current measures the rate of charge flow: I = Q/t. A current of one ampere means one coulomb passes a point each second. In a simple steady circuit, charge does not pile up inside a resistor or disappear after a lamp. Mobile charge is already present throughout the conducting path; closing the circuit establishes the conditions for a flow around that path.
Potential difference measures energy transferred per unit charge: V = E/Q. A lamp transfers electrical energy to light and thermal energy as charge passes, but it does not consume the charge. A cell transfers energy into the electrical system from a chemical store. Keep these accounts separate: current concerns how much charge passes per second, while voltage concerns how much energy is transferred by each coulomb.
Understand it · 02
Solve the network before substituting numbers
In a series circuit there is one path, so the current is the same through each component. The supply potential difference is shared between the components, and their resistances add. In parallel, each branch connects across the same two junctions and therefore has the same potential difference. The total current is the sum of the branch currents. These rules follow charge and energy conservation, not the physical appearance of the wires on the page.
Trace the junctions carefully before classifying a circuit. Two resistors drawn side by side are not automatically parallel. Adding a branch across an ideal fixed-voltage supply gives charge another path, reducing total resistance and increasing the total supply current. It does not require the current in an unchanged existing branch to decrease, because that branch’s voltage and resistance remain the same.
Understand it · 03
Know when resistance is constant
Resistance is defined by R = V/I for a particular operating point. An ohmic conductor has an approximately constant resistance when physical conditions such as temperature are constant, giving a straight current–voltage graph through the origin. A filament lamp heats as current rises; the metal’s resistance increases, so the graph curves. You can still calculate V/I at a point, but should not assume one constant value fits the whole curve.
The graph axes matter. On an I-against-V graph, the gradient of a straight line equals 1/R; on a V-against-I graph it equals R. For a curved graph, a tangent gradient is not generally the same as the ratio V/I. A diode also behaves differently in forward and reverse directions, so reversing a component can change the circuit substantially.
Understand it · 04
Connect calculations to safe measurement
Electrical power is the rate of energy transfer: P = IV. Combining this with the resistance relationship gives P = I²R or P = V²/R for an appropriate resistor. Decide which quantities are fixed before predicting a change. At fixed current, a larger resistance gives larger power; at fixed voltage, a larger resistance gives smaller power. The apparently opposite conclusions describe different controlled conditions.
Connect an ammeter in series so the branch current passes through it. Connect a voltmeter in parallel across the component whose potential difference you want. A current meter has a small resistance and must not simply be placed across a supply. For household safety questions, a fuse or circuit breaker interrupts excessive current; an earth connection provides a protective path for a fault in suitable metal-cased appliances.
Another view of the physics
Parallel branches share the same potential difference
Key definitions
- 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).
Equations, units & conditions
Current
I = Q ÷ t
Units: A
Potential difference
V = I × R
Units: V
Ohm’s law: V ∝ I when resistance stays constant.
Series resistance
R = R₁ + R₂
Parallel resistance (two resistors)
1/R = 1/R₁ + 1/R₂
Electrical power
P = I × V
Units: W
Electrical energy
E = I × V × t
Units: J
Worked examples, step by step
Example 1
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.
Follow the method
- 1/R = 1/6.0 + 1/3.0 = 1/6 + 2/6 = 3/6, so R = 2.0 Ω.
- Check: 2.0 Ω is smaller than the smallest branch (3.0 Ω) — as parallel must be.
- I = V ÷ R = 6.0 ÷ 2.0.
R = 2.0 Ω, I = 3.0 A
Find two potential differences in series
A 4.0 Ω resistor and an 8.0 Ω resistor are in series across an ideal 9.0 V supply. Calculate the current and the potential difference across each resistor.
Follow the method
- Total resistance = 4.0 + 8.0 = 12 Ω.
- The common current is I = V/R = 9.0/12 = 0.75 A.
- Across 4.0 Ω: V₁ = IR₁ = 0.75 × 4.0 = 3.0 V. Across 8.0 Ω: V₂ = 0.75 × 8.0 = 6.0 V.
- Check: 3.0 + 6.0 = 9.0 V, the supply potential difference.
Current = 0.75 A; potential differences = 3.0 V and 6.0 V.
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Starter · 4 suggested marks
1. Share voltage in series
Two resistors, 8 Ω and 4 Ω, are connected in series across a 6 V supply. Calculate the current and the voltage across the 8 Ω resistor.
Show solution and self-check: question 1
- Total resistance = 8 + 4 = 12 Ω.
- Current I = 6/12 = 0.50 A.
- Voltage across 8 Ω = IR = 0.50 × 8 = 4.0 V.
Current 0.50 A; voltage across 8 Ω is 4.0 V.
Self-check · one mark per point
- Add resistances to obtain 12 Ω.
- Use the supply voltage with the total resistance.
- Calculate 0.50 A.
- Calculate a 4.0 V drop across 8 Ω.
Starter · 3 suggested marks
2. Convert minutes before finding charge
A steady current of 0.24 A flows through a lamp for 3.0 minutes. Calculate the charge passing through it. Is less charge leaving the lamp than entering it during this time?
Show solution and self-check: question 2
- Time = 3.0 × 60 = 180 s.
- Q = It = 0.24 × 180 = 43.2 C.
- In steady operation, charge does not accumulate in the lamp: the same charge enters and leaves. Electrical energy is transferred, not charge consumed.
Q = 43.2 C (about 43 C). The same charge enters and leaves.
Self-check · one mark per point
- Converts the time to 180 s.
- Applies Q = It to obtain the charge with unit C.
- States that charge is conserved through the lamp.
Build · 4 suggested marks
3. Add parallel currents
A 15 Ω resistor and a 30 Ω resistor are connected in parallel across a 9 V supply. Calculate the total current and the power supplied.
Show solution and self-check: question 3
- Each branch has 9 V.
- Branch currents are 9/15 = 0.60 A and 9/30 = 0.30 A.
- Total current = 0.90 A.
- Total power = VI = 9 × 0.90 = 8.1 W.
Total current 0.90 A; supply power 8.1 W.
Self-check · one mark per point
- Apply 9 V to each branch.
- Calculate 0.60 A and 0.30 A.
- Add currents to obtain 0.90 A.
- Calculate 8.1 W.
Build · 4 suggested marks
4. Check energy transfer in a series circuit
Resistors of 5.0 Ω and 7.0 Ω are in series across an ideal 6.0 V supply. Find the circuit current, the voltage across the 7.0 Ω resistor and the power dissipated in that resistor.
Show solution and self-check: question 4
- Total resistance = 5.0 + 7.0 = 12 Ω.
- I = 6.0/12 = 0.50 A.
- Voltage across 7.0 Ω = IR = 0.50 × 7.0 = 3.5 V.
- Power in that resistor = IV = 0.50 × 3.5 = 1.75 W.
I = 0.50 A; V = 3.5 V; P = 1.75 W (about 1.8 W).
Self-check · one mark per point
- Adds the series resistances.
- Finds current 0.50 A.
- Uses that current to find 3.5 V.
- Uses the resistor voltage, not the full supply, to find 1.75 W.
Build · 4 suggested marks
5. Solve parallel resistance through branch currents
An ideal 12 V supply is connected to 6.0 Ω and 12 Ω resistors in parallel. Find the current in each branch, the total current and the equivalent resistance.
Show solution and self-check: question 5
- Both branches have 12 V across them.
- Branch currents are 12/6.0 = 2.0 A and 12/12 = 1.0 A.
- Total current = 2.0 + 1.0 = 3.0 A.
- Equivalent resistance = 12/3.0 = 4.0 Ω, smaller than either branch resistance.
Branch currents = 2.0 A and 1.0 A; total = 3.0 A; equivalent resistance = 4.0 Ω.
Self-check · one mark per point
- Finds current 2.0 A in the 6.0 Ω branch.
- Finds current 1.0 A in the 12 Ω branch.
- Adds branch currents to obtain 3.0 A.
- Finds equivalent resistance 4.0 Ω.
Stretch · 5 suggested marks
6. Separate a mixed circuit into stages
A 4.0 Ω resistor is in series with a parallel pair of 6.0 Ω and 3.0 Ω resistors. An ideal 12 V supply is connected across the whole arrangement. Find the total current and the current in each parallel branch.
Show solution and self-check: question 6
- The parallel pair has resistance Rₚ = (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω.
- Total resistance = 4.0 + 2.0 = 6.0 Ω, so total current = 12/6.0 = 2.0 A.
- Voltage across the series resistor = 2.0 × 4.0 = 8.0 V.
- Voltage across the parallel pair = 12 − 8.0 = 4.0 V.
- Branch currents are 4.0/6.0 = 0.667 A and 4.0/3.0 = 1.333 A. They add to 2.0 A.
Total current = 2.0 A. Branch currents ≈ 0.67 A through 6.0 Ω and 1.33 A through 3.0 Ω.
Self-check · one mark per point
- Finds parallel resistance 2.0 Ω.
- Finds total resistance 6.0 Ω and total current 2.0 A.
- Finds parallel-section voltage 4.0 V.
- Finds the 6.0 Ω branch current.
- Finds the 3.0 Ω branch current and checks the sum.
Common mistakes & how to avoid them
- 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.
- Reversing the meter connections: ammeters go in series, voltmeters in parallel across the component. Drawn wrong, the circuit mark is gone.
- 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.
- Wire resistance rules half-remembered: resistance increases with length (double length, double resistance) and decreases with cross-sectional area (double area, half resistance).
- 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.
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:
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
What are the series and parallel circuit rules?
In series, current is the same through each component and potential differences add to the supply value. In parallel, each branch has the same potential difference and branch currents add to the supply current. Identify the actual junctions before deciding which rules apply.
A circuit’s appearance can be misleading because a wire can bend without changing its electrical connections. Follow each uninterrupted wire to establish which points are at the same potential in the ideal model. Components connected between the same two junctions are parallel, even if drawn far apart. Components on a single unbranched path are series. Conservation of charge explains current continuity and junction addition, while energy per unit charge explains the voltage relationships. Applying these ideas is safer than choosing a rule from how the resistors are arranged visually.
Apply both kinds of rule
A 12 V source supplies two parallel branches carrying 0.4 A and 0.6 A. State each branch voltage and the source current.
- Both branches span the source terminals, so each has 12 V.
- Source current = 0.4 + 0.6 = 1.0 A.
Each branch has 12 V; total current is 1.0 A.
Why does adding resistors in parallel decrease total resistance?
Adding a resistive branch across an ideal fixed-voltage supply creates an extra current path, increasing total current. Since total resistance is supply voltage divided by total current, the equivalent resistance decreases. Existing branch currents remain unchanged if their voltage and resistance remain unchanged.
The phrase “the current splits” sometimes leads to the wrong prediction that every new branch steals current from the others. A fixed-voltage source can supply more current overall, so each original branch still follows its own I = V/R relationship. A real source with appreciable internal resistance can behave differently because its terminal voltage may fall as total current rises. State the source assumption when predicting unchanged branch currents. For ordinary positive resistors in parallel, the equivalent resistance is smaller than the smallest individual branch resistance.
What is the difference between e.m.f. and potential difference?
E.m.f. describes energy supplied by a source per unit charge, while potential difference describes energy transferred per unit charge between two points. Both are measured in volts. E.m.f. is an energy-per-charge quantity, despite the historical word “force” in its name.
Following one coulomb around a simple circuit provides a clear account. The source transfers energy into the electrical system, and components transfer energy to other stores or forms. In the ideal steady circuit, the gains and losses per coulomb balance around the loop. If the source has internal resistance, some energy per coulomb is transferred inside it, so its terminal potential difference under load can be lower than its e.m.f. Do not confuse either quantity with current, which counts charge passing each second.
Why is a filament lamp’s current–voltage graph curved?
A filament lamp’s current–voltage graph curves because its filament heats as current increases and its resistance rises. For an I-against-V graph, the curve becomes less steep. Calculate resistance at a point using V/I; a tangent gradient on a curved graph is a different quantity.
Ohm’s law requires appropriate physical conditions, especially temperature, to remain constant. A lamp does not satisfy that condition over a wide range of operating currents. At higher temperature, the metal’s increased lattice motion makes electron flow more strongly opposed. Read both current and voltage at the same point when calculating resistance. Check the axis orientation before interpreting steepness: on V-against-I axes, increased resistance is associated with a steeper curve instead. Quoting “gradient equals resistance” without naming the axes can therefore give the wrong conclusion.
Compare lamp resistance
A lamp carries 0.20 A at 2.0 V and 0.35 A at 6.0 V. Find its resistance at each point.
- At 2.0 V: R = 2.0/0.20 = 10 Ω.
- At 6.0 V: R = 6.0/0.35 ≈ 17.1 Ω.
The resistance rises from 10 Ω to approximately 17 Ω as the filament heats.
Why are ammeters connected in series and voltmeters in parallel?
An ammeter belongs in series so the current being measured passes through it. A voltmeter belongs in parallel across the two points whose potential difference is measured. Their resistances are designed for those roles: low for an ammeter and high for a voltmeter.
Adding a large resistance in series would significantly change the current being investigated, so an ideal ammeter has zero resistance. A meter connected across a component should draw very little extra current, so an ideal voltmeter has infinite resistance. Real meters approximate these behaviours. Putting an ammeter directly across a supply can create a very large current rather than measure the intended branch current. Trace the circuit with the power off, check the meter function and range, and use the correct terminals in a supervised practical setup.
Does increasing resistance increase or decrease electrical power?
Whether a larger resistance produces more power depends on what is held constant. At fixed current, P = I²R increases with resistance; at fixed voltage, P = V²/R decreases. State the supply condition before applying either relationship or predicting a hotter component.
The two formulas are not contradictory because they describe different controlled situations. Replacing a resistor across an ideal voltage supply changes the current, so assuming that current stays fixed would be incorrect. A constant-current source instead changes its voltage as the load changes, within its operating limits. For a numerical problem, list the known quantities and select P = IV, I²R or V²/R accordingly. The eventual temperature also depends on energy transfer to the surroundings, so greater electrical power does not independently specify a final temperature.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- 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
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Electric circuits. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/electric-circuits
Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.