What is the relationship between energy and voltage in a circuit?

In a circuit, voltage is the potential difference that drives the flow of electric charge, while energy is the capacity to do work. The relationship between energy and voltage is given by the formula: Energy (E) = Voltage (V) x Charge (Q). This means that the energy transferred in a circuit is directly proportional to the voltage and the amount of charge that flows through the circuit.

How does voltage affect the current in a circuit according to Ohm's Law?

Ohm's Law states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. This relationship is expressed as I = V/R. Therefore, an increase in voltage will result in an increase in current, provided the resistance remains constant.

Why is voltage sometimes referred to as electrical potential energy?

Voltage is often called electrical potential energy because it represents the potential energy per unit charge. It is the energy required to move a charge between two points in an electric field. This potential energy is what causes charges to move through a circuit, doing work in the process, such as lighting a bulb or powering a motor.

What is the role of a battery in providing energy and voltage in a circuit?

A battery provides the necessary energy and voltage to a circuit by converting chemical energy into electrical energy. The voltage of a battery determines the potential difference it can provide, which in turn drives the flow of electric current through the circuit. This flow of current allows the circuit to perform work, such as powering electronic devices.

How do you calculate the total energy used in a circuit over time?

The total energy used in a circuit over time can be calculated using the formula: Energy (E) = Power (P) x Time (t). Power is the rate at which energy is transferred and is calculated as the product of voltage (V) and current (I), i.e., P = V x I. Therefore, the total energy used is E = V x I x t, where t is the time in seconds.

Why is "Saying the current 'gets used up' as it flows through resistors in series" a common mistake in Energy and Voltage in Circuits?

Why it happens: Misinterpreting where the energy goes. How to avoid it: Current is the same through every component in a series loop (charge is conserved). ENERGY is transferred at each resistor (the voltage across that resistor) — but the rate of charge flow is unchanged. Source: 4PH1 Examiner Reports 2022-2024

Why is "Saying total resistance INCREASES when a parallel branch is added" a common mistake in Energy and Voltage in Circuits?

Why it happens: Confusing series and parallel. How to avoid it: Adding a parallel branch provides an EXTRA route for current → MORE current flows for the same voltage → total resistance DECREASES.

Why is "Applying $V = IR$ to a filament lamp with constant $R$" a common mistake in Energy and Voltage in Circuits?

Why it happens: Treating Ohm's law as universal. How to avoid it: Filament lamps, thermistors and diodes are NON-OHMIC: R varies with V. The I-V graph is non-linear. Use the gradient of the I-V graph AT THE POINT OF INTEREST to find R there.

Why is "Saying a thermistor's resistance INCREASES with temperature" a common mistake in Energy and Voltage in Circuits?

Why it happens: Confusing with a thermocouple. How to avoid it: For NTC thermistors (which are what Edexcel means by 'thermistor'), R DECREASES as T increases. Same for an LDR with light intensity. Easy mnemonic: 'more stimulus → less resistance'. Source: 4PH1 Examiner Reports 2022-2024

Edexcel IGCSE Physics (4PH1)

Energy and Voltage in Circuits

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Short Notes - Energy and Voltage in Circuits

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Detailed Study Notes

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

Energy and Voltage in Circuits — Pearson Edexcel International GCSE Physics 4PH1 Study Notes (2026 onwards, Spec Issue 4)

How current, voltage, resistance and charge are related. Series vs parallel circuits. I-V characteristics for different components. LDR and thermistor behaviour. $V = IR$ , $Q = It$ , $E = QV$ .

At a glance

Series: same current; voltages add; resistance adds.
Parallel: same voltage; currents add (spec 2.17, 2.18).
I-V curves: resistor = straight; lamp = S-curve; diode = one-way.
LDR: $R$ decreases when light increases (spec 2.11).
Thermistor (NTC): $R$ decreases when temperature increases (spec 2.11).
$V = IR$ (spec 2.13); $Q = It$ (spec 2.15); $E = QV$ (spec 2.21).
Volt = J/C (spec 2.20).

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4PH1 syllabus (2026 onwards).

2.7 — Explain why a series or parallel circuit is more appropriate for particular applications.
2.8 — Understand how current in a series circuit depends on the voltage and the components.
2.9 — Describe how current varies with voltage in wires, resistors, lamps and diodes and how to investigate this experimentally.
2.10 — Describe the qualitative effect of changing resistance on the current in a circuit.
2.11 — Describe qualitatively LDR (light dependence) and thermistor (temperature dependence).
2.12 — Know that lamps and LEDs indicate the presence of current.
2.13 — Know and use $V = I \times R$ .
2.14 — Know that current is the rate of flow of charge.
2.15 — Know and use $Q = I \times t$ .
2.16 — Know that current in solid metallic conductors is a flow of negatively charged electrons.
2.17 — Understand why current is conserved at a junction in a circuit.
2.18 — Know that the voltage across two components connected in parallel is the same.
2.19 — Calculate currents, voltages and resistances of two resistive components in series.
2.20 — Know that voltage = energy per unit charge passed; the volt is a joule per coulomb.
2.21 — Know and use $E = Q \times V$ .

Series vs parallel circuits (spec 2.7, 2.17 - 2.19)

Two ways to wire components.

Series circuit. Components form a single loop. There is one path for the current.

The current is the SAME at every point in the loop.
The voltages across components ADD to give the supply voltage. (Kirchhoff's voltage law.)
Total resistance: $R_{\text{total}} = R_1 + R_2 + R_3 + \ldots$ (spec 2.19).

Parallel circuit. Components are on separate branches. There are multiple paths for the current.

The voltage across each branch is the SAME and equals the supply voltage (spec 2.18).
The branch currents ADD to give the total current — current is CONSERVED at a junction (spec 2.17, Kirchhoff's current law).
Adding a parallel branch DECREASES the total resistance because there are now more routes for current.

When series is better. Christmas-tree fairy lights (one breaks → whole string off, but uses less wire), variable-brightness lamp + variable resistor controller.

When parallel is better. Domestic lighting (each lamp can be switched independently; if one fails the others stay on), domestic appliances (each gets the full mains voltage).

Series = one path (current shared, voltage split); parallel = branches (voltage shared, current splits).

Series: same I, V splits, R adds.
Parallel: same V, I splits, R decreases.
Domestic = parallel; failure isolation.

See the full worked example for energy and voltage in circuits →

I-V characteristics of components (spec 2.9, 2.10)

Different components give different graph shapes.

How the experiment is done. A variable supply (or battery + variable resistor) drives current through the component under test. An ammeter measures current $I$ and a voltmeter measures p.d. $V$ across the component. Vary the supply, record matched pairs of $V$ and $I$ , plot $I$ (y-axis) against $V$ (x-axis).

(a) Fixed resistor at constant temperature — STRAIGHT LINE through the origin. Resistance constant → current proportional to voltage → obeys Ohm's law. Same shape in forward and reverse direction.

(b) Filament lamp — S-CURVE through the origin. Steep at low V (low temperature → low resistance), curve flattens at high V (filament HOT → atoms vibrate more → electrons collide more → resistance INCREASES). Symmetric about the origin.

(c) Semiconductor diode — ASYMMETRIC. Forward biased (above ~0.6 V): current rises rapidly. Reverse biased: virtually no current (a diode lets current flow in ONE direction only).

(d) Wire — same as a fixed resistor (straight line, constant gradient = $1/R$ ).

Effect of changing resistance (spec 2.10). Adding a higher-resistance component to a circuit DECREASES the current (for the same supply voltage). Decreasing resistance INCREASES the current. Use a VARIABLE resistor (rheostat) to control current/brightness without changing the supply voltage.

Each component has a signature I-V shape: resistor straight, lamp S-curve, diode one-way.

Resistor: straight, symmetric.
Lamp: S-curve, R rises with T.
Diode: forward conducts, reverse blocks.

See the full worked example for energy and voltage in circuits →

LDR and thermistor (spec 2.11, 2.12)

Sensor components whose R changes with conditions.

LDR (light-dependent resistor). A semiconductor whose RESISTANCE DECREASES as the light intensity on it INCREASES.

In darkness: very high resistance (MΩ).
In bright light: low resistance (a few hundred Ω).

Used in: street lamps that switch on at dusk, camera light meters, automatic blinds.

Thermistor (NTC type). A semiconductor whose RESISTANCE DECREASES as TEMPERATURE INCREASES.

Cold: high resistance.
Hot: low resistance.

Used in: thermostats, fire alarms, electronic thermometers.

Indicator components (spec 2.12). Lamps and LEDs (light-emitting diodes) light up when a current flows. They are used as INDICATORS — to show that a circuit is active. LEDs are preferred because they use less power and last longer.

LDR: light up → R down.
Thermistor (NTC): T up → R down.
Lamps/LEDs: indicate current.

Key equations: $V = IR$ , $Q = It$ , $E = QV$ (spec 2.13 - 2.21)

The three relationships you must memorise.

Current = rate of flow of charge (spec 2.14). Linking the two:

$Q = I \times t \quad\text{(spec 2.15)}$

$Q$ in coulombs, $I$ in amperes, $t$ in seconds.
1 ampere = 1 coulomb of charge passing a point per second.

In a metallic conductor (spec 2.16) the charge carriers are NEGATIVELY CHARGED ELECTRONS. (Conventional current — defined historically as from + to − — runs OPPOSITE to the actual electron flow. Both are equally valid.)

Voltage / potential difference (spec 2.20). The ENERGY TRANSFERRED per unit charge passed between two points:

$V = \dfrac{E}{Q} \quad\Leftrightarrow\quad E = Q \times V \quad\text{(spec 2.21)}$

The unit of voltage — the volt — is therefore a joule per coulomb (1 V = 1 J/C). A 12 V battery delivers 12 J of electrical energy for every coulomb of charge it pumps around the circuit.

Resistance (spec 2.13). Linking voltage, current and resistance:

$V = I \times R$

$V$ in volts, $I$ in amperes, $R$ in ohms.
For an Ohmic conductor at constant temperature, $R$ is a constant → I-V plot is a straight line through the origin (Ohm's law).
For non-Ohmic components (lamps, thermistors, diodes), $R$ varies. Use the gradient of the I-V plot at a specific point to find $R$ at that point.

$Q = It$ (spec 2.15).
$V = E/Q$ → $E = QV$ (spec 2.20, 2.21).
$V = IR$ (spec 2.13).

See the full worked example for energy and voltage in circuits →

Quick recap

Series: I constant, V splits, R adds.
Parallel: V constant, I splits, R decreases.
I-V graphs: resistor = straight; lamp = S-curve; diode = one-way.
LDR/thermistor: more stimulus → less resistance.
$V = IR$ , $Q = It$ , $E = QV$ — and the volt is a J/C.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$V = I \times R$ (spec 2.13).
$Q = I \times t$ (spec 2.15).
$E = Q \times V$ (spec 2.21).
1 V = 1 J/C (spec 2.20).
Series: I same, V adds, $R = R_1 + R_2$ (spec 2.19).
Parallel: V same on each branch, currents add at junction (spec 2.17, 2.18).
LDR: $R$ ↓ as light↑; thermistor (NTC): $R$ ↓ as $T$ ↑ (spec 2.11).

How it’s examined

Energy-and-voltage circuits is THE highest-yielding section of Topic 2. Expect 10-15 marks across Paper 1: $V = IR$ in single-resistor or series circuits (3-4 marks), I-V graph interpretation (5-6 marks), $Q = It$ and $E = QV$ short calculations (3-4 marks each), and a 4-5 mark question involving LDRs or thermistors as sensor components.

Sources: Pearson Edexcel International GCSE Physics 4PH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4PH1 Examiner Reports 2022-2024; 4PH1/1P May/Jun 2024 question paper and mark scheme; 4PH1/1P January 2024 question paper and mark scheme. Last reviewed 2026-05-12.

Take this whole topic with you

Step-by-step worked examples — Energy and Voltage in Circuits

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

1 $V = IR$ in a simple circuit (3 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• V=IR, spec-2.13
Question
A 12 Ω resistor has a current of 0.50 A flowing through it. Calculate the potential difference across the resistor. (3 marks)
Step-by-step solution
1. Step 1
  Spec 2.13: $V = I \times R$ .
  $V = 0.50 \times 12$
2. Step 2
  Evaluate with unit.
  $V = 6.0\ \text{V}$
Answer
Potential difference = 6.0 V.
2Series resistor combination + branch current (5 marks, Paper 1)
Core• Adapted from 4PH1/1P January 2024• series, V=IR, spec-2.19
Question
A 6.0 V cell is connected in series with two resistors of 4.0 Ω and 8.0 Ω. (a) Calculate the total resistance. (b) Calculate the current through the circuit. (c) Calculate the p.d. across the 8.0 Ω resistor. (5 marks)
Step-by-step solution
1. Step 1
  (a) In series, total resistance = sum of individual resistances (spec 2.19).
  $R_{\text{total}} = 4.0 + 8.0 = 12\ \text{Ω}$
2. Step 2
  (b) $I = V_{\text{total}}/R_{\text{total}}$ .
  $I = 6.0/12 = 0.50\ \text{A}$
3. Step 3
  (c) Current is the SAME through every series component. Use $V = IR$ for the 8.0 Ω.
  $V_{8\Omega} = 0.50 \times 8.0 = 4.0\ \text{V}$
Answer
(a) 12 Ω. (b) 0.50 A. (c) 4.0 V across the 8.0 Ω resistor.
Examiner tip
Note (b) + (c) sum to the supply voltage: $4.0 + 0.50 \times 4.0 = 6.0$ V — Kirchhoff's voltage law as a sanity check. Edexcel mark scheme awards ECF marks throughout.
3Charge from current and time (3 marks, Paper 1)
Core• Q=It, spec-2.15
Question
A current of 2.0 A flows through a wire for 30 s. Calculate the charge that passes a point. (3 marks)
Step-by-step solution
1. Step 1
  Spec 2.15: $Q = I \times t$ .
  $Q = 2.0 \times 30$
2. Step 2
  Evaluate with unit.
  $Q = 60\ \text{C}$
Answer
Charge = 60 C.
4Identifying I-V curves (5 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• I-V, spec-2.9
Question
Sketch and explain the current-voltage (I-V) characteristics of (a) a fixed resistor at constant temperature, (b) a filament lamp, (c) a diode. (5 marks)
Step-by-step solution
1. Step 1
  (a) Fixed resistor. Straight line through the origin → current is directly proportional to voltage → resistance constant. (Symmetric in both directions.)
2. Step 2
  (b) Filament lamp. S-shaped curve through the origin. At low V the gradient is steep (low resistance). As V (and hence I) rises, the filament HEATS UP — its resistance INCREASES — so the curve flattens.
3. Step 3
  (c) Diode. Asymmetric. In the FORWARD direction (above ~0.6 V), the diode conducts and the current rises sharply. In the REVERSE direction, almost no current flows (until breakdown, which is outside the syllabus).
Answer
(a) Straight line through origin. (b) S-curve flattening at high V due to heating. (c) Asymmetric: forward conducts above ~0.6 V; reverse blocks.
Examiner tip
Edexcel diagram-questions reward CLEAR sketches with labelled axes, origin marked, and the SHAPE described in words. Mention the PHYSICS (heating, p-n junction) only if asked.
5 $E = QV$ — energy transferred by charge (4 marks, Paper 1)
Core• E=QV, spec-2.21
Question
A charge of 25 C passes through a 12 V battery. Calculate the energy transferred. (4 marks)
Step-by-step solution
1. Step 1
  Spec 2.21: $E = Q \times V$ .
  $E = 25 \times 12$
2. Step 2
  Evaluate with unit.
  $E = 300\ \text{J}$
3. Step 3
  Physical meaning (1 mark). The volt is a joule per coulomb (spec 2.20). 12 V means each coulomb of charge gains 12 J of electrical energy from the battery.
Answer
Energy transferred = 300 J.
Examiner tip
Edexcel adds a 'state physical meaning of voltage' mark in many 4-mark $E = QV$ questions. The exact phrasing examiners want is 'volt = joule per coulomb' (or 'energy transferred per unit charge') — verbatim from spec 2.20.

Key Formulae — Energy and Voltage in Circuits

The formulae you need to memorise for energy and voltage in circuits on the Pearson Edexcel IGCSE 4PH1 paper, with every variable defined in plain English and a note on when to use it.

Voltage, current and resistance (spec 2.13)
$V = I \times R$
When to use
$V$ in volts, $I$ in amperes, $R$ in ohms. Applies to OHMIC conductors (constant temperature). For non-ohmic components (filament lamps, thermistors) $R$ varies with conditions.
Charge (spec 2.15)
$Q = I \times t$
When to use
$Q$ in coulombs (C), $I$ in A, $t$ in s. Current is the RATE of flow of charge (spec 2.14).
Electrical energy from charge and voltage (spec 2.21)
$E = Q \times V$
When to use
$E$ in joules, $Q$ in coulombs, $V$ in volts. Direct restatement of spec 2.20: voltage = energy transferred per unit charge.
Series resistance (spec 2.19)
$R_{\text{total}} = R_1 + R_2$
When to use
Resistors in series: resistance ADDS. Current is the SAME through each; voltages ADD to the supply voltage.

Key Definitions and Keywords — Energy and Voltage in Circuits

Definitions to memorise and the exact keywords mark schemes credit for energy and voltage in circuits answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4PH1 sitting.

Current (spec 2.14)
Examiner keyword
The RATE OF FLOW OF CHARGE. $I = Q/t$ . SI unit: ampere (A) = 1 C/s.
Voltage / potential difference (spec 2.20)
Examiner keyword
The ENERGY TRANSFERRED PER UNIT CHARGE passed between two points. SI unit: volt (V) = 1 J/C. A higher voltage means each coulomb of charge delivers more energy.
Resistance
Examiner keyword
The opposition to the flow of current. $R = V/I$ . SI unit: ohm (Ω) = 1 V/A.
Series circuit
Examiner keyword
Components connected end-to-end forming a SINGLE LOOP. The SAME CURRENT flows through every component; voltages across components ADD UP to the supply voltage. Resistance adds: $R_{\text{total}} = R_1 + R_2$ .
Parallel circuit
Examiner keyword
Components connected on SEPARATE BRANCHES so the current can split. SAME VOLTAGE across each branch (spec 2.18); currents in the branches ADD to give the total current (spec 2.17). Adding a parallel branch DECREASES total resistance.
Light-dependent resistor (LDR) (spec 2.11)
Examiner keyword
A semiconductor component whose RESISTANCE DECREASES as light intensity INCREASES. Used in light-sensing circuits (street lamps, camera meters).
Thermistor (spec 2.11)
Examiner keyword
A semiconductor component whose RESISTANCE DECREASES as temperature INCREASES (negative temperature coefficient, NTC). Used in temperature-sensing circuits (thermostats, fire alarms).
Diode (spec 2.9)
Examiner keyword
A component that allows current to flow in ONE direction only — forward (above ~0.6 V) conducts; reverse blocks.

Common Mistakes and Misconceptions — Energy and Voltage in Circuits

The traps other students keep falling into on energy and voltage in circuits questions — taken from recent Pearson Edexcel IGCSE 4PH1 examiner reports and mark schemes — and how to avoid them.

✕Saying the current 'gets used up' as it flows through resistors in series
4PH1 Examiner Reports 2022-2024
Why it happens
Misinterpreting where the energy goes.
How to avoid it
Current is the same through every component in a series loop (charge is conserved). ENERGY is transferred at each resistor (the voltage across that resistor) — but the rate of charge flow is unchanged.
✕Saying total resistance INCREASES when a parallel branch is added
Why it happens
Confusing series and parallel.
How to avoid it
Adding a parallel branch provides an EXTRA route for current → MORE current flows for the same voltage → total resistance DECREASES.
✕Applying $V = IR$ to a filament lamp with constant $R$
Why it happens
Treating Ohm's law as universal.
How to avoid it
Filament lamps, thermistors and diodes are NON-OHMIC: $R$ varies with $V$ . The I-V graph is non-linear. Use the gradient of the I-V graph AT THE POINT OF INTEREST to find $R$ there.
✕Saying a thermistor's resistance INCREASES with temperature
4PH1 Examiner Reports 2022-2024
Why it happens
Confusing with a thermocouple.
How to avoid it
For NTC thermistors (which are what Edexcel means by 'thermistor'), $R$ DECREASES as $T$ increases. Same for an LDR with light intensity. Easy mnemonic: 'more stimulus → less resistance'.

Energy and Voltage in Circuits — frequently asked questions

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

Edexcel IGCSE Physics (4PH1)

Energy and Voltage in Circuits

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Short Notes - Energy and Voltage in Circuits

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Detailed Study Notes

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

Energy and Voltage in Circuits — Pearson Edexcel International GCSE Physics 4PH1 Study Notes (2026 onwards, Spec Issue 4)

How current, voltage, resistance and charge are related. Series vs parallel circuits. I-V characteristics for different components. LDR and thermistor behaviour. $V = IR$ , $Q = It$ , $E = QV$ .

At a glance

Series: same current; voltages add; resistance adds.
Parallel: same voltage; currents add (spec 2.17, 2.18).
I-V curves: resistor = straight; lamp = S-curve; diode = one-way.
LDR: $R$ decreases when light increases (spec 2.11).
Thermistor (NTC): $R$ decreases when temperature increases (spec 2.11).
$V = IR$ (spec 2.13); $Q = It$ (spec 2.15); $E = QV$ (spec 2.21).
Volt = J/C (spec 2.20).

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4PH1 syllabus (2026 onwards).

2.7 — Explain why a series or parallel circuit is more appropriate for particular applications.
2.8 — Understand how current in a series circuit depends on the voltage and the components.
2.9 — Describe how current varies with voltage in wires, resistors, lamps and diodes and how to investigate this experimentally.
2.10 — Describe the qualitative effect of changing resistance on the current in a circuit.
2.11 — Describe qualitatively LDR (light dependence) and thermistor (temperature dependence).
2.12 — Know that lamps and LEDs indicate the presence of current.
2.13 — Know and use $V = I \times R$ .
2.14 — Know that current is the rate of flow of charge.
2.15 — Know and use $Q = I \times t$ .
2.16 — Know that current in solid metallic conductors is a flow of negatively charged electrons.
2.17 — Understand why current is conserved at a junction in a circuit.
2.18 — Know that the voltage across two components connected in parallel is the same.
2.19 — Calculate currents, voltages and resistances of two resistive components in series.
2.20 — Know that voltage = energy per unit charge passed; the volt is a joule per coulomb.
2.21 — Know and use $E = Q \times V$ .

Series vs parallel circuits (spec 2.7, 2.17 - 2.19)

Two ways to wire components.

Series circuit. Components form a single loop. There is one path for the current.

The current is the SAME at every point in the loop.
The voltages across components ADD to give the supply voltage. (Kirchhoff's voltage law.)
Total resistance: $R_{\text{total}} = R_1 + R_2 + R_3 + \ldots$ (spec 2.19).

Parallel circuit. Components are on separate branches. There are multiple paths for the current.

The voltage across each branch is the SAME and equals the supply voltage (spec 2.18).
The branch currents ADD to give the total current — current is CONSERVED at a junction (spec 2.17, Kirchhoff's current law).
Adding a parallel branch DECREASES the total resistance because there are now more routes for current.

When series is better. Christmas-tree fairy lights (one breaks → whole string off, but uses less wire), variable-brightness lamp + variable resistor controller.

When parallel is better. Domestic lighting (each lamp can be switched independently; if one fails the others stay on), domestic appliances (each gets the full mains voltage).

Series = one path (current shared, voltage split); parallel = branches (voltage shared, current splits).

Series: same I, V splits, R adds.
Parallel: same V, I splits, R decreases.
Domestic = parallel; failure isolation.

See the full worked example for energy and voltage in circuits →

I-V characteristics of components (spec 2.9, 2.10)

Different components give different graph shapes.

(c) Semiconductor diode — ASYMMETRIC. Forward biased (above ~0.6 V): current rises rapidly. Reverse biased: virtually no current (a diode lets current flow in ONE direction only).

(d) Wire — same as a fixed resistor (straight line, constant gradient = $1/R$ ).

Each component has a signature I-V shape: resistor straight, lamp S-curve, diode one-way.

Resistor: straight, symmetric.
Lamp: S-curve, R rises with T.
Diode: forward conducts, reverse blocks.

See the full worked example for energy and voltage in circuits →

LDR and thermistor (spec 2.11, 2.12)

Sensor components whose R changes with conditions.

LDR (light-dependent resistor). A semiconductor whose RESISTANCE DECREASES as the light intensity on it INCREASES.

In darkness: very high resistance (MΩ).
In bright light: low resistance (a few hundred Ω).

Used in: street lamps that switch on at dusk, camera light meters, automatic blinds.

Thermistor (NTC type). A semiconductor whose RESISTANCE DECREASES as TEMPERATURE INCREASES.

Cold: high resistance.
Hot: low resistance.

Used in: thermostats, fire alarms, electronic thermometers.

LDR: light up → R down.
Thermistor (NTC): T up → R down.
Lamps/LEDs: indicate current.

Key equations: $V = IR$ , $Q = It$ , $E = QV$ (spec 2.13 - 2.21)

The three relationships you must memorise.

Current = rate of flow of charge (spec 2.14). Linking the two:

$Q = I \times t \quad\text{(spec 2.15)}$

$Q$ in coulombs, $I$ in amperes, $t$ in seconds.
1 ampere = 1 coulomb of charge passing a point per second.

Voltage / potential difference (spec 2.20). The ENERGY TRANSFERRED per unit charge passed between two points:

$V = \dfrac{E}{Q} \quad\Leftrightarrow\quad E = Q \times V \quad\text{(spec 2.21)}$

The unit of voltage — the volt — is therefore a joule per coulomb (1 V = 1 J/C). A 12 V battery delivers 12 J of electrical energy for every coulomb of charge it pumps around the circuit.

Resistance (spec 2.13). Linking voltage, current and resistance:

$V = I \times R$

$V$ in volts, $I$ in amperes, $R$ in ohms.
For an Ohmic conductor at constant temperature, $R$ is a constant → I-V plot is a straight line through the origin (Ohm's law).
For non-Ohmic components (lamps, thermistors, diodes), $R$ varies. Use the gradient of the I-V plot at a specific point to find $R$ at that point.

$Q = It$ (spec 2.15).
$V = E/Q$ → $E = QV$ (spec 2.20, 2.21).
$V = IR$ (spec 2.13).

See the full worked example for energy and voltage in circuits →

Quick recap

Series: I constant, V splits, R adds.
Parallel: V constant, I splits, R decreases.
I-V graphs: resistor = straight; lamp = S-curve; diode = one-way.
LDR/thermistor: more stimulus → less resistance.
$V = IR$ , $Q = It$ , $E = QV$ — and the volt is a J/C.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$V = I \times R$ (spec 2.13).
$Q = I \times t$ (spec 2.15).
$E = Q \times V$ (spec 2.21).
1 V = 1 J/C (spec 2.20).
Series: I same, V adds, $R = R_1 + R_2$ (spec 2.19).
Parallel: V same on each branch, currents add at junction (spec 2.17, 2.18).
LDR: $R$ ↓ as light↑; thermistor (NTC): $R$ ↓ as $T$ ↑ (spec 2.11).

How it’s examined

Take this whole topic with you

Step-by-step worked examples — Energy and Voltage in Circuits

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

1 $V = IR$ in a simple circuit (3 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• V=IR, spec-2.13
Question
A 12 Ω resistor has a current of 0.50 A flowing through it. Calculate the potential difference across the resistor. (3 marks)
Step-by-step solution
1. Step 1
  Spec 2.13: $V = I \times R$ .
  $V = 0.50 \times 12$
2. Step 2
  Evaluate with unit.
  $V = 6.0\ \text{V}$
Answer
Potential difference = 6.0 V.
2Series resistor combination + branch current (5 marks, Paper 1)
Core• Adapted from 4PH1/1P January 2024• series, V=IR, spec-2.19
Question
A 6.0 V cell is connected in series with two resistors of 4.0 Ω and 8.0 Ω. (a) Calculate the total resistance. (b) Calculate the current through the circuit. (c) Calculate the p.d. across the 8.0 Ω resistor. (5 marks)
Step-by-step solution
1. Step 1
  (a) In series, total resistance = sum of individual resistances (spec 2.19).
  $R_{\text{total}} = 4.0 + 8.0 = 12\ \text{Ω}$
2. Step 2
  (b) $I = V_{\text{total}}/R_{\text{total}}$ .
  $I = 6.0/12 = 0.50\ \text{A}$
3. Step 3
  (c) Current is the SAME through every series component. Use $V = IR$ for the 8.0 Ω.
  $V_{8\Omega} = 0.50 \times 8.0 = 4.0\ \text{V}$
Answer
(a) 12 Ω. (b) 0.50 A. (c) 4.0 V across the 8.0 Ω resistor.
Examiner tip
Note (b) + (c) sum to the supply voltage: $4.0 + 0.50 \times 4.0 = 6.0$ V — Kirchhoff's voltage law as a sanity check. Edexcel mark scheme awards ECF marks throughout.
3Charge from current and time (3 marks, Paper 1)
Core• Q=It, spec-2.15
Question
A current of 2.0 A flows through a wire for 30 s. Calculate the charge that passes a point. (3 marks)
Step-by-step solution
1. Step 1
  Spec 2.15: $Q = I \times t$ .
  $Q = 2.0 \times 30$
2. Step 2
  Evaluate with unit.
  $Q = 60\ \text{C}$
Answer
Charge = 60 C.
4Identifying I-V curves (5 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• I-V, spec-2.9
Question
Sketch and explain the current-voltage (I-V) characteristics of (a) a fixed resistor at constant temperature, (b) a filament lamp, (c) a diode. (5 marks)
Step-by-step solution
1. Step 1
  (a) Fixed resistor. Straight line through the origin → current is directly proportional to voltage → resistance constant. (Symmetric in both directions.)
2. Step 2
  (b) Filament lamp. S-shaped curve through the origin. At low V the gradient is steep (low resistance). As V (and hence I) rises, the filament HEATS UP — its resistance INCREASES — so the curve flattens.
3. Step 3
  (c) Diode. Asymmetric. In the FORWARD direction (above ~0.6 V), the diode conducts and the current rises sharply. In the REVERSE direction, almost no current flows (until breakdown, which is outside the syllabus).
Answer
(a) Straight line through origin. (b) S-curve flattening at high V due to heating. (c) Asymmetric: forward conducts above ~0.6 V; reverse blocks.
Examiner tip
Edexcel diagram-questions reward CLEAR sketches with labelled axes, origin marked, and the SHAPE described in words. Mention the PHYSICS (heating, p-n junction) only if asked.
5 $E = QV$ — energy transferred by charge (4 marks, Paper 1)
Core• E=QV, spec-2.21
Question
A charge of 25 C passes through a 12 V battery. Calculate the energy transferred. (4 marks)
Step-by-step solution
1. Step 1
  Spec 2.21: $E = Q \times V$ .
  $E = 25 \times 12$
2. Step 2
  Evaluate with unit.
  $E = 300\ \text{J}$
3. Step 3
  Physical meaning (1 mark). The volt is a joule per coulomb (spec 2.20). 12 V means each coulomb of charge gains 12 J of electrical energy from the battery.
Answer
Energy transferred = 300 J.
Examiner tip
Edexcel adds a 'state physical meaning of voltage' mark in many 4-mark $E = QV$ questions. The exact phrasing examiners want is 'volt = joule per coulomb' (or 'energy transferred per unit charge') — verbatim from spec 2.20.

Key Formulae — Energy and Voltage in Circuits

The formulae you need to memorise for energy and voltage in circuits on the Pearson Edexcel IGCSE 4PH1 paper, with every variable defined in plain English and a note on when to use it.

Voltage, current and resistance (spec 2.13)
$V = I \times R$
When to use
$V$ in volts, $I$ in amperes, $R$ in ohms. Applies to OHMIC conductors (constant temperature). For non-ohmic components (filament lamps, thermistors) $R$ varies with conditions.
Charge (spec 2.15)
$Q = I \times t$
When to use
$Q$ in coulombs (C), $I$ in A, $t$ in s. Current is the RATE of flow of charge (spec 2.14).
Electrical energy from charge and voltage (spec 2.21)
$E = Q \times V$
When to use
$E$ in joules, $Q$ in coulombs, $V$ in volts. Direct restatement of spec 2.20: voltage = energy transferred per unit charge.
Series resistance (spec 2.19)
$R_{\text{total}} = R_1 + R_2$
When to use
Resistors in series: resistance ADDS. Current is the SAME through each; voltages ADD to the supply voltage.

Key Definitions and Keywords — Energy and Voltage in Circuits

Current (spec 2.14)
Examiner keyword
The RATE OF FLOW OF CHARGE. $I = Q/t$ . SI unit: ampere (A) = 1 C/s.
Voltage / potential difference (spec 2.20)
Examiner keyword
The ENERGY TRANSFERRED PER UNIT CHARGE passed between two points. SI unit: volt (V) = 1 J/C. A higher voltage means each coulomb of charge delivers more energy.
Resistance
Examiner keyword
The opposition to the flow of current. $R = V/I$ . SI unit: ohm (Ω) = 1 V/A.
Series circuit
Examiner keyword
Components connected end-to-end forming a SINGLE LOOP. The SAME CURRENT flows through every component; voltages across components ADD UP to the supply voltage. Resistance adds: $R_{\text{total}} = R_1 + R_2$ .
Parallel circuit
Examiner keyword
Components connected on SEPARATE BRANCHES so the current can split. SAME VOLTAGE across each branch (spec 2.18); currents in the branches ADD to give the total current (spec 2.17). Adding a parallel branch DECREASES total resistance.
Light-dependent resistor (LDR) (spec 2.11)
Examiner keyword
A semiconductor component whose RESISTANCE DECREASES as light intensity INCREASES. Used in light-sensing circuits (street lamps, camera meters).
Thermistor (spec 2.11)
Examiner keyword
A semiconductor component whose RESISTANCE DECREASES as temperature INCREASES (negative temperature coefficient, NTC). Used in temperature-sensing circuits (thermostats, fire alarms).
Diode (spec 2.9)
Examiner keyword
A component that allows current to flow in ONE direction only — forward (above ~0.6 V) conducts; reverse blocks.

Common Mistakes and Misconceptions — Energy and Voltage in Circuits

The traps other students keep falling into on energy and voltage in circuits questions — taken from recent Pearson Edexcel IGCSE 4PH1 examiner reports and mark schemes — and how to avoid them.

✕Saying the current 'gets used up' as it flows through resistors in series
4PH1 Examiner Reports 2022-2024
Why it happens
Misinterpreting where the energy goes.
How to avoid it
Current is the same through every component in a series loop (charge is conserved). ENERGY is transferred at each resistor (the voltage across that resistor) — but the rate of charge flow is unchanged.
✕Saying total resistance INCREASES when a parallel branch is added
Why it happens
Confusing series and parallel.
How to avoid it
Adding a parallel branch provides an EXTRA route for current → MORE current flows for the same voltage → total resistance DECREASES.
✕Applying $V = IR$ to a filament lamp with constant $R$
Why it happens
Treating Ohm's law as universal.
How to avoid it
Filament lamps, thermistors and diodes are NON-OHMIC: $R$ varies with $V$ . The I-V graph is non-linear. Use the gradient of the I-V graph AT THE POINT OF INTEREST to find $R$ there.
✕Saying a thermistor's resistance INCREASES with temperature
4PH1 Examiner Reports 2022-2024
Why it happens
Confusing with a thermocouple.
How to avoid it
For NTC thermistors (which are what Edexcel means by 'thermistor'), $R$ DECREASES as $T$ increases. Same for an LDR with light intensity. Easy mnemonic: 'more stimulus → less resistance'.

Energy and Voltage in Circuits — frequently asked questions

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

Energy and Voltage in Circuits

Short Notes - Energy and Voltage in 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

1V=IRV = IRV=IR in a simple circuit (3 marks, Paper 1)

2Series resistor combination + branch current (5 marks, Paper 1)

3Charge from current and time (3 marks, Paper 1)

4Identifying I-V curves (5 marks, Paper 1)

5E=QVE = QVE=QV — energy transferred by charge (4 marks, Paper 1)

Voltage, current and resistance (spec 2.13)

Charge (spec 2.15)

Electrical energy from charge and voltage (spec 2.21)

Series resistance (spec 2.19)

Current (spec 2.14)

Voltage / potential difference (spec 2.20)

Resistance

Series circuit

Parallel circuit

Light-dependent resistor (LDR) (spec 2.11)

Thermistor (spec 2.11)

Diode (spec 2.9)

✕Saying the current 'gets used up' as it flows through resistors in series

✕Saying total resistance INCREASES when a parallel branch is added

✕Applying V=IRV = IRV=IR to a filament lamp with constant RRR

✕Saying a thermistor's resistance INCREASES with temperature

Energy and Voltage in Circuits — frequently asked questions

Energy and Voltage in Circuits

Short Notes - Energy and Voltage in 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

1V=IRV = IRV=IR in a simple circuit (3 marks, Paper 1)

2Series resistor combination + branch current (5 marks, Paper 1)

3Charge from current and time (3 marks, Paper 1)

4Identifying I-V curves (5 marks, Paper 1)

5E=QVE = QVE=QV — energy transferred by charge (4 marks, Paper 1)

Voltage, current and resistance (spec 2.13)

Charge (spec 2.15)

Electrical energy from charge and voltage (spec 2.21)

Series resistance (spec 2.19)

Current (spec 2.14)

Voltage / potential difference (spec 2.20)

Resistance

Series circuit

Parallel circuit

Light-dependent resistor (LDR) (spec 2.11)

Thermistor (spec 2.11)

Diode (spec 2.9)

✕Saying the current 'gets used up' as it flows through resistors in series

✕Saying total resistance INCREASES when a parallel branch is added

✕Applying V=IRV = IRV=IR to a filament lamp with constant RRR

✕Saying a thermistor's resistance INCREASES with temperature

Energy and Voltage in Circuits — frequently asked questions

1 $V = IR$ in a simple circuit (3 marks, Paper 1)

5 $E = QV$ — energy transferred by charge (4 marks, Paper 1)

✕Applying $V = IR$ to a filament lamp with constant $R$

1 $V = IR$ in a simple circuit (3 marks, Paper 1)

5 $E = QV$ — energy transferred by charge (4 marks, Paper 1)

✕Applying $V = IR$ to a filament lamp with constant $R$