Edexcel IGCSE Physics 2(c) Energy and Voltage in Circuits — Notes
Written by an examiner for Pearson Edexcel International GCSE Physics (4PH1), Science Double Award (4SD0) and Science Single Award (4SS0), linear and modular. In the Single Award specification this sub-topic is called “Current and voltage in circuits”.
Which parts do I need?
Single Award candidates need current, \(V=I\times R\), series circuits, the current–voltage graphs for wires, resistors and filament lamps (not diodes), and lamps and LEDs as indicators. They skip every section tagged Double · Physics: \(Q=I\times t\), voltage as energy per charge and \(E=Q\times V\), parallel circuits and choosing series or parallel, the diode graph, and LDRs and thermistors.
What this page covers
| Spec | You need to be able to… | Route |
|---|---|---|
| 2.7 | explain when a series or a parallel circuit is the better choice, including for lighting in a house | Double · Physics |
| 2.8 | explain what decides the size of the current in a single loop: the supply voltage, and how many components there are and what kind they are | Single · Double · Physics |
| 2.9 | describe and investigate how current changes with voltage for wires, resistors, filament lamps and diodes (diodes: Double · Physics only) | Single · Double · Physics |
| 2.10 | describe, without numbers, how changing the resistance changes the current | Single · Double · Physics |
| 2.11 | describe how the resistance of an LDR changes with light and of a thermistor with temperature | Double · Physics |
| 2.12 | know that a lamp or an LED can show whether there is a current | Single · Double · Physics |
| 2.13 | recall and use voltage = current × resistance | Single · Double · Physics |
| 2.14 | know that current is the rate at which charge flows | Single · Double · Physics |
| 2.15 | recall and use charge = current × time | Double · Physics |
| 2.16 | know that the current in a metal wire is a flow of electrons, which are negatively charged | Single · Double · Physics |
| 2.17 | explain why the current into a junction equals the current out of it | Double · Physics |
| 2.18 | know that components in parallel have the same voltage across them | Double · Physics |
| 2.19 | calculate current, voltage and resistance for two resistive components in series | Single · Double · Physics |
| 2.20 | know that voltage tells you how much energy each coulomb of charge transfers, so one volt means one joule for every coulomb | Double · Physics |
| 2.21 | recall and use energy transferred = charge × voltage | Double · Physics |
Modular candidates: these are statements 2.7–2.21 of 4XPH1 Unit 1 (4WPH1) and 4XSD1 Physics Unit 5 (4WSD5).
In one minute
- Current is the rate of flow of charge. In metals the charge carriers are free electrons. Conventional current is shown flowing from + to −; the electrons actually move from − to +.
- \(V=I\times R\). For a fixed voltage, more resistance means less current.
- Series: the same current everywhere; the voltages add up to the supply voltage; total resistance \(R=R_1+R_2\).
- Parallel: the same voltage across each branch; the branch currents add up to the current from the supply. House lights are wired in parallel. Double · Physics
- \(Q=I\times t\) and \(E=Q\times V\): a volt is a joule per coulomb. Double · Physics
- I–V graphs: resistor — straight line through the origin; filament lamp — curve (resistance rises as it heats up); diode — current in one direction only Double · Physics.
- LDR: more light → less resistance. Thermistor: hotter → less resistance. Double · Physics
Which circuit symbols do I need? Single · Double · Physics
Pearson papers draw circuits with the symbols below. You must be able to recognise them and draw them in your own circuit diagrams — use a ruler for the wires and leave no gaps.
- An ammeter measures current and is connected in series, in the loop with the component.
- A voltmeter measures voltage and is connected in parallel, across the component.
- A battery is two or more cells joined together.
What is electric current? Single · Double · Physics
- In a solid metal, such as a copper wire, the current is a flow of free electrons, which are negatively charged.
- The electrons are repelled by the negative terminal and attracted to the positive terminal, so they drift from − to + round the external circuit.
- By convention, current is drawn from the positive terminal to the negative terminal (the direction a positive charge would move). This conventional current is opposite to the electron flow. Circuit diagrams on this page show conventional current with red arrows.
- A current needs a complete circuit. A break anywhere in a series loop stops the current everywhere in that loop.
How are charge, current and time related? Double · Physics
Not required for Single Award (4SS0).
\[ \text{charge}=\text{current}\times\text{time}\qquad Q=I\times t \]- \(Q\) in coulombs (C), \(I\) in amps (A), \(t\) in seconds (s).
- A current of 1 A means 1 coulomb of charge passes a point every second.
What does voltage mean, and how is it linked to energy? Double · Physics
Not required for Single Award (4SS0).
- A 12 V battery transfers 12 J of energy to each coulomb of charge that passes through it.
- A voltage of 3 V across a lamp means each coulomb transfers 3 J of energy to the lamp as it passes through.
- \(E\) in joules (J), \(Q\) in coulombs (C), \(V\) in volts (V).
A current of 0.50 A passes through a 6.0 V lamp for 4.0 minutes. Calculate (a) the charge that passes through the lamp and (b) the energy transferred to the lamp.
(a) \(t=4.0\times60=240\ \text{s}\); \(Q=I\times t=0.50\times240=\) 120 C
(b) \(E=Q\times V=120\times6.0=\) 720 J
Check: \(E=I\times V\times t=0.50\times6.0\times240=720\ \text{J}\) — the same answer, as it must be.
What is resistance and how does it affect the current? Single · Double · Physics
Resistance opposes the current. For a given voltage, a component with a larger resistance lets a smaller current through it.
\[ \text{voltage}=\text{current}\times\text{resistance}\qquad V=I\times R \]- \(V\) in volts (V), \(I\) in amps (A), \(R\) in ohms (Ω).
- Rearranged: \(I=\dfrac{V}{R}\) and \(R=\dfrac{V}{I}\).
- Qualitatively: if the resistance in a circuit increases and the voltage stays the same, the current decreases; if the resistance decreases, the current increases. This is how a variable resistor dims a lamp or slows a motor.
(a) The current in a 47 Ω resistor is 0.20 A. Calculate the voltage across it.
(b) A 6.0 V supply drives a current of 0.25 A through a heating coil. Calculate its resistance.
(a) \(V=I\times R=0.20\times47=\) 9.4 V
(b) \(R=\dfrac{V}{I}=\dfrac{6.0}{0.25}=\) 24 Ω
What are the rules for a series circuit? Single · Double · Physics
In a series circuit the components are joined one after another in a single loop.
| Quantity | Rule in series |
|---|---|
| current | the same at every point in the loop |
| voltage | the supply voltage is shared: \(V=V_1+V_2\); the larger resistance gets the larger share |
| resistance | the total resistance is the sum: \(R=R_1+R_2\) |
What sets the current in a series circuit?
- Applied voltage: increasing the supply voltage (for example, adding another cell the same way round) increases the current.
- Number of components: adding another lamp or resistor in series increases the total resistance, so the current decreases — each lamp is dimmer.
- Nature of the components: a component with a larger resistance gives a smaller current. Some components change their resistance — a filament lamp's resistance rises as it heats up, and an LDR or thermistor changes with conditions.
A 30 Ω resistor and a 50 Ω resistor are connected in series to a 12 V supply, as in the diagram. Calculate (a) the total resistance, (b) the reading on the ammeter and (c) the reading on the voltmeter.
(a) \(R=R_1+R_2=30+50=\) 80 Ω
(b) \(I=\dfrac{V}{R}=\dfrac{12}{80}=\) 0.15 A
(c) \(V=I\times R=0.15\times50=\) 7.5 V
Check: the voltage across the 30 Ω resistor is 0.15 × 30 = 4.5 V, and 4.5 + 7.5 = 12 V, the supply voltage.
What are the rules for a parallel circuit? Double · Physics
Not required for Single Award (4SS0).
In a parallel circuit the components are on separate branches, each connected directly across the supply.
- Why current is conserved: current is a flow of charge, and charge cannot be created or destroyed at a junction or stored there. Every coulomb that arrives per second must leave per second, so the total current out equals the current in: \(I=I_1+I_2\).
- Why the voltage is the same: each branch is connected between the same two points, so each coulomb transfers the same energy whichever branch it goes through.
- Each branch works independently: the branch current is \(I=\dfrac{V}{R}\) for that branch. A smaller resistance in a branch takes a larger current.
- Adding another branch gives the charge another path, so the current from the supply increases.
In the circuit shown, a 12 Ω resistor and a 20 Ω resistor are connected in parallel to a 6.0 V battery. Calculate the readings on ammeters A₂, A₃ and A₁, and explain how you found A₁.
Each resistor has the full 6.0 V across it.
A₂: \(I=\dfrac{6.0}{12}=\) 0.50 A A₃: \(I=\dfrac{6.0}{20}=\) 0.30 A
A₁: current is conserved at the junction, so A₁ = 0.50 + 0.30 = 0.80 A.
When is a series or a parallel circuit more suitable? Double · Physics
Not required for Single Award (4SS0).
| Feature | Series | Parallel |
|---|---|---|
| voltage across each lamp | a share of the supply voltage | the full supply voltage |
| if one lamp fails (breaks the circuit) | all the lamps go out | the other lamps stay on |
| switching | one switch controls everything | each branch can have its own switch |
| adding more lamps | every lamp gets dimmer | the brightness of the others does not change |
Domestic lighting is wired in parallel
- Each lamp is connected directly across the mains, so it gets the full 230 V it is designed for and works at its normal brightness.
- Each lamp can be switched on and off on its own, using a switch in its own branch.
- If one lamp fails, the others stay lit, because their branches are still complete circuits.
When series is the better choice
- Switches and fuses are always placed in series with the device they control or protect, so that opening them breaks the circuit to that device.
- A string of small low-voltage lamps (decorative lights) can be run in series from a higher supply voltage, because the voltage is shared between them.
- A variable resistor goes in series with a lamp or motor so that changing its resistance changes the current.
How does the current vary with voltage for different components? Single · Double · Physics
Single Award: you do not need diodes — the 4SS0 statement covers wires, resistors and filament lamps only.
| Component | Shape of the I–V graph | What it shows | Route |
|---|---|---|---|
| wire or resistor (at constant temperature) | straight line through the origin | current is directly proportional to voltage; the resistance is constant | All |
| filament lamp | curve through the origin that gets less steep as the voltage increases (same shape for reversed voltage) | as the current increases, the filament gets hotter and its resistance increases, so the current rises less for each extra volt | All |
| diode | zero current for reversed voltage; in the forward direction almost no current until a small voltage, then the current rises steeply | a diode lets current flow in one direction only — very high resistance in reverse, low resistance forward | Double · Physics |
- The resistance at any point is \(R=\dfrac{V}{I}\), using the values read from the graph at that point.
- Why does a hot filament have more resistance? The metal ions vibrate more, so the electrons collide with them more often.
How do you investigate how current varies with voltage? Single · Double · Physics
The specification asks you to know how to investigate current–voltage relationships experimentally. Questions ask you to draw the circuit, describe the method and process the results.
Investigating how the current in a component depends on the voltage across it
- Aim: find how the current in a resistor, a length of wire or a filament lamp (or a diode — Double · Physics) changes as the voltage across it changes.
- Apparatus: battery or low-voltage d.c. supply, variable resistor, ammeter, voltmeter, the component, switch, connecting leads.
- Independent variable: voltage across the component (changed with the variable resistor).
- Dependent variable: current in the component (ammeter reading).
- Control variables: the same component; for a wire or resistor, the same temperature (keep the current small).
Method
- Set up the circuit shown, with the ammeter in series and the voltmeter across the component.
- Adjust the variable resistor to give a small voltage. Record the voltage and the current.
- Change the variable resistor to obtain at least six different voltages, recording the current each time.
- Reverse the connections to the component (or to the supply) and repeat to obtain negative values.
- Repeat the set of readings and calculate mean currents.
- Plot current (vertical axis) against voltage (horizontal axis) and draw a line or smooth curve of best fit.
Processing, errors and improvements
- Calculate \(R=\dfrac{V}{I}\) for each pair of readings to see whether the resistance is constant.
- A wire or resistor heats up if the current is large, which increases its resistance. Switch off between readings and keep the current small.
- Check that both meters read zero before you start (no zero error). Use meters with a suitable range so that readings are not too small to read precisely.
- For a diode, include a fixed resistor in series to stop the forward current becoming large enough to damage it. (Double · Physics)
- Safety: use a low-voltage supply; a lamp filament and a thin wire can become very hot — do not touch them while the current is on.
A 300 Ω resistor is connected in series with a component X and a 9.0 V battery. A voltmeter across the resistor reads 3.0 V.
(a) Calculate the current in the circuit. (b) Determine the voltage across X. (c) Calculate the resistance of X. (d) The 300 Ω resistor is replaced by a 150 Ω resistor and X keeps the same resistance. State and explain what happens to the current.
(a) \(I=\dfrac{V}{R}=\dfrac{3.0}{300}=\) 0.010 A (10 mA)
(b) In series the voltages add up to the supply: \(9.0-3.0=\) 6.0 V
(c) The current is the same everywhere in a series circuit: \(R=\dfrac{6.0}{0.010}=\) 600 Ω
(d) The total resistance decreases (from 900 Ω to 750 Ω), so the current increases — to \(9.0\div750=0.012\ \text{A}\). The voltage across X increases to \(0.012\times600=7.2\ \text{V}\).
How do LDRs and thermistors behave? Double · Physics
Not required for Single Award (4SS0). The Double Award wording of this statement differs by one word from the Physics wording; what you must know is identical.
- Light-dependent resistor (LDR): in bright light its resistance is low; in the dark its resistance is high. Use: switching on street lights or a night light when it gets dark.
- Thermistor: when it is hot its resistance is low; when it is cold its resistance is high. Use: temperature sensors in fire alarms, ovens and thermostats.
- In a series circuit with a fixed supply, more light on an LDR (or a higher temperature for a thermistor) means less resistance, so a larger current.
How can a lamp or LED show that there is a current? Single · Double · Physics
- A filament lamp placed in series lights up when there is a current in the circuit; the brighter it is, the larger the current.
- A light-emitting diode (LED) lights up when a current passes through it. It needs only a small current, so it is used as an indicator light on chargers, televisions and other appliances to show that they are switched on.
- An LED lights only when it is connected the right way round, because it lets current flow in one direction only.
Examiner tips for 2(c)
When you draw a circuit, the ammeter goes in series and the voltmeter in parallel across the component. A voltmeter drawn in series, or across the cell and the component together, loses the mark.
In a two-resistor series question, write down what is the same (current) and what adds up (voltage, resistance). Then use \(V=I\times R\) for one resistor at a time with the values that belong to that resistor.
“Describe the I–V graph of a filament lamp” needs the shape (curve through the origin, getting less steep) and the link to resistance (resistance increases as the filament gets hotter). Saying only “it curves” earns one mark at most.
Learn the definitions exactly: current is the rate of flow of charge; voltage is energy transferred per unit charge. “Current is the flow of electricity” or “voltage is the push” is not credited.
For “why are house lights in parallel?”, give reasons with the physics: each lamp has the full mains voltage, can be switched independently, and keeps working if another lamp fails.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Current is used up by each component, so it is smaller after a lamp. | Charge cannot be destroyed; it is energy that is transferred. | In series the current is the same everywhere. |
| Electrons flow from + to −. | Electrons are negative, so they are attracted to the positive terminal. | Electrons flow from − to +; conventional current is drawn from + to −. |
| Adding a resistor in parallel makes the supply current smaller. | An extra branch gives charge another path. | Adding a branch in parallel increases the supply current. |
| A filament lamp's graph curves because the lamp “uses up” voltage. | The curve is caused by a change in resistance. | The filament gets hotter and its resistance increases. |
| An LDR's resistance increases in bright light. | It is the other way round. | More light → lower resistance; similarly, hotter thermistor → lower resistance. |
Key equations
| Equation | Use it to find | Route | On the Equation List? |
|---|---|---|---|
| voltage = current × resistance \(V=I\times R\) | voltage, current or resistance | All | yes — all papers |
| total resistance in series \(R=R_1+R_2\) | the combined resistance of two components in series | All | no — learn it |
| voltages in series \(V=V_1+V_2\) | the share of the supply voltage across each component | All | no — learn it |
| charge = current × time \(Q=I\times t\) | charge passing a point | Double · Physics | yes — all papers |
| energy transferred = charge × voltage \(E=Q\times V\) | energy transferred by a charge | Double · Physics | yes — all papers |
| currents at a junction \(I=I_1+I_2\) | the supply current in a parallel circuit | Double · Physics | no — learn it |
Statements 2.13, 2.15 and 2.21 ask you to know these relationships, so learn them even though they are on the Equation List.
Check your understanding
1. State what is meant by electric current, and name the particles that carry the current in a metal wire. All
Current is the rate of flow of charge. In a metal it is carried by (free) electrons, which are negatively charged.
2. A 230 V heater has a resistance of 46 Ω. Calculate the current in it. All
\(I=\dfrac{V}{R}=\dfrac{230}{46}=\) 5.0 A
3. A 4.0 Ω and an 8.0 Ω resistor are in series with a 6.0 V battery. Calculate the current and the voltage across the 8.0 Ω resistor. All
\(R=12\ \Omega\); \(I=6.0\div12=\) 0.50 A; \(V=0.50\times8.0=\) 4.0 V
4. Sketch the I–V graph for a filament lamp and explain its shape. All
A curve through the origin that gets less steep at higher voltages, in both directions (1). As the current increases the filament gets hotter (1), so its resistance increases (1).
5. A 12 V motor takes a current of 2.0 A for 10 s. Calculate the charge that passes through it and the energy transferred. State what “12 V” means. Double · Physics
\(Q=I\times t=2.0\times10=\) 20 C; \(E=Q\times V=20\times12=\) 240 J. 12 V means 12 J of energy is transferred for each coulomb of charge (12 J/C).
6. A current of 0.75 A enters a junction that splits into two branches. One branch carries 0.30 A. Find the current in the other branch and explain your answer. Double · Physics
0.45 A. Charge is conserved at a junction, so the total current out (0.30 + 0.45) equals the current in (0.75 A).
7. An LDR is in series with a lamp and a battery. The room gets darker. Describe what happens to the brightness of the lamp. Double · Physics
The resistance of the LDR increases (1), so the current in the circuit decreases (1) and the lamp becomes dimmer (1).
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