# Energy and voltage in circuits

> Edexcel International GCSE Physics · 4PH1 (2017)
> Source: https://www.owlsprep.com/study/edexcel-igcse-physics-s2-energy-and-voltage-in-circuits/

This guide covers all core Edexcel IGCSE Physics (4PH1) content for energy and voltage in circuits, including circuit types, I-V characteristics, core formulae, and component behavior for your exam.

**Prerequisites:** [Basic circuit symbol and component identification](https://www.owlsprep.com/study/edexcel-igcse-physics-s2-circuit-fundamentals/)

## Learning objectives

- Explain suitability of series and parallel circuits for applications including domestic lighting
- Describe how current varies with voltage for wires, resistors, filament lamps, and diodes, including experimental method for I-V characteristics
- Recall and use core formulae: V=IR, Q=It, E=QV for circuit calculations
- Describe resistance variation of LDRs with illumination and NTC thermistors with temperature
- Calculate current, voltage, and resistance for two resistive components in series
- Explain conservation of current at junctions and parallel circuit voltage rules

## Series and Parallel Circuit Properties and Applications

Circuits are classified as series (components connected in a single loop) or parallel (components connected across separate branches). Each type has unique properties that make it suitable for different uses.

**Series vs Parallel Circuits** — Series: Single loop, current identical through all components, supply voltage shared between components. Parallel: Multiple branches, voltage identical across all branches, supply current split between branches.

- Series circuits are used for low-power fairy lights, where multiple bulbs share the supply voltage to reduce power use
- Parallel circuits are used for domestic lighting, so each bulb can be switched independently and receives full mains voltage

**Worked example:** A 2Ω resistor and a 3Ω resistor are connected in series to a 10V battery. Calculate the total resistance, total current, and voltage across each resistor.

1. Total resistance for series circuits is the sum of individual resistances:

   $$R_{total} = R_1 + R_2 = 2 + 3 = 5\Omega$$
2. Use Ohm's Law to calculate total current:

   $$I = \frac{V}{R} = \frac{10}{5} = 2A$$
3. Current is identical across all series components, so calculate voltage across each resistor:

   $$V_1 = I \times R_1 = 2 \times 2 = 4V$$
4. $$V_2 = I \times R_2 = 2 \times 3 = 6V$$
5. Verify: 4V + 6V = 10V, matching the supply voltage.

> **Exam tip:** When asked to explain domestic lighting circuit design, always mention independent switching and full voltage per bulb as key benefits of parallel layouts.

*Calculator:* allowed

## Core Circuit Formulae and Quantities

Three core formulae must be memorized for this topic, as they are not provided on the exam formula sheet. Always convert all units to SI base units before calculation.

**Voltage** — Energy transferred per unit charge passed between two points in a circuit. 1 volt = 1 joule per coulomb (1V = 1J/C).

$$V = I \times R \text{ (Ohm's Law: Voltage = Current } \times \text{ Resistance)}$$

$$Q = I \times t \text{ (Charge = Current } \times \text{ Time)}$$

$$E = Q \times V \text{ (Energy Transferred = Charge } \times \text{ Voltage)}$$

> **tip**
>
> Current is the rate of flow of charge. In metallic conductors, current is a flow of negatively charged electrons, moving from the negative to positive terminal of a cell.

**Worked example:** A 12V battery supplies a current of 2A for 30 seconds. Calculate the total charge flow and energy transferred by the battery.

1. Calculate charge flow using Q = I × t:

   $$Q = 2 \times 30 = 60C$$
2. Calculate energy transferred using E = Q × V:

   $$E = 60 \times 12 = 720J$$

*Calculator:* allowed

## I-V Characteristics of Common Components

I-V graphs plot current against voltage for a component, showing how resistance changes with voltage. You need to know the shape of these graphs for four common components, and the experimental method to measure them.

**Ohmic Conductor** — A component (e.g. fixed resistor, wire at constant temperature) where current is directly proportional to voltage, producing a straight-line I-V graph through the origin.

| Component | I-V Graph Shape | Reason for Shape |
| --- | --- | --- |
| Fixed Resistor (Ohmic) | Straight line through origin | Resistance constant at constant temperature |
| Filament Lamp | Curved line, gradient decreases at high voltage | Resistance increases as filament temperature rises with current |
| Diode | Zero current for negative voltage, sharp rise in current for positive voltage | Only conducts in forward bias (positive direction) |

**Worked example:** Describe an experiment to measure the I-V characteristic of a filament lamp.

1. Set up a series circuit with a cell, variable resistor, ammeter, filament lamp, and a voltmeter connected in parallel across the lamp.
2. Adjust the variable resistor to change the voltage across the lamp, recording pairs of current (ammeter) and voltage (voltmeter) readings for at least 5 settings.
3. Reverse the cell connections to take negative voltage and current readings.
4. Plot a graph of current (y-axis) against voltage (x-axis) to produce the I-V characteristic.

> **Exam tip:** If asked to explain the filament lamp curve, always link the shape to increasing temperature causing increasing resistance, which reduces the rate of current rise at high voltages.

*Calculator:* allowed

## Variable Resistance Components: LDRs, Thermistors and Diodes

Specialist resistors change their resistance in response to external conditions, making them useful for sensor circuits. Lamps and LEDs can also be used to indicate the presence of current, as they light up when current flows through them.

**LDR and NTC Thermistor** — LDR (Light Dependent Resistor): Resistance decreases as light intensity increases. NTC Thermistor: Resistance decreases as temperature increases.

> **mnemonic**
>
> LDR: Lighter = Lower Resistance; Thermistor (NTC): Hotter = Lower Resistance. This pair of rules is frequently tested in 1-mark application questions.

**Worked example:** An LDR is used in an automatic porch light circuit. Explain why the light turns on when it gets dark.

1. When light intensity is low (dark), the resistance of the LDR increases.
2. Higher resistance reduces the current in the control circuit.
3. This triggers a relay switch to close and turn on the porch light.

*Calculator:* allowed

## Current Conservation and Junction Rules

Current is conserved in all circuits, meaning no charge is lost at junctions in parallel circuits. Voltage across components in parallel is identical to the supply voltage.

**Junction Rule** — The total current flowing into a junction in a parallel circuit is equal to the total current flowing out of the junction, as charge cannot be created or destroyed.

**Worked example:** At a junction in a parallel circuit, 3A of current flows into the junction, and 1A flows out through one branch. Calculate the current flowing out through the second branch.

1. Apply the junction rule: total current in = total current out

   $$I_{in} = I_{1} + I_{2}$$
2. Rearrange to solve for the second branch current:

   $$I_{2} = 3A - 1A = 2A$$

> **Exam tip:** For parallel circuits, you only need to describe trends qualitatively: do not attempt to calculate combined parallel resistance, as this is out of scope for the Edexcel IGCSE specification.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Assuming current is identical across all parallel branches
  - Why it fails: Current splits at junctions, only voltage is identical across parallel branches
  - Correct: Use the junction rule: total current into a junction equals total current out
- **Wrong:** Using the parallel combined resistance formula for calculations
  - Why it fails: Edexcel IGCSE only requires series resistance calculations for two components; parallel resistance calculations are out of scope
  - Correct: Only add resistances directly for series circuits, describe parallel behavior qualitatively
- **Wrong:** Drawing the filament lamp I-V graph as a straight line
  - Why it fails: Filament temperature rises with current, increasing resistance, so the graph curves
  - Correct: Draw an S-shaped curve for the filament lamp, with decreasing gradient at high voltages
- **Wrong:** Defining voltage as a 'push of current' in exam questions
  - Why it fails: Mark schemes require the precise definition: energy transferred per unit charge, 1V = 1J/C
  - Correct: Use the exact syllabus definition for all 2+ mark voltage explanation questions
- **Wrong:** Mixing up LDR and thermistor resistance trends
  - Why it fails: Both components have resistance that falls with an increase in their trigger stimulus, but students often mix up the stimulus for each
  - Correct: Use the mnemonic: Lighter = Lower LDR resistance, Hotter = Lower thermistor resistance

## Cheatsheet

| Quantity/Rule | Formula/Statement | Key Detail | Exam Tip |
| --- | --- | --- | --- |
| Ohm's Law | V = I × R | Units: V (V), I (A), R (Ω) | Memorize, not provided on formula sheet |
| Charge Flow | Q = I × t | Units: Q (C), t (s) | Convert time to seconds before calculation |
| Energy Transferred | E = Q × V | 1V = 1 J/C | Use exact definition for voltage explain questions |
| Series Circuits | I same, V_total = V1+V2, R_total = R1+R2 | Only for 2 resistive components | Add resistances directly only for series |
| Parallel Circuits | V same across branches, I_in = I_out at junctions | No parallel resistance calculations needed | Domestic lighting uses parallel for independent control |
| I-V Graphs | Ohmic: straight line, Filament: curved, Diode: forward only | Diode only conducts with positive voltage | Draw correct shapes for graph questions |
| Variable Resistors | LDR: higher light = lower R, Thermistor: higher temp = lower R | Only NTC thermistors are assessed | Use the mnemonic to avoid mixing trends |

## What's next

Now you have mastered core energy and voltage in circuits content for Edexcel IGCSE Physics, you are ready to progress to mains electricity and circuit safety, which builds on your current and voltage knowledge to explain fuse operation and the P=IV power formula. You can also move to the next electricity sub-topic, electrostatic charge. This content is assessed in both Paper 1 and Paper 2, as well as the Double Award (4SD0) specification. Make sure you practice past paper questions focusing on series circuit calculations and I-V characteristic descriptions, and memorize the three core formulae as they are not provided on the exam formula sheet.

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