# Electric Circuits

> Physics · Edexcel IAL Physics (2018 Spec)
> Source: https://www.owlsprep.com/study/edexcel-ial-physics-u2-electric-circuits/

This guide covers all Edexcel IAL Physics Unit 2 (WPH12) electric circuits content, including circuit laws, resistance combinations, I-V graphs, potential dividers, emf and internal resistance, plus required core practicals.

**Prerequisites:** Basic algebra and graph interpretation skills; [Edexcel IAL Physics Unit 1 physical quantities and units](https://www.owlsprep.com/study/edexcel-ial-physics-u1-physical-quantities/)

## Learning objectives

- Calculate current, potential difference, resistance and power using standard circuit formulae
- Derive and apply series and parallel resistance rules from charge and energy conservation laws
- Interpret I-V graphs for ohmic conductors, filament bulbs, NTC thermistors and diodes
- Solve resistivity and current (I = nqvA) questions, including Core Practical 7 procedures
- Perform potential divider calculations, including circuits with thermistors and LDRs
- Distinguish emf from terminal pd, solve internal resistance problems, including Core Practical 8 methods

## Fundamental Circuit Quantities & Ohm's Law

The three core circuit quantities are defined by simple algebraic relationships: current as the rate of charge flow, potential difference as energy transferred per unit charge, and resistance as the ratio of pd to current. Ohm's Law is a special case where current is directly proportional to pd, provided temperature (and other physical conditions) remain constant.

**Ohm's Law** — For an ohmic conductor at constant temperature, the current through the component is directly proportional to the potential difference across it, so $R = V/I$ is constant.

Power is the rate of energy transfer, given by $P=VI$. You can derive two alternative power formulae by substituting $V=IR$ or $I=V/R$ into the base equation, giving $P=I^2R$ and $P=V^2/R$ respectively. Total energy transferred is $W=VIt$.

**Worked example:** A 12V battery is connected to a 4Ω fixed resistor for 5 minutes. Calculate (a) the current in the circuit, (b) total charge that flows through the resistor, (c) total energy transferred to the resistor.

1. Step 1: Calculate current using Ohm's Law:

   $$I = \frac{V}{R} = \frac{12}{4} = 3A$$
2. Step 2: Convert time to seconds (5 mins = 300s), then calculate charge:

   $$\Delta Q = I\Delta t = 3 \times 300 = 900C$$
3. Step 3: Calculate energy transferred:

   $$W = VIt = 12 \times 3 \times 300 = 10800J$$

## Series & Parallel Circuits + Conservation Laws

Circuit rules are derived from two fundamental conservation laws: charge conservation (current into a junction equals current out of the junction) and energy conservation (sum of emfs around a closed loop equals sum of potential drops around the loop).

For series circuits: current is the same through all components, total pd is the sum of individual pds. For parallel circuits: pd is the same across all branches, total current is the sum of individual branch currents.

**Derivation:** Derive series resistance formula

*Starting from:* Two resistors $R_1$ and $R_2$ in series with current $I$ flowing through both

1. Total pd across the combination is sum of individual pds:

   $$V_{total} = V_1 + V_2$$
2. Substitute $V=IR$ for each term:

   $$IR_{total} = IR_1 + IR_2$$
3. Cancel $I$ (same for all components):

   $$R_{total} = R_1 + R_2$$

*Conclusion:* Total series resistance equals the sum of individual resistances.

**Derivation:** Derive parallel resistance formula

*Starting from:* Two resistors $R_1$ and $R_2$ in parallel with pd $V$ across both

1. Total current through the combination is sum of individual branch currents:

   $$I_{total} = I_1 + I_2$$
2. Substitute $I=V/R$ for each term:

   $$\frac{V}{R_{total}} = \frac{V}{R_1} + \frac{V}{R_2}$$
3. Cancel $V$ (same for all branches):

   $$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$$

*Conclusion:* Reciprocal of total parallel resistance equals the sum of reciprocals of individual resistances.

**Worked example:** A 2Ω and 3Ω resistor are connected in series, then the combination is connected in parallel with a 5Ω resistor. Calculate the total resistance of the circuit.

1. Step 1: Calculate the series resistance of the 2Ω and 3Ω resistors:

   $$R_{series} = 2 + 3 = 5\Omega$$
2. Step 2: Calculate the parallel resistance of 5Ω and 5Ω:

   $$\frac{1}{R_{total}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$$
3. Step 3: Take the reciprocal to get total resistance:

   $$R_{total} = 2.5\Omega$$

## I-V Characteristics & Resistivity

I-V graphs show how current through a component changes with applied pd. Ohmic conductors have a straight line through the origin (constant resistance). Filament bulbs have a decreasing gradient (resistance increases with temperature, as lattice vibrations impede electron flow). NTC thermistors have an increasing gradient (resistance decreases with temperature, as more conduction electrons are released). Diodes only conduct above a ~0.6V threshold pd, with very high resistance below this value.

**Resistivity** — An intrinsic property of a material that determines its resistance for a given size: $R = \rho l / A$, where $l$ is length of the material, $A$ is cross-sectional area.

*Example:* Copper has a low resistivity ($1.7 \times 10^{-8}\Omega m$) so is used for electrical wiring.

The current equation $I = nqvA$ explains resistivity differences between materials: $n$ (number density of free electrons) is very high for metals (low resistivity), lower for semiconductors, and near zero for insulators (very high resistivity).

> **tip**
>
> Core Practical 7 (resistivity measurement): Use a ruler to measure wire length, a micrometer to measure diameter (to calculate area), and an ammeter/voltmeter to measure resistance via $R=V/I$. Calculate $\rho$ using $\rho = RA/l$.

**Worked example:** A 2m long copper wire has a cross-sectional area of $1 \times 10^{-7} m^2$. Copper has a resistivity of $1.7 \times 10^{-8} \Omega m$. Calculate the resistance of the wire.

1. Substitute values into the resistivity formula:

   $$R = \frac{\rho l}{A} = \frac{1.7 \times 10^{-8} \times 2}{1 \times 10^{-7}}$$
2. Simplify to get resistance:

   $$R = 0.34\Omega$$

## Potential Dividers

A potential divider uses two or more resistors in series to split an input pd into a smaller output pd. For a uniform current-carrying wire, potential is proportional to length, so you can adjust output pd by changing the length of wire across which output is measured.

The standard potential divider formula is $V_{out} = V_{in} \times (R_{out}/R_{total})$, where $R_{out}$ is the resistance across the output terminals. You can use variable components like thermistors (resistance decreases with temperature) or LDRs (resistance decreases with increasing light intensity) to make output pd respond to environmental changes.

**Worked example:** A potential divider circuit uses a 10kΩ fixed resistor and an LDR in series with a 9V battery. In dark conditions, the LDR has a resistance of 100kΩ. Calculate the output pd measured across the LDR.

1. Step 1: Calculate total series resistance:

   $$R_{total} = 10k\Omega + 100k\Omega = 110k\Omega$$
2. Step 2: Substitute into potential divider formula:

   $$V_{out} = 9 \times \frac{100k\Omega}{110k\Omega}$$
3. Step 3: Calculate output pd:

   $$V_{out} \approx 8.2V$$

## EMF & Internal Resistance

**Electromotive Force (emf)** — The total energy transferred per unit charge by a power supply, equal to the terminal pd when no current flows. The relationship between emf, terminal pd and internal resistance is $\varepsilon = V + Ir = I(R + r)$, where $Ir$ is the 'lost volts' (energy dissipated per unit charge in the internal resistance of the supply).

To measure emf and internal resistance experimentally (Core Practical 8), vary the external resistance, measure terminal pd $V$ and current $I$, then plot a graph of $V$ against $I$. The y-intercept equals emf $\varepsilon$, and the negative gradient equals internal resistance $r$.

**Worked example:** A cell has an emf of 1.5V and internal resistance of 0.2Ω. It is connected to a 2.8Ω external resistor. Calculate the terminal pd of the cell when current flows.

1. Step 1: Calculate total circuit resistance:

   $$R_{total} = R + r = 2.8 + 0.2 = 3\Omega$$
2. Step 2: Calculate current in the circuit:

   $$I = \frac{\varepsilon}{R_{total}} = \frac{1.5}{3} = 0.5A$$
3. Step 3: Calculate terminal pd, either as $V=IR$ or $V=\varepsilon - Ir$:

   $$V = 0.5 \times 2.8 = 1.4V$$

## Common pitfalls

- **Wrong:** Using Ohm's Law for non-ohmic components with a fixed resistance value
  - Why it fails: Ohm's Law only applies at constant temperature; filament bulbs, diodes and thermistors have variable resistance that changes with current/temperature.
  - Correct: Use $R = V/I$ at the specific current or pd value given for the component, rather than a fixed resistance value.
- **Wrong:** Adding parallel resistors directly instead of summing reciprocals
  - Why it fails: Parallel circuits have the same pd across all branches, so current splits and total resistance is always less than the smallest individual resistor.
  - Correct: Use $1/R_{total} = 1/R_1 + 1/R_2 + ...$ for parallel combinations, then take the reciprocal to get total resistance.
- **Wrong:** Assuming emf and terminal pd are always equal
  - Why it fails: Terminal pd is the pd available to the external circuit, and is lower than emf when current flows due to energy lost across the supply's internal resistance.
  - Correct: Use $\varepsilon = V + Ir$, where $V$ is terminal pd and $Ir$ is the lost volts when current $I$ flows.
- **Wrong:** Swapping the resistance ratio in potential divider calculations
  - Why it fails: Output pd is proportional to the resistance across which it is measured, so using the wrong resistor will give an incorrect value.
  - Correct: Always use $V_{out} = V_{in} \times (R_{out}/R_{total})$, where $R_{out}$ is the resistance across the output terminals.
- **Wrong:** Using diameter directly instead of radius when calculating cross-sectional area for resistivity
  - Why it fails: Cross-sectional area $A = \pi r^2$, so using diameter will give an area 4x larger than the true value.
  - Correct: Measure diameter with a micrometer, divide by 2 to get radius, then calculate area before substituting into $R = \rho l / A$.

## Cheatsheet

| Quantity | Formula | Units | Key Notes |
| --- | --- | --- | --- |
| Current | $I = \Delta Q/\Delta t$ | A | 1A = 1C/s |
| Resistance | $R = V/I$ | Ω | Ohm's Law only for constant T |
| Series Resistance | $R_{total} = R_1 + R_2 + ...$ | Ω | Same current through all components |
| Parallel Resistance | $1/R_{total} = 1/R_1 + 1/R_2 + ...$ | Ω | Same pd across all branches |
| Power | $P=VI = I^2R = V^2/R$ | W | 1W = 1J/s |
| Resistivity | $R = \rho l/A$ | Ω·m | Intrinsic material property |
| Potential Divider | $V_{out} = V_{in} \times R_{out}/R_{total}$ | V | Proportional to resistance ratio |
| EMF & Internal Resistance | $\varepsilon = V + Ir = I(R + r)$ | V | Y-intercept of V vs I = $\varepsilon$, gradient = $-r$ |

## What's next

Now that you have mastered electric circuits for Edexcel IAL Physics Unit 2, you can move on to revising the remaining content for the WPH12 exam: waves and quantum phenomena, which make up 50% of the unit marks. You should also practice past paper circuit questions, focusing on core practical assessments and extended response questions explaining resistance changes for thermistors, LDRs and filament bulbs. Once you complete all Unit 2 content, you can start preparing for Unit 4 which introduces more advanced circuit concepts like capacitors and electromagnetism.

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