Electric Circuits

Physics· 2.4 (Statements 64-80, WPH12)· 45 min read

1. Fundamental Circuit Quantities & Ohm's Law★★☆☆☆⏱ 8 min

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.

📘 Definition

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 $P = V I$ . You can derive two alternative power formulae by substituting $V = I R$ or $I = V / R$ into the base equation, giving $P = I^{2} R$ and $P = V^{2} / R$ respectively. Total energy transferred is $W = V I t$ .

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

2. Series & Parallel Circuits + Conservation Laws★★★☆☆⏱ 10 min

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

Goal:

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 $V = I R$ for each term:
$IR_{total} = IR_1 + IR_2$
3
Cancel $I$ (same for all components):
$R_{total} = R_1 + R_2$

Result:

Total series resistance equals the sum of individual resistances.

🔬 Derivation

Goal:

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

Result:

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:
$R_{total} = 2.5\Omega$

3. I-V Characteristics & Resistivity★★★☆☆⏱ 10 min

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.

📘 Definition

Resistivity

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

Example:

Copper has a low resistivity ( $1.7 \times 1 0^{- 8} Ω m$ ) so is used for electrical wiring.

The current equation $I = n q v A$ 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).

📐 Worked Example

A 2m long copper wire has a cross-sectional area of $1 \times 1 0^{- 7} m^{2}$ . Copper has a resistivity of $1.7 \times 1 0^{- 8} Ω m$ . 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:
$R = 0.34\Omega$

4. Potential Dividers★★★★☆⏱ 9 min

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 $V_{o u t} = V_{in} \times (R_{o u t} / R_{t o t a l})$ , where $R_{o u t}$ 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:
$V_{out} \approx 8.2V$

5. EMF & Internal Resistance★★★★☆⏱ 8 min

📘 Definition

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 $ε = V + I r = I (R + r)$ , where $I r$ 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 $ε$ , 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 $V = I R$ or $V = ε - I r$ :
$V = 0.5 \times 2.8 = 1.4V$

6. Common Pitfalls

Wrong move:

Using Ohm's Law for non-ohmic components with a fixed resistance value

Why:

Ohm's Law only applies at constant temperature; filament bulbs, diodes and thermistors have variable resistance that changes with current/temperature.

Correct move:

Use $R = V / I$ at the specific current or pd value given for the component, rather than a fixed resistance value.

Wrong move:

Adding parallel resistors directly instead of summing reciprocals

Why:

Parallel circuits have the same pd across all branches, so current splits and total resistance is always less than the smallest individual resistor.

Correct move:

Use $1/ R_{t o t a l} = 1/ R_{1} + 1/ R_{2} + ...$ for parallel combinations, then take the reciprocal to get total resistance.

Wrong move:

Assuming emf and terminal pd are always equal

Why:

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 move:

Use $ε = V + I r$ , where $V$ is terminal pd and $I r$ is the lost volts when current $I$ flows.

Wrong move:

Swapping the resistance ratio in potential divider calculations

Why:

Output pd is proportional to the resistance across which it is measured, so using the wrong resistor will give an incorrect value.

Correct move:

Always use $V_{o u t} = V_{in} \times (R_{o u t} / R_{t o t a l})$ , where $R_{o u t}$ is the resistance across the output terminals.

Wrong move:

Using diameter directly instead of radius when calculating cross-sectional area for resistivity

Why:

Cross-sectional area $A = π r^{2}$ , so using diameter will give an area 4x larger than the true value.

Correct move:

Measure diameter with a micrometer, divide by 2 to get radius, then calculate area before substituting into $R = ρl / A$ .

7. Quick Reference Cheatsheet

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

Going deeper

official_documentEdexcel IAL Physics 2018 SpecificationSee Unit 2 Section 2.4 for full circuit requirements

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.