Recall and use the equation for electrical energy E = I V t

4.2.4 Resistance

Learning Objectives

• Recall and use Ohm’s law: R = V ⁄ I
• Derive and apply the resistivity formula R = ρ L ⁄ A
• Sketch and interpret the straight‑line I‑V graph of an ideal resistor.
• Design a simple experiment with a voltmeter and ammeter to determine an unknown resistance.
• Relate resistance to electrical power and energy:
   P = IV = I²R = V²⁄R and E = Pt = IVt = I²Rt = V²t⁄R
• Explain qualitatively and quantitatively how length, cross‑sectional area and material affect resistance.
• Identify common practical uses of resistors (current limiting, voltage division, heating elements, etc.) and note relevant safety points.

Key Concepts & Formulas

Symbol	Quantity	Unit	Key Formula(s)
R	Resistance	Ω (ohm)	R = V ⁄ I  R = ρ L ⁄ A
ρ	Resistivity (intrinsic property)	Ω·m	Material constant (see table below)
I	Current	A (ampere)
V	Potential difference (voltage)	V (volt)
t	Time	s (second)
P	Power	W (watt)	P = IV = I²R = V²⁄R
E	Energy transferred	J (joule)	E = Pt = IVt = I²Rt = V²t⁄R

Resistivity of Common Materials

Material	Resistivity ρ (Ω·m)	Typical use
Copper	1.7 × 10⁻⁸	Electrical wiring
Aluminium	2.8 × 10⁻⁸	Power‑line conductors
Nichrome (heating element)	1.1 × 10⁻⁶	Toasters, hair dryers
Glass (dry)	≈ 10¹⁴	Insulation
Rubber	≈ 10¹³	Cable coating

Quantitative Dependence of Resistance

From R = ρ L ⁄ A:

• Length (L): Doubling the length doubles the resistance (R ∝ L).
• Cross‑sectional area (A): Doubling the area halves the resistance (R ∝ 1/A).
• Resistivity (ρ): Materials with larger ρ give larger R for the same geometry.

Example: A copper wire (ρ = 1.7 × 10⁻⁸ Ω·m) 2 m long and 1 mm² in cross‑section has
R = ρL/A = (1.7 × 10⁻⁸ Ω·m · 2 m) ⁄ (1 × 10⁻⁶ m²) ≈ 0.034 Ω. If the length is increased to 4 m, R becomes 0.068 Ω (double). If the area is increased to 2 mm², R becomes 0.017 Ω (half).

I‑V Graph of an Ideal Resistor

Simple Experiment to Determine an Unknown Resistance

1. Connect the unknown resistor (Rₓ) in series with a known supply, an ammeter (A) and a voltmeter (V) as shown below.
2. Read the current (I) from the ammeter and the voltage across the resistor (V) from the voltmeter.
3. Calculate Rₓ using Ohm’s law: Rₓ = V ⁄ I.
4. Repeat with at least two different supply voltages to check that the I‑V graph is linear (constant R).

Power and Energy in Resistive Circuits

1. Start with the definition of power: P = IV.
2. Using Ohm’s law (V = IR) you can rewrite power as:
   • P = I²R (useful when current is known)
   • P = V²⁄R (useful when voltage is known)
3. Energy is power multiplied by the time for which it is delivered:
   E = Pt = IVt = I²Rt = V²t⁄R [units J = W·s].
4. The product I·t alone has units of coulomb (C) and represents electric charge, **not** energy.

Worked Example – Heating Element

Problem: A heating element of resistance R = 20 Ω is connected to a 240 V mains supply. Calculate the energy it uses in 5 minutes.

1. Current from Ohm’s law: I = V⁄R = 240 V ⁄ 20 Ω = 12 A
2. Time in seconds: t = 5 min × 60 s ⁄ min = 300 s
3. Energy: E = IVt = (12 A)(240 V)(300 s) = 8.64 × 10⁵ J
4. In kilojoules: E = 864 kJ

Units check: A·V·s = W·s = J, so the result is in joules.

Voltage Divider – Using Resistors to Obtain a Desired Output Voltage

Two resistors R₁ and R₂ in series across a supply Vₛ give an output voltage across R₂:

\[ V_{\text{out}} = V_s \frac{R_2}{R_1+R_2} \]

Example: With Vₛ = 12 V, R₁ = 4 kΩ and R₂ = 2 kΩ, the output is

\[ V_{\text{out}} = 12 \times \frac{2}{4+2}= 4\text{ V} \]

Current‑Limiting Resistor (LED Example)

To protect a 2 V, 20 mA LED from a 9 V battery:

• Desired current: I = 20 mA = 0.02 A
• Voltage that must be dropped by the resistor: V_R = 9 V – 2 V = 7 V
• Required resistance: R = V_R⁄I = 7 V ⁄ 0.02 A = 350 Ω
• Power dissipated in the resistor: P = I²R = (0.02)² × 350 ≈ 0.14 W → use a 0.25 W (or larger) resistor.

Safety Considerations When Working with Resistors

• Resistors that dissipate significant power become hot; allow them to cool before handling.
• Never touch a resistor that is glowing or feels warm after operation.
• Use insulated tools and keep fingers away from exposed leads when the circuit is powered.
• Check the resistor’s power rating; exceeding it can cause burns or fire.

Practice Questions

1. A lamp has a resistance of 30 Ω and is connected to a 120 V** supply.
   • Current: I = V⁄R = 4 A
   • Power: P = IV = 480 W
   • Energy used in 2 h**: t = 7200 s, E = Pt = 480 × 7200 = 3.46 × 10⁶ J ≈ 0.96 kWh.
2. A circuit contains a resistor of 10 Ω. A current of 3 A** flows for **45 s**.
   • Voltage across the resistor: V = IR = 30 V.
   • Energy transferred: E = I²Rt = (3)² × 10 × 45 = 4 050 J.
3. Explain why a material with a high resistivity (ρ) is a good insulator, using R = ρ L⁄A. Give two everyday examples.
   • High ρ makes R extremely large for any realistic dimensions, so only a negligible current can flow.
   • Examples: glass windows (ρ ≈ 10¹⁴ Ω·m) and rubber coating on electrical cables (ρ ≈ 10¹³ Ω·m).
4. In a heating element the resistance is 15 Ω** and it is supplied with 230 V**.
   • Current: I = V⁄R = 15.33 A; Power: P = V²⁄R ≈ 3 523 W.
   • Running for 10 min** (600 s):
     E = Pt = 3 523 × 600 ≈ 2.11 × 10⁶ J ≈ 0.59 kWh.

Suggested Diagram

Summary

Resistance links voltage, current and the physical dimensions of a conductor. Mastery of the following relationships enables students to:

• Use R = V⁄I and R = ρ L⁄A to predict how a conductor behaves.
• Interpret the straight‑line I‑V graph of an ideal resistor.
• Calculate power (P = IV = I²R = V²⁄R) and energy (E = IVt) for real‑world devices such as heaters, lights and LEDs.
• Design simple experiments with voltmeters and ammeters to measure unknown resistances.
• Apply resistors safely in circuits for current limiting, voltage division, and heating.

With these tools, students can analyse a wide variety of IGCSE‑level circuits and understand why different materials are chosen for specific electrical roles.