recall and use V = IR

Resistance and Resistivity

In this topic we will recall the fundamental relationship between voltage, current and resistance, and explore how the intrinsic property of a material – its resistivity – determines the resistance of a conductor.

Learning Objective

Students should be able to recall and apply the equation

$$ V = I R $$

to solve problems involving resistors in series and parallel, and to calculate resistance from the material’s resistivity.

Key Concepts

• Resistance ($R$) – opposition to the flow of electric charge, measured in ohms (Ω).
• Resistivity ($\rho$) – a material‑specific constant that quantifies how strongly the material opposes current, measured in Ω·m.
• Relationship between resistance and resistivity: $$ R = \rho \frac{L}{A} $$ where $L$ is the length of the conductor and $A$ its cross‑sectional area.
• Temperature dependence of resistivity (for metals): $$ \rho(T) = \rho_0[1+\alpha (T-T_0)] $$ where $\alpha$ is the temperature coefficient, $T_0$ a reference temperature, and $\rho_0$ the resistivity at $T_0$.

Resistivity of Common Materials

Material	Resistivity, $\rho$ (Ω·m)	Typical Uses
Copper	1.68 × 10⁻⁸	Electrical wiring
Aluminium	2.82 × 10⁻⁸	Power lines
Silver	1.59 × 10⁻⁸	High‑frequency contacts
Constantan (Ni‑Cu alloy)	4.9 × 10⁻⁷	Thermocouples
Glass (dry)	≈ 10¹⁴	Insulators

Series and Parallel Combinations

When resistors are connected together, the total resistance depends on the configuration.

1. Series: $R_{\text{total}} = R_1 + R_2 + \dots + R_n$
2. Parallel: $\displaystyle \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$

Worked Example 1 – Calculating Resistance from Resistivity

A copper wire has a length of $2.0\ \text{m}$ and a cross‑sectional area of $0.5\ \text{mm}^2$. Find its resistance.

1. Convert area to square metres: $0.5\ \text{mm}^2 = 0.5 \times 10^{-6}\ \text{m}^2$.
2. Use $R = \rho \dfrac{L}{A}$ with $\rho_{\text{Cu}} = 1.68 \times 10^{-8}\ \Omega\!\cdot\!\text{m}$.
3. Calculate: $$ R = (1.68 \times 10^{-8}) \frac{2.0}{0.5 \times 10^{-6}} = 6.72 \times 10^{-2}\ \Omega $$

Worked Example 2 – Using $V = IR$ in a Mixed Circuit

In the circuit below, a $12\ \text{V}$ battery supplies three resistors: $R_1 = 4\ \Omega$ (series), and $R_2 = 6\ \Omega$, $R_3 = 12\ \Omega$ (parallel). Find the current supplied by the battery.

1. Find the equivalent resistance of the parallel part: $$ \frac{1}{R_{23}} = \frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} $$ $$ R_{23} = \frac{12}{3} = 4\ \Omega $$
2. Total resistance: $$ R_{\text{total}} = R_1 + R_{23} = 4\ \Omega + 4\ \Omega = 8\ \Omega $$
3. Apply $V = IR$: $$ I = \frac{V}{R_{\text{total}}} = \frac{12\ \text{V}}{8\ \Omega} = 1.5\ \text{A} $$

Temperature Effects – Example

A nichrome wire has $\rho_0 = 1.10 \times 10^{-6}\ \Omega\!\cdot\!\text{m}$ at $20^\circ\text{C}$ and a temperature coefficient $\alpha = 0.00017\ \text{K}^{-1}$. What is its resistivity at $100^\circ\text{C}$?

1. Use $\rho(T) = \rho_0[1+\alpha (T-T_0)]$ with $T_0 = 20^\circ\text{C}$.
2. Calculate: $$ \rho(100^\circ\text{C}) = 1.10 \times 10^{-6}[1+0.00017(100-20)] $$ $$ = 1.10 \times 10^{-6}[1+0.00017 \times 80] $$ $$ = 1.10 \times 10^{-6}[1+0.0136] $$ $$ = 1.10 \times 10^{-6} \times 1.0136 \approx 1.115 \times 10^{-6}\ \Omega\!\cdot\!\text{m} $$

Summary Checklist

• Remember $V = IR$ and be able to rearrange for $I$ or $R$.
• Use $R = \rho L/A$ to link material properties to resistance.
• Apply series and parallel formulas correctly.
• Consider temperature effects when dealing with metals or alloys.
• Check units carefully – convert mm² to m², cm to m, etc.

Practice Questions

1. A steel rod ($\rho = 1.0 \times 10^{-7}\ \Omega\!\cdot\!\text{m}$) is $0.5\ \text{m}$ long and has a diameter of $2\ \text{mm}$. Find its resistance.
2. Three resistors of $2\ \Omega$, $3\ \Omega$, and $6\ \Omega$ are connected in parallel. What is the total resistance?
3. A circuit contains a $9\ \text{V}$ battery and two resistors in series: $R_1 = 1\ \Omega$ and $R_2 = 3\ \Omega$. Determine the voltage drop across each resistor.
4. Given a copper wire with $R = 0.1\ \Omega$ at $20^\circ\text{C}$, estimate its resistance at $80^\circ\text{C}$ using $\alpha = 0.0039\ \text{K}^{-1}$.