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) = \rho0[1+\alpha (T-T0)] \$

  where \$\alpha\$ is the temperature coefficient, \$T0\$ a reference temperature, and \$\rho0\$ the resistivity at \$T_0\$.

Resistivity of Common Materials

Series and Parallel Combinations

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

1. Series: \$R{\text{total}} = R1 + R2 + \dots + Rn\$
2. Parallel: \$\displaystyle \frac{1}{R{\text{total}}} = \frac{1}{R1} + \frac{1}{R2} + \dots + \frac{1}{Rn}\$

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: \$R1 = 4\ \Omega\$ (series), and \$R2 = 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}} = R1 + 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) = \rho0[1+\alpha (T-T0)]\$ 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: \$R1 = 1\ \Omega\$ and \$R2 = 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}\$.