recall and use V = IR

Resistance, Resistivity and Ohm’s Law (Cambridge 9702 – 9.2 & 9.3)

This unit covers the relationships between voltage, current and resistance, introduces the material property resistivity, and shows how to analyse simple DC circuits using series‑parallel rules, Kirchhoff’s laws and the voltage‑divider concept. All formulas are presented with SI units the first time they appear.

Learning Objectives

• State and rearrange Ohm’s law \(V = I R\).
• Distinguish resistance \(R\) (Ω) from resistivity \(\rho\) (Ω·m).
• Calculate resistance from a material’s resistivity using \(\displaystyle R = \rho \frac{L}{A}\).
• Explain how temperature affects resistivity and the internal resistance of a source.
• Analyse series and parallel networks; apply Kirchhoff’s laws and the voltage‑divider formula.
• Use the power relations \(\displaystyle P = IV = I^{2}R = \frac{V^{2}}{R}\) to assess heating loss.
• Interpret I–V graphs for ohmic and non‑ohmic devices.

Key Symbols & Units

Symbol	Name	Definition	SI Unit
\(V\)	Potential difference (voltage)	Energy per unit charge between two points	V (volt)
\(I\)	Current	Rate of charge flow	A (ampere)
\(R\)	Resistance	Opposition to current in a component	Ω (ohm)
\(\rho\)	Resistivity	Intrinsic property of a material; independent of shape	Ω·m (ohm‑metre)
\(L\)	Length of a conductor	Physical length of the material	m (metre)
\(A\)	Cross‑sectional area	Area through which current flows	m² (square metre)
\(\alpha\)	Temperature coefficient of resistivity	Fractional change in \(\rho\) per kelvin for a metal	K⁻¹
\(r\)	Internal resistance of a source	Resistance “seen” inside a battery or cell	Ω
\(\mathcal{E}\)	Electromotive force (emf)	Ideal terminal voltage of a source when no current flows	V

Fundamental Relationships

1. Ohm’s Law

\[ V = I R \]

• For an ohmic device, \(R\) is constant over the range of applied voltage.
• Rearrangements: \[ I = \frac{V}{R},\qquad R = \frac{V}{I} \]

2. Resistivity vs. Resistance

• Resistivity (\(\rho\)) is a material constant – the same for any piece of that material.
• Resistance (\(R\)) depends on both the material and the geometry of the conductor.
• Relationship: \[ R = \rho \frac{L}{A} \] where \(L\) is the length and \(A\) the cross‑sectional area.

3. Temperature Dependence (Metals)

\[ \rho(T) = \rho_{0}\bigl[1+\alpha\,(T-T_{0})\bigr] \]

• \(\rho_{0}\) – resistivity at the reference temperature \(T_{0}\) (usually \(20^{\circ}\text{C}\)).
• \(\alpha\) – temperature coefficient (e.g. copper \(3.9\times10^{-3}\,\text{K}^{-1}\)).

4. Internal‑Resistance Model of a Battery

\[ V = \mathcal{E} - I r \]

• The terminal voltage falls below the emf by the amount \(I r\) when current flows.

5. Power Relations

\[ P = IV = I^{2}R = \frac{V^{2}}{R} \]

• Useful for estimating heating loss and checking the consistency of answers.

Series and Parallel Networks

Configuration	Equivalent Resistance
Series: \(R_{1},R_{2},\dots,R_{n}\)	\(R_{\text{eq}} = R_{1}+R_{2}+ \dots + R_{n}\)
Parallel: \(R_{1},R_{2},\dots,R_{n}\)	\(\displaystyle\frac{1}{R_{\text{eq}}}= \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \dots + \frac{1}{R_{n}}\)

Kirchhoff’s Laws (9.3 DC Circuits)

Potential (Voltage) Divider

Two resistors \(R_{1}\) and \(R_{2}\) in series across a supply voltage \(V_{s}\) give a voltage across \(R_{2}\) of

\[ V_{R_{2}} = V_{s}\,\frac{R_{2}}{R_{1}+R_{2}} \]

The same expression applies for the voltage across \(R_{1}\) by swapping the indices.

Worked Examples

Example 1 – Resistance from Resistivity

A copper wire is 2.0 m long with a cross‑sectional area of \(0.5\;\text{mm}^{2}\). Find its resistance.

1. Convert area: \(0.5\;\text{mm}^{2}=0.5\times10^{-6}\;\text{m}^{2}\).
2. Use \(R = \rho \dfrac{L}{A}\) with \(\rho_{\text{Cu}} = 1.68\times10^{-8}\;\Omega\!\cdot\!\text{m}\): \[ R = (1.68\times10^{-8})\frac{2.0}{0.5\times10^{-6}} = 6.72\times10^{-2}\;\Omega. \]

Example 2 – Mixed Series‑Parallel Circuit

A 12 V battery supplies three resistors: \(R_{1}=4\;\Omega\) (series) and a parallel branch containing \(R_{2}=6\;\Omega\) and \(R_{3}=12\;\Omega\). Find the total current.

1. Parallel equivalent: \[ \frac{1}{R_{23}} = \frac{1}{6} + \frac{1}{12}= \frac{3}{12}\;\Rightarrow\;R_{23}=4\;\Omega. \]
2. Total resistance: \[ R_{\text{tot}} = R_{1}+R_{23}=4+4=8\;\Omega. \]
3. Current from the battery: \[ I = \frac{V_{s}}{R_{\text{tot}}}= \frac{12}{8}=1.5\;\text{A}. \]

Example 3 – Temperature Effect on Resistivity

A nichrome wire has \(\rho_{0}=1.10\times10^{-6}\;\Omega\!\cdot\!\text{m}\) at \(20^{\circ}\text{C}\) and \(\alpha =1.7\times10^{-4}\;\text{K}^{-1}\). Find \(\rho\) at \(100^{\circ}\text{C}\).

\[ \rho(100^{\circ}\text{C}) = 1.10\times10^{-6}\bigl[1+1.7\times10^{-4}(100-20)\bigr] =1.10\times10^{-6}\bigl[1+0.0136\bigr] \approx 1.115\times10^{-6}\;\Omega\!\cdot\!\text{m}. \]

Example 4 – Battery with Internal Resistance

A 9 V battery has \(\mathcal{E}=9\;\text{V}\) and internal resistance \(r=0.5\;\Omega\). It supplies two series resistors \(R_{1}=2\;\Omega\) and \(R_{2}=3\;\Omega\).

1. Total resistance: \(R_{\text{tot}} = r+R_{1}+R_{2}=0.5+2+3=5.5\;\Omega\).
2. Current: \(I = \dfrac{\mathcal{E}}{R_{\text{tot}}}= \dfrac{9}{5.5}=1.64\;\text{A}\).
3. Terminal voltage: \(V = \mathcal{E} - I r = 9 - (1.64)(0.5)=8.18\;\text{V}\).

Example 5 – Designing a Voltage Divider

Reduce a 15 V supply to 5 V using two resistors in series. Find the required ratio \(R_{2}:R_{1}\).

\[ 5 = 15\;\frac{R_{2}}{R_{1}+R_{2}}\;\Longrightarrow\;\frac{R_{2}}{R_{1}+R_{2}}=\frac{1}{3} \] \[ \Rightarrow\;R_{2}= \frac{1}{2}R_{1}\quad\text{or}\quad R_{1}=2R_{2}. \] Any pair satisfying this ratio (e.g., \(R_{1}=20\;\Omega,\;R_{2}=10\;\Omega\)) will give the desired 5 V output.

Resistivity of Common Materials

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

Summary Checklist

• Write and rearrange \(V = IR\) correctly.
• Distinguish \(\rho\) (Ω·m) from \(R\) (Ω) and use \(R = \rho L/A\) with proper unit conversion.
• Apply series (\(R_{\text{eq}}=\sum R\)) and parallel (\(1/R_{\text{eq}}=\sum 1/R\)) formulas.
• Include temperature effects: \(\rho(T)=\rho_{0}[1+\alpha(T-T_{0})]\).
• Model a real battery with \(V = \mathcal{E} - I r\).
• Use Kirchhoff’s laws for circuits that are not simple series/parallel.
• Calculate voltage division with \(V_{out}=V_{s}\dfrac{R_{2}}{R_{1}+R_{2}}\).
• Apply power formulas to find heating loss and check consistency.
• Always check units – convert mm² to m², cm to m, etc.

Practical Skills (Syllabus 10)

• Measure resistance with a multimeter or Wheatstone bridge; record uncertainties.
• Measure voltage and current simultaneously (voltmeter in parallel, ammeter in series) and propagate errors.
• Plot I–V graphs for a resistor; determine the slope (resistance) and comment on linearity.
• Investigate temperature effects on resistivity using a calibrated thermometer or a thermistor.
• Construct and test a voltage divider; verify the output voltage with a multimeter.

Practice Questions

1. A steel rod (\(\rho = 1.0\times10^{-7}\;\Omega\!\cdot\!\text{m}\)) is 0.5 m long and has a diameter of 2 mm. Find its resistance.
2. Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel. What is the total resistance?
3. A circuit contains a 9 V battery and two series resistors: \(R_{1}=1\;\Omega\) and \(R_{2}=3\;\Omega\). Determine the voltage drop across each resistor.
4. Given a copper wire with \(R = 0.10\;\Omega\) at \(20^{\circ}\text{C}\), estimate its resistance at \(80^{\circ}\text{C}\) using \(\alpha = 3.9\times10^{-3}\;\text{K}^{-1}\).
5. For a battery of emf \(\mathcal{E}=6\;\text{V}\) and internal resistance \(r=0.2\;\Omega\), a load of 4 Ω is connected. Calculate:
   • Terminal voltage.
   • Current supplied.
   • Power dissipated in the internal resistance.
6. Design a voltage divider that reduces a 15 V supply to 5 V. Choose standard resistor values and state the resulting output voltage (to two‑significant figures).