state Ohm’s law

Resistance, Resistivity & Ohm’s Law

Learning objectives (AO1, AO2, AO3)

• Define resistance, resistivity and conductivity.
• State Ohm’s law for an ohmic conductor and distinguish it from non‑ohmic behaviour.
• Derive and use the relationship R = ρ L/A, including algebraic rearrangement.
• Apply the temperature‑coefficient formula for metals, semiconductors and insulators.
• Calculate power dissipation in a resistor.
• Combine resistors in series and parallel.
• Plan and evaluate a simple experiment to determine ρ and α (AO3).

Key definitions

Ohm’s law

For an ohmic conductor:

or equivalently

where V is the potential difference (V), I the current (A) and R the resistance (Ω).

Non‑ohmic example

A filament lamp’s filament heats as current increases, so its resistance rises. The I‑V curve is therefore convex upwards. This illustrates the syllabus point that “the resistance of a filament lamp increases with temperature”.

Power dissipation in a resistor

Power (P) is measured in watts (W). Use the form most convenient for the given data.

Resistance of a uniform cylindrical conductor

• \( \rho \) – resistivity of the material (Ω·m)
• \( L \) – length of the conductor (m)
• \( A \) – cross‑sectional area (m²)

Re‑arranging the formula

Depending on what is unknown you may need:

• \( \displaystyle \rho = R \dfrac{A}{L} \)
• \( \displaystyle L = R \dfrac{A}{\rho} \)
• \( \displaystyle A = \rho \dfrac{L}{R} \)

Unit‑check tip

Confirm that the units reduce to ohms:

Temperature dependence of resistivity

For most metals (positive temperature coefficient):

• \( \alpha \) – temperature coefficient of resistivity (K⁻¹). Typical for copper: \(3.9\times10^{-3}\,\text{K}^{-1}\).
• \( \rho{0} \) – resistivity at reference temperature \(T{0}\) (usually 20 °C).

For semiconductors the opposite trend occurs (ρ decreases as T increases) because more charge carriers are thermally generated. Insulators have very large, often non‑linear temperature coefficients; the simple linear formula is only an approximation over a limited range.

Typical temperature‑coefficient values

Typical resistivity values (20 °C)

Series and parallel combinations (later syllabus use)

• Series: \( R{\text{eq}} = R{1}+R{2}+…+R{n} \)
• Parallel: \( \displaystyle \frac{1}{R{\text{eq}}}= \frac{1}{R{1}}+\frac{1}{R{2}}+…+\frac{1}{R{n}} \)

These formulas are needed when analysing D.C. circuits and when designing heating elements or wiring layouts.

Practical: Determining resistivity and the temperature coefficient

Apparatus

• Known‑length copper wire (e.g., 1.00 m) with uniform cross‑section.
• Four‑wire (Kelvin) measurement set‑up: two leads for current, two leads for voltage.
• Digital voltmeter (V), ammeter (A), and a calibrated resistor for current source.
• Thermostatically controlled water bath (±0.1 °C) and a thermometer.
• Ruler or caliper for measuring diameter → area.

Method (AO3)

1. Measure the wire’s diameter at several points; calculate average area \(A = \pi d^{2}/4\).
2. Set the bath to a series of temperatures (e.g., 20 °C, 30 °C, 40 °C, 50 °C). Allow the wire to equilibrate.
3. For each temperature record the current (I) and voltage (V) across the measured length using the four‑wire technique.
4. Calculate resistance at each temperature: \(R = V/I\).
5. Plot \(R\) against \(T\); the slope gives \(R{0}\alpha\). From the intercept obtain \(R{0}\) (resistance at \(T_{0}=20 °C\)).
6. Compute resistivity using \( \rho = R A/L \) and then α from the linear fit.

Sources of error & safety

• Systematic: contact resistance (minimised by four‑wire method), thermometer calibration, non‑uniform cross‑section.
• Random: fluctuations in bath temperature, reading uncertainties.
• Safety: avoid overheating the wire; use insulated leads; never touch the wire when the circuit is live.

Worked example (with unit‑check & significant figures)

Problem: A copper wire 2.00 m long has a cross‑sectional area of 0.500 mm². Find its resistance at 20 °C.

Given

• \( \rho_{\text{Cu}} = 1.68 \times 10^{-8}\ \text{Ω·m} \)
• \( L = 2.00\ \text{m} \)
• \( A = 0.500\ \text{mm}^2 = 0.500 \times 10^{-6}\ \text{m}^2 \)

Solution

Unit‑check: (Ω·m · m) / m² → Ω.

Result to two significant figures (as the data are given to three SF, the final answer is limited by the area): 0.067 Ω.

Suggested diagram

Summary checklist

• Ohm’s law for an ohmic conductor: V = IR.
• Resistance of a uniform conductor: R = ρ L/A (and the rearranged forms).
• Resistivity ρ is intrinsic; resistance varies with length, area and temperature.
• Temperature coefficient: \( \rhoT = \rho0[1+\alpha(T-T_0)] \) (positive for metals, negative for semiconductors).
• Power in a resistor: \( P = VI = I^{2}R = V^{2}/R \).
• Series: \(R{\text{eq}} = \sum R\); Parallel: \(1/R{\text{eq}} = \sum 1/R\).
• Non‑ohmic devices (e.g., filament lamp) show a changing resistance with temperature.
• Practical skills: four‑wire measurement, temperature‑controlled bath, error analysis (systematic vs random).
• Low‑ρ materials (copper, silver) are used for wiring; high‑ρ alloys (nichrome) are chosen for heating elements.