Know that the p.d. across an electrical conductor increases as its resistance increases for a constant current

Published by Patrick Mutisya · 14 days ago

IGCSE Physics 0625 – Action and Use of Circuit Components

4.3.3 Action and Use of Circuit Components

Learning Objective

Understand that, for a constant current, the potential difference (p.d.) across an electrical conductor increases as its resistance increases.

Key Concepts

  • Potential Difference (p.d.) – the energy per unit charge transferred between two points, measured in volts (V).
  • Current (I) – the rate of flow of charge, measured in amperes (A).
  • Resistance (R) – the opposition to the flow of current, measured in ohms (Ω).
  • Ohm’s Law – the fundamental relationship between p.d., current and resistance: \$V = I R\$

Why Does p.d. Increase with Resistance?

From Ohm’s law, if the current I is kept constant, the voltage \cdot across a component is directly proportional to its resistance R. Therefore, increasing R while I remains unchanged must increase V.

Mathematical Illustration

Consider a circuit with a constant current of \$I = 2\ \text{A}\$.

  1. When \$R = 5\ \Omega\$, the p.d. is \$V = I R = 2 \times 5 = 10\ \text{V}.\$
  2. If the resistance is doubled to \$R = 10\ \Omega\$, the p.d. becomes \$V = 2 \times 10 = 20\ \text{V}.\$
  3. The p.d. has increased in proportion to the increase in resistance.

Factors That Influence the Resistance of a Conductor

  • Material (resistivity \$\rho\$)
  • Length (\$L\$) – \$R \propto L\$
  • Cross‑sectional area (\$A\$) – \$R \propto \dfrac{1}{A}\$
  • Temperature – for most metals, \$R\$ increases with temperature.

Resistivity of Common Materials

MaterialResistivity \$\rho\$ (Ω·m)
Copper\$1.68 \times 10^{-8}\$
Aluminium\$2.82 \times 10^{-8}\$
Iron\$9.71 \times 10^{-8}\$
Constantan (alloy)\$4.9 \times 10^{-7}\$

Practical Example

A simple circuit consists of a 12 V battery, a resistor, and an ammeter. The ammeter reads \$I = 0.5\ \text{A}\$.

Using Ohm’s law, the resistance of the resistor is:

\$R = \frac{V}{I} = \frac{12}{0.5} = 24\ \Omega\$

If a second resistor of \$48\ \Omega\$ is added in series, the total resistance becomes \$72\ \Omega\$ and the p.d. across the new resistor is:

\$V_{\text{new}} = I \times 48 = 0.5 \times 48 = 24\ \text{V}\$

The p.d. across the added resistor is twice that across the original \$24\ \Omega\$ resistor, confirming the direct proportionality.

Summary

  • For a constant current, \$V\$ and \$R\$ are directly proportional.
  • Increasing the resistance of a conductor raises the p.d. required to maintain the same current.
  • Ohm’s law provides the quantitative link: \$V = I R\$.
  • Understanding material properties and dimensions helps predict how resistance (and thus p.d.) will change.

Practice Questions

  1. A wire of length \$2\ \text{m}\$ and cross‑section \$1.0 \times 10^{-6}\ \text{m}^2\$ is made of copper. Calculate its resistance. (Use \$\rho_{\text{Cu}} = 1.68 \times 10^{-8}\ \Omega\!\cdot\!m\$.)
  2. If a constant current of \$3\ \text{A}\$ flows through the wire in (1), what p.d. appears across it?
  3. Explain qualitatively what happens to the p.d. across a heating element if its temperature rises, assuming the current remains constant.

Suggested diagram: A simple series circuit showing a battery, an ammeter, a variable resistor, and a voltmeter across the resistor.