Resistance and Ohm's law, the equation linking voltage, current and resistance, and the current-voltage (I-V) characteristics of an ohmic resistor, a filament lamp and a silicon diode.

What this topic is asking

WJEC Eduqas wants you to understand resistance, use Ohm's law ($V = IR$) to link voltage, current and resistance, and recognise and explain the current-voltage (I-V) characteristics of the three key components: an ohmic resistor, a filament lamp and a silicon diode. Reading and sketching these graphs is a frequently examined skill.

Resistance and Ohm's law

Rearranging gives $R = \dfrac{V}{I}$ (to find a resistance) and $I = \dfrac{V}{R}$ (to find a current). A high resistance lets only a small current flow for a given voltage. Resistance arises because the moving electrons collide with the fixed ions in the material, transferring energy and heating it. The resistance of a wire depends on its length, its cross-sectional area, the material and its temperature.

The ohmic resistor

The straight line through the origin is the signature of an ohmic component. The steeper the line, the smaller the resistance (a larger current for the same voltage). Because the resistance is constant, you can read any pair of values from the line and they always give the same $\dfrac{V}{I}$. Most fixed resistors are ohmic provided they do not get hot enough to change their resistance.

The filament lamp

The curve bends over because the resistance is not constant. At low voltages the filament is cool and has a low resistance, so the line is steep. As more current flows, the metal filament heats up, the ions vibrate more strongly, and the resistance rises, so the graph flattens. The shape is symmetrical: reversing the voltage gives the same curve reflected through the origin, because the lamp behaves the same way whichever direction the current flows.

The silicon diode

The diode characteristic is very different from the symmetrical resistor and lamp curves. For forward voltages below about $0.7\,\text{V}$ the diode is effectively off; above it, the current increases sharply for a tiny rise in voltage, so the diode is treated as having a roughly constant $0.7\,\text{V}$ across it when conducting. In reverse, it acts like an open switch. This one-way behaviour is the basis of rectification and of using a diode to protect a circuit.

Try this

Q1. A $330\,\Omega$ resistor has a voltage of $6.6\,\text{V}$ across it. Calculate the current. [2 marks]

• Cue. $I = \dfrac{V}{R} = \dfrac{6.6}{330} = 0.020\,\text{A} = 20\,\text{mA}$.

Q2. State the approximate forward voltage at which a silicon diode begins to conduct. [1 mark]

• Cue. About $0.7\,\text{V}$.