Physics 10.18 - 10.20 - Component characteristics

Physics 10.18 - 10.20 - Component characteristics

A fixed resistor, filament lamp and diode respond differently as the potential difference changes. Read their characteristic graphs, calculate resistance at a point, and explain how light and temperature change two useful sensors.

Equation support is provided for GCSE Combined Science Physics examinations in 2025–2027. Practise choosing, interpreting and rearranging the supplied relationships; these activities do not assume unaided equation memorisation.

Reading current-potential difference characteristics

A current-potential difference characteristic, often shortened to an I-V characteristic, shows how the current through one component changes as the potential difference across that component changes.

  • Potential difference, VV, is plotted on the horizontal axis.
  • Current, II, is plotted on the vertical axis.
  • Positive and negative values show opposite directions of potential difference and current. A negative current is not “less current”; it is current in the opposite direction.

Resistance at a point

potential difference=current×resistance\text{potential difference}=\text{current}\times\text{resistance}

V=I×RV = I \times R, so R=VIR = \frac{V}{I}

VV is in volts (V), II is in amperes (A), and RR is in ohms (Ω).

The resistance at a non-zero point on an I-V graph comes from that point's ratio VI\frac{V}{I}. Do not calculate resistance at the origin: 00\frac{0}{0} is undefined. Also, do not call resistance “the gradient” of any curved I-V graph. With these axes, the gradient of a straight fixed-resistor line is IV\frac{I}{V}, which is 1R\frac{1}{R}.

Worked example

A filament lamp has these two readings:

Potential difference, VVCurrent, II
1.0 V1.0\,\mathrm{V}0.25 A0.25\,\mathrm{A}
4.0 V4.0\,\mathrm{V}0.50 A0.50\,\mathrm{A}

Solution

At 1.0 V1.0\,\mathrm{V}:

R=VI=1.0 V/0.25 A=4.0 ΩR = \frac{V}{I} = 1.0\,\mathrm{V} / 0.25\,\mathrm{A} = 4.0\,\Omega

At 4.0 V4.0\,\mathrm{V}:

R=VI=4.0 V/0.50 A=8.0 ΩR = \frac{V}{I} = 4.0\,\mathrm{V} / 0.50\,\mathrm{A} = 8.0\,\Omega

The potential difference is four times larger while the current is only twice as large. The calculated resistance has increased from 4.0 Ω4.0\,\Omega to 8.0 Ω8.0\,\Omega, which is the expected behaviour of a filament lamp as it becomes hotter.

Three component characteristics

[DIAGRAM: asset_name: Physics 10.18-10.21 - Component characteristics and circuit design - diagram 01; asset_slug: physics_10_18_10_21_component_characteristics_and_circuit_design__diagram_01; recommended_method: matplotlib; description: Three aligned monochrome illustrative I-V plots with potential difference V on each horizontal axis and current I on each vertical axis: a straight fixed-resistor line through the origin, a symmetric filament-lamp curve that becomes less steep as the magnitude of potential difference increases, and a diode curve with approximately zero reverse current and a steep forward rise after a threshold region. Direct panel titles, no numerical scale, no reverse-breakdown branch, and a note that curves are illustrative and not to scale.]
Diagram

Fixed resistor

For a fixed resistor kept at constant temperature, current is directly proportional to potential difference. Doubling the p.d. doubles the current, so VI\frac{V}{I} stays constant. Its I-V characteristic is a straight line through the origin. Reversing the p.d. reverses the current, giving the same pattern in the opposite quadrant.

For two straight fixed-resistor I-V lines, the steeper line has the smaller resistance because its gradient is IV=1R\frac{I}{V} = \frac{1}{R}. This shortcut is not valid for a curved component characteristic.

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