This topic accounts for approximately 12% of your exam marks.
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Calculating resistance, current and voltage in series and parallel circuits tested every series.
What an I-V graph shows
An I-V graph plots the current through a component (y-axis) against the voltage across it (x-axis), with the voltage varied from negative through zero to positive
The shape of the line tells you how the component's resistance changes with the voltage applied:
A straight line through the origin means the current is directly proportional to the voltage. The resistance is constant (this is ohmic behaviour)
A curved line through the origin means the current is not directly proportional to the voltage. The resistance is variable (this is non-ohmic behaviour)
The resistance at any point on an I-V graph is given by R = V / I, not by the gradient. (For an ohmic conductor R = 1/gradient because the line is straight, but in general the V/I ratio at each point is what matters.)
I-V graph for a fixed resistor (or a metal wire at constant temperature)
A straight line through the origin, because current and voltage are directly proportional, in both polarities
The slope is the same everywhere, so the resistance is the same everywhere; fixed resistors and wires at a steady temperature are ohmic conductors
I-V graph for a filament lamp
The curve passes through the origin, but it flattens off as the voltage and current grow; the same shape appears in reverse on the negative side
A flatter slope corresponds to a higher resistance (because R = V / I and V grows faster than I)
The physics: a larger current heats the filament; the metal ions then vibrate more vigorously, scattering the drifting electrons more often, so the resistance climbs
A filament lamp is therefore a non-ohmic component, because its resistance is not constant across its working range
I-V graph for a semiconductor diode
The diode is a one-way conductor: it lets current pass in the direction of its arrowhead symbol only, which is called forward bias
In forward bias, almost no current flows until the voltage exceeds about 0.6–0.7 V; past that threshold, the current climbs very sharply with very little extra voltage. The resistance is effectively infinite below the threshold and very small above it
In reverse bias, no current flows at all (over the normal operating range); the diode's resistance is enormous
The I-V graph therefore sits flat on the V axis for negative V, hugs the V axis from 0 to about 0.6 V, then rises almost vertically
Exam tip
Explain why the I-V graph of a filament lamp is a curve, not a straight line
What comes up: the exam asks you to explain why the current-voltage graph for a filament lamp curves and flattens at higher voltages, rather than being a straight line through the origin.
Write (two marks): (1) As the current increases, it heats the filament. (2) The resistance of the filament increases with temperature, so the current rises more slowly than the voltage — producing a curve that flattens as voltage grows.
Watch out: do not simply say "the filament heats up" and stop there. The mark scheme requires you to link the heating to a change in resistance, and to show how that changing resistance produces the non-linear shape. One mark for the heating, one mark for stating the resistance changes with temperature (or that this causes the reduced rate of current increase).
Investigating an I-V curve in the lab
Apparatus:
the component under test (resistor, filament lamp or diode)
an ammeter wired in series with the component to read the current through it
a voltmeter wired in parallel across the component to read the voltage across it
a variable resistor in series with the component to vary the current
a cell or low-voltage supply
Method:
Build the circuit and set the variable resistor to its highest resistance (so the initial current is small)
Record the voltmeter and ammeter readings
Reduce the resistance of the variable resistor by a small step; record V and I again
Repeat until a wide range of voltages has been swept (do not exceed the rated voltage of the component, or it may burn out)
To capture the negative side of the curve, reverse the cell's connections and repeat
Plot I against V with the recorded data points and join them with a smooth curve