The current through a conductor is proportional to the pd across it, provided physical conditions such as temperature are constant.

What shape is the I–V graph for an ohmic conductor?

A straight line through the origin.

What is the gradient of an I–V graph for an ohmic conductor?

1 / R, if I is on the vertical axis.

Describe the I–V characteristic of a filament lamp.

A curve through the origin, symmetrical about it, with the current rising less than proportionally as the pd increases.

Why does the resistance of a filament lamp increase with pd?

The filament heats up, so the lattice ions vibrate more and collide more often with the electrons.

Describe the forward characteristic of a diode.

Almost no current below a threshold of about 0.6 V, then the current increases rapidly.

Describe the reverse characteristic of a diode.

Almost no current, so the resistance is very high.

Is Ohm's law the same as V = I R?

No. V = I R defines resistance for any component; Ohm's law says R is constant, so I is proportional to V.

How do you find the resistance at a point on a curved I–V graph?

Read V and I at that point and calculate R = V / I.

Current-voltage characteristicsAQA A-Level Physics: Flashcards

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State Ohm's law.

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State Ohm's law.: The current through a conductor is proportional to the pd across it, provided physical conditions such as temperature are constant.
What shape is the I–V graph for an ohmic conductor?: A straight line through the origin.
What is the gradient of an I–V graph for an ohmic conductor?: 1 / R, if I is on the vertical axis.
What is the gradient of a V–I graph (V vertical) for an ohmic conductor?: The resistance R.
Describe the I–V characteristic of a filament lamp.: A curve through the origin, symmetrical about it, with the current rising less than proportionally as the pd increases.
Why does the resistance of a filament lamp increase with pd?: The filament heats up, so the lattice ions vibrate more and collide more often with the electrons.
Describe the forward characteristic of a diode.: Almost no current below a threshold of about 0.6 V, then the current increases rapidly.
Describe the reverse characteristic of a diode.: Almost no current, so the resistance is very high.
Is Ohm's law the same as V = I R?: No. V = I R defines resistance for any component; Ohm's law says R is constant, so I is proportional to V.
How do you find the resistance at a point on a curved I–V graph?: Read V and I at that point and calculate R = V / I.
What is the resistance of an ideal ammeter?: Zero.
What is the resistance of an ideal voltmeter?: Infinite.

Exam questions on Current-voltage characteristics

A student investigates a fixed resistor kept at a constant temperature, using an ideal ammeter and an ideal voltmeter. She records a pd of 2.0 V with a current of 4.0 mA, a pd of 4.0 V with a current of 8.0 mA, and a pd of 6.0 V with a current of 12.0 mA. She plots a graph with pd on the vertical axis and current on the horizontal axis.
Calculate the resistance of the resistor at 2.0 V and at 6.0 V, and state what the results show about the resistor.2 marks
Two components, X and Y, are connected in turn to a variable power supply, with an ideal ammeter and an ideal voltmeter. For each component a graph is plotted of pd (vertical axis) against current (horizontal axis). The graph for X is a straight line through the origin with a gradient of 40 Ω. The graph for Y is a curve through the origin whose gradient increases as the current increases.
X is a metal wire. State the condition needed for X to obey Ohm's law, and explain what would happen to the graph for X if the current became large enough to make the wire very hot.2 marks
A technician tests a semiconductor diode with an ideal ammeter and an ideal voltmeter. In the forward direction the current is negligible at 0.50 V, 2.0 mA at 0.70 V and 20 mA at 0.75 V. In the reverse direction, with 5.0 V across the diode, the current is 0.10 μA.
Describe how the current in the diode varies with pd in the forward and reverse directions.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).