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I–V graph of a metallic conductor

Key idea: At constant temperature, current in a metallic conductor is proportional to p.d. Its I-against-V graph is a straight line through the origin with gradient 1/R.

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Pause after each idea, then explain it in your own words before moving on.

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Enlarged diagram

An I–V graph shows how the current I through a component changes as the p.d. V across it changes. For a metal wire or a fixed resistor at constant temperature, it is the simplest possible graph: a straight line through the origin.

A straight line through the origin

At constant temperature, the ions in a metal vibrate by the same amount whatever the current, so the resistance stays the same. With R constant, I = V/R means the current is directly proportional to the p.d.: double the p.d. and the current doubles. That is Ohm’s law, and a component that obeys it is called ohmic.

Reversing the p.d. reverses the current by the same amount, so the line continues through the origin into the negative quarter of the graph. Sketch both halves unless the axes show only one.

I–V graph for a metallic conductor (ohmic)

A straight-line I–V graph through the origin for an ohmic conductor at constant temperature.

If the figure is wider than its frame, scroll across to read all labels.

A straight-line I–V graph through the origin for an ohmic conductor at constant temperature.A straight-line I–V graph through the origin for an ohmic conductor at constant temperature.
The complete straight line passes through quadrants I and III: reversing p.d. reverses current while V/I remains constant at non-zero points.
Open full-size graphOpen full-size graph
What to noticeConstructed example coordinates, not reported measurements. Temperature is held constant. The origin shows zero current at zero p.d.; use another point to calculate resistance.
View figure data

This table is included in print while it is open.

Figure data: I–V graph for a metallic conductor (ohmic)
Potential difference (V)Ohmic conductor
-5-1.0
-4-0.8
-3-0.6
-2-0.4
-1-0.2
00.0
10.2
20.4
30.6
40.8
51.0

The temperature condition matters. A wire that carries a large current heats up, its resistance rises, and the graph starts to bend, as a filament lamp’s does. In an experiment, keep the current small or switch off between readings so the wire stays cool.

Guided practice 1

Predict the current

About 4 min

Problem

A metal resistor has resistance 5.0 Ω and is kept at constant temperature. Find the current at 2.0 V.

Use the constant resistance

Unit: A

Hints

Hint 1

use the graph condition

At constant temperature the resistance stays 5.0 Ω.

Hint 2

calculate current

Divide 2.0 V by 5.0 Ω.

Show full solution

  1. Predict the current

    Method

    Divide the p.d. by the constant resistance.

    Reason

    The straight-line graph obeys I = V/R.

    Working

    Answer

    I = (2.0 V)/(5.0 Ω) = 0.40 A

Find the resistance from the graph

From a point. Choose a point on the line well away from the origin, read its p.d. and current, and calculate R = V/I. The origin itself is no use: there V/I is 0/0.

Worked example 1

Resistance from a point

Problem

A point on the straight line of an I–V graph is (V,I) = (3.0 V, 0.60 A). Find the resistance.

Show full solution

  1. Read the coordinates in axis order

    Method

    Take V = 3.0 V and I = 0.60 A.

    Reason

    Resistance is p.d. divided by current, so check which value is which.

    Working

    Answer

    The point is (V,I), not (I,V).

  2. Calculate the resistance

    Method

    Divide the p.d. by the current.

    Reason

    For an ohmic conductor every point on the line gives the same ratio.

    Working

    Answer

    R = (3.0 V)/(0.60 A) = 5.0 Ω

From the gradient. On a graph of I against V, the gradient is Δ I/Δ V. For a straight line through the origin this equals I/V, which is 1/R. So the resistance is 1/gradient, not the gradient itself. A steeper line means more current for each volt, so a smaller resistance.

Try it yourself 1

Resistance from the gradient

Minimal support

Problem

A straight I-against-V line through the origin passes through (1.6 V,0.20 A) and (6.4 V,0.80 A). Use the gradient between these points to find the resistance.

Find the gradient, then the resistance

Show the steps Optional working
Unit: A V⁻¹
Unit: Ω

Hints

Hint 1

calculate rise over run

Use Δ I/Δ V for these axes.

Hint 2

then invert

The gradient is 1/R.

Show solution step by step

  1. Find the gradient

    Method

    Divide the change in current by the change in p.d.

    Reason

    Two points far apart give a more reliable gradient than one point close to the origin.

    Working

    Answer

    (Δ I)/(Δ V) = (0.80 A-0.20 A)/(6.4 V-1.6 V) = 0.125 A/V

  2. Invert the gradient

    Method

    Take the reciprocal.

    Reason

    With I plotted against V, the gradient is 1/R.

    Working

    Answer

    R = 1/(0.125 A/V) = 8.0 Ω

Check your understanding 1 Swap the axes

If you plot V against I instead, what does the gradient of the straight line represent?

Gradient when V is vertical and I is horizontal
Show answer

The resistance. The gradient is now Δ V/Δ I = V/I = R.

Test whether a conductor is ohmic

Calculate V/I from each measured pair. Matching values are consistent with constant resistance at those points. To check the behaviour between two readings, take separated readings across the proposed range and check the shape as well as the ratios. Keep temperature fixed.

Spot the mistake 1

Do two readings prove it?

About 5 min

Learner response

At constant temperature, a student records (2.0 V,0.20 A) and (4.0 V,0.40 A), then says these two points prove the conductor is ohmic at every p.d. What is wrong with the claim?

Test the measured ratios first

What do these two readings tell you?
Unit: Ω
Show solution step by step

  1. Check the two readings

    Method

    Both points give the same resistance.

    Reason

    Doubling the p.d. doubled the current, as Ohm’s law predicts.

    Working

    Answer

    (2.0 V)/(0.20 A) = (4.0 V)/(0.40 A) = 10 Ω

  2. Limit the conclusion

    Method

    The readings are consistent with Ohm’s law at the two measured points. Readings between and beyond them still need testing.

    Reason

    Two points always lie on some straight line. To show the graph is a straight line through the origin, take several readings across the whole range, including negative p.d.

    Working

    Answer

    Consistent with the same resistance at these two measured points

Measure the characteristic

Try it in the lab below: vary the p.d. across the fixed resistor, record the current at several values, and check whether V/I stays the same.

Check your understanding 2 Predict before testing

The lab’s fixed resistor is 20 Ω. For p.d.s of +2.0 V and −2.0 V across this resistor, predict both currents in amperes. Also convert the positive current to milliamperes for the recorded table. Write your predictions before using the controls.

Show the steps Optional working
Unit: A
Unit: mA
Unit: A
Show answer

I = 2.0/20 = 0.10 A = 100 mA. At −2.0 V, current is −0.10 A. The current and voltage reverse together, so V/I stays positive.

Testing a fixed 20 ohm resistor with a 2.0 volt supply. The voltmeter reads 1.99 volts and the ammeter 99.7 milliamperes.

V
p.d. across it, V
1.99 V
Current, I
0.0997 A
Resistance, V / I
20.0 Ω

Try this

0 of 4 done

  1. Record at least five readings for the fixed resistor, from 0 V to at least 5 V. (not done yet)

  2. Do the same for the filament lamp and compare the two graphs. (not done yet)

  3. Test the diode in both directions, including readings above 0.6 V forwards. (not done yet)

  4. Record the thermistor at the same supply p.d. at 20 °C and at 80 °C. (not done yet)

Your readings

#V / VI / mARemove
No readings yet. Set up a measurement, then record it.

Check your understanding 3 Compare the recorded characteristic

Choose Fixed resistor (20 Ω) in Component under test. Use Supply p.d. to record separated readings, including 2 V and 4 V, then use Reverse the supply. Read p.d. across it, V and Current, I, and use Record V and I. Compare the table’s milliamperes with the amperes readout. Sketch both halves of I against V and explain its slope and resistance. You may use the written model below if interaction is unavailable.

Show answer

The readings are approximately (+2 V, +0.10 A), (+4 V, +0.20 A) and (−2 V, −0.10 A); the table currents are about +100, +200 and −100 mA. Use measured p.d. across the resistor when calculating V/I. At each non-zero point it is about 20 Ω. Both signs lie on one straight line through the origin, with gradient 1/R = 0.050 A/V. Convert a table current from mA to A before dividing. Label voltage horizontally and current vertically, with units and both signs; at the origin 0/0 is undefined.

Common mistakes

  • Leaving out “at constant temperature” when stating Ohm’s law.
  • Calling the gradient of an I-against-V graph the resistance; it is 1/R.
  • Using the origin, or a point read off the line carelessly, to find V/I.
  • Sketching only the positive half when the axes include negative values.

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