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I/V Graph Of Metallic Conductor

Key idea: Learn the I–V characteristic of a metallic (ohmic) conductor at constant temperature, how to sketch it, and how to find resistance from the graph.

  • SEC G3 Physics 2027
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Learning objectives

  • State current as rate of charge flow measured in amperes
  • Distinguish conventional current from electron flow
  • Apply charge equals current multiplied by time
  • Define source e.m.f. as work done per unit charge around a circuit
  • Calculate total e.m.f. for sources in series
  • Define component potential difference as work done per unit charge
  • State resistance as potential difference divided by current
  • Apply resistance equals potential difference divided by current
  • Apply wire-resistance proportionalities for length and cross-sectional area
  • Describe the effect of temperature on metallic resistance
  • Sketch and interpret required current–voltage characteristics

1. Definition

For a metallic conductor at constant temperature, current is directly proportional to voltage (Ohm’s law), so its I–V graph is a straight line through the origin.

2. Key Ideas

  • Ohm’s law condition: constant temperature.
  • For an ohmic conductor: I ∝ V and R is constant.
  • The complete characteristic continues through the origin into the negative region because reversing p.d. reverses current.
  • On an I vs V graph:
    • straight line through origin ⇒ constant R
    • gradient = I/V = 1/R
  • To find resistance from a point on the graph: R = V/I.
What you need for this course

You should be able to sketch and interpret the I–V characteristic of a metallic conductor at constant temperature. Draw the complete straight line through quadrants I and III unless the axes restrict the range.

3. Detailed Explanations

A. What the straight line means

If you double V, the current I doubles. This shows I ∝ V.

B. Why “constant temperature” matters

If the wire heats up, its resistance increases and the graph may stop being a straight line. In experiments, use small currents or take readings quickly to reduce heating.

C. Reading resistance from the I–V graph

Pick any convenient point on the line (not too close to the origin), then:

R = V/I

I–V graph for a metallic conductor (ohmic)

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

Scroll across the graph 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.
Open full-size graph
View figure data
Values for I–V graph for a metallic conductor (ohmic)
Potential difference (V)Ohmic conductor
-5-1
-4-0.8
-3-0.6
-2-0.4
-1-0.2
00
10.2
20.4
30.6
40.8
51
Link

Using R = V/I is covered here: Resistance (R = V/I).

4. Common Mistakes

  • Forgetting “constant temperature” when stating Ohm’s law.
  • Saying “the gradient is the resistance” for an I vs V graph (here gradient = 1/R).
  • Using points that are not on the straight line (read the graph carefully).

5. Exam Tips

  1. For a metallic conductor (ohmic): draw a straight line through the origin.
  2. If asked for resistance, either:
    • use a point: R = V/I, or
    • use slope: R = 1/gradient (for I vs V).
  3. Always include units: V (V), I (A), R (Ω).

6. Worked Examples

Modelled example 1

Finding resistance from a point

Core

Problem

On an I–V graph, a point on the straight line is (V,I) = (3.0 V,0.60 A). Find the resistance.
Study the worked solution

  1. Read the coordinates in axis order

    Method

    Take V = 3.0 V and I = 0.60 A.

    Reason

    The resistance definition requires voltage divided by current.

    Working

    Point: (V,I), not (I,V).

  2. Calculate resistance

    Method

    Divide voltage by current.

    Reason

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

    Working

    R = 3.0/0.60 = 5.0 Ω

Guided practice 2

Predicting current (ohmic)

About 4 min

Problem

A metallic 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
Constant temperature keeps the ohmic resistance constant.
Hint 2: calculate current
Divide 2.0 V by 5.0 Ω.
View solution step by step

  1. Predict the current

    Method

    Divide voltage by the constant resistance.

    Reason

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

    Working

    I = 2.0/5.0 = 0.40 A

Common misconception 3

Checking if data is ohmic

Find and correct the mistake

Learner response

At constant temperature, a student records (2.0 V,0.20 A) and (4.0 V,0.40 A), then says two points prove the conductor is ohmic at every voltage. Diagnose the claim.

Test the measured ratios first

Unit: ohm

View solution step by step

  1. Evaluate the evidence

    Method

    Show that doubling V doubles I and both points give R = 10 Ω.

    Reason

    The readings are consistent with direct proportionality over the measured range.

    Working

    2.0/0.20 = 4.0/0.40 = 10 Ω

  2. Limit the conclusion

    Method

    Say the data support, but do not prove, an ohmic relationship across every voltage.

    Reason

    More points including the origin and controlled temperature are needed to establish the full straight-line characteristic.

    Working

    Consistent with ohmic behaviour in the tested range.

Examiner practice 4

Finding resistance from gradient

3 marks

Examination question

An I-against-V graph is a straight line through the origin with gradient 0.25 A V⁻¹. Find the resistance. [3 marks]

Interpret the axes before using the gradient

View solution step by step

  1. Relate gradient to resistance

    2 marks

    Method

    Use gradient = Δ I/Δ V = 1/R.

    Reason

    Current is on the vertical axis and voltage on the horizontal axis.

    Working

    R = 1/gradient

  2. Calculate resistance

    1 mark

    Method

    Take the reciprocal of 0.25 A V⁻¹.

    Reason

    The reciprocal unit is V A⁻¹ = Ω.

    Working

    R = 1/0.25 = 4.0 Ω

Challenge 5

Using two points to find resistance

Minimal support

Two-point graph transfer

A straight I-against-V line passes through (1.0 V,0.20 A) and (4.0 V,0.80 A). Use both points to find the resistance.

Separate gradient calculation from resistance conversion

Hints

Hint 1: calculate rise over run
Use Δ I/Δ V for these axes.
Hint 2: then invert
The graph gradient is conductance, 1/R.
View solution step by step

  1. Find the line gradient

    Method

    Divide the change in current by the change in voltage.

    Reason

    Using two separated points reduces reliance on a single coordinate.

    Working

    (Δ I)/(Δ V) = (0.80-0.20)/(4.0-1.0) = 0.20 A V⁻¹

  2. Convert gradient to resistance

    Method

    Take the reciprocal.

    Reason

    For I plotted against V, gradient is 1/R.

    Working

    R = 1/0.20 = 5.0 Ω

7. Mind Stretchers

Mind stretcher 1: Gradient vs resistanceExtension

An I–V graph is a straight line through the origin. Line A is steeper than Line B. Which conductor has larger resistance?

Show Answer

Line A has a larger gradient (I/V), so it has smaller resistance because gradient = 1/R. Therefore Line B has the larger resistance.

Mind stretcher 2: Changing axesExtension

If you plot a V vs I graph instead, what does the gradient represent?

Show Answer

For a V vs I graph, gradient = V/I = R (so the gradient is the resistance).

8. Practice and next step

Sketch the complete line without copying, label both axes and state the constant-temperature condition. Check your reasoning in the I–V Characteristics Lab, then compare it with the filament lamp.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027