Mini Physics
Part of Mini EducationSwitch subject
Search

I/V Graph Of Filament Lamp

Key idea: Learn why a filament lamp is non-ohmic, how heating makes its I–V graph curve, and how to calculate resistance at a point on the graph (O Level).

  • SEC G3 Physics 2027
On this page

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

A filament lamp is non-ohmic: its resistance changes as current changes, so its I–V graph is a curve (not a straight line).

2. Key Ideas

  • The filament heats up when current flows, so its resistance increases as it gets hotter.
  • On an I vs V graph for a filament lamp:
    • the curve becomes less steep as V increases (because R increases)
    • R is not constant
    • the complete curve has the same shape in the negative region because reversing p.d. reverses current
  • Resistance at a point: R = V/I (use that point’s V and I).
What you need for this course

You should be able to sketch and interpret the complete I–V characteristic of a filament lamp. Show the symmetric curve through the origin unless the axes restrict the range.

3. Detailed Explanations

A. Why the graph is curved

  1. Increase V across the lamp.
  2. Current I increases, so the filament heats up.
  3. Hotter filament → higher resistance.
  4. Higher resistance means current does not increase proportionally with voltage, so the graph curves.

I–V graph for a filament lamp (non-ohmic)

A curved I–V graph that becomes less steep at higher voltages because the filament heats up and resistance increases.

Scroll across the graph to read all labels.

A curved I–V graph that becomes less steep at higher voltages because the filament heats up and resistance increases.A curved I–V graph that becomes less steep at higher voltages because the filament heats up and resistance increases.
The complete curve is symmetric about the origin. Greater |V| heats the filament, so resistance rises and the curve becomes less steep in both directions.
Open full-size graph
View figure data
Values for I–V graph for a filament lamp (non-ohmic)
Potential difference (V)Filament lamp
-10-1.25
-8-1.15
-6-1
-4-0.8
-2-0.5
-1-0.35
00
10.35
20.5
40.8
61
81.15
101.25

B. How to compare resistance at different points

Pick two points on the curve and calculate R = V/I at each point. The point with the larger V/I has the larger resistance.

4. Common Mistakes

  • Saying a filament lamp is ohmic (it is non-ohmic).
  • Explaining the curve without mentioning heating and increasing resistance.
  • Treating resistance as constant and using one R value for all points.

5. Exam Tips

  1. Sketch: a curve through the origin that becomes less steep as V increases.
  2. Explanation keywords: “current increases → filament heats → resistance increases”.
  3. If asked for resistance at a point: use R = V/I at that point.

6. Worked Examples

Modelled example 1

Resistance at a point

Core

Problem

At V = 2.0 V, a filament lamp carries I = 0.50 A. Find its resistance at that operating point.
Study the worked solution

  1. Use the point coordinates

    Method

    Divide the p.d. by the current at the same point.

    Reason

    A non-ohmic device still has an operating-point resistance V/I.

    Working

    R = 2.0/0.50 = 4.0 Ω

Guided practice 2

Comparing resistance at two points

About 5 min

Problem

At A, (V,I) = (2.0 V,0.50 A); at B, (6.0 V,1.0 A). Which point has the larger resistance?

Calculate both V/I ratios

Larger resistance

Hints

Hint 1: do not compare voltage alone
Calculate V/I at each point.
Hint 2: connect to temperature
The higher-current point corresponds to a hotter filament.
View solution step by step

  1. Calculate and compare

    Method

    Find R_A = 4.0 Ω and R_B = 6.0 Ω.

    Reason

    The hotter filament at B has greater lattice vibration and resistance.

    Working

    R_A = 2.0/0.50 = 4.0 Ω; R_B = 6.0/1.0 = 6.0 Ω

Common misconception 3

Explaining the shape

Find and correct the mistake

Learner response

A learner says the filament lamp curve becomes less steep because current falls as voltage rises. Diagnose the explanation.

Track the direction of every change

Current as voltage rises

View solution step by step

  1. Build the heating chain

    Method

    State that greater voltage produces greater current and heats the filament.

    Reason

    Higher temperature increases ion vibration and electron scattering.

    Working

    V↑ ⇒ I↑ ⇒ T↑ ⇒ R↑.

  2. Explain the decreasing gradient

    Method

    State that current continues to rise, but less per additional volt.

    Reason

    The increasing resistance reduces the I-against-V gradient.

    Working

    Curve becomes less steep; current does not reverse.

Examiner practice 4

Another resistance calculation

3 marks

Examination question

At a point on the curve, V = 4.0 V and I = 0.80 A. Find the resistance at that point and state why it is an operating-point value. [3 marks]

Calculate and qualify the result

View solution step by step

  1. Calculate resistance

    2 marks

    Method

    Use R = V/I at the stated coordinates.

    Reason

    Both values describe the same operating point.

    Working

    R = 4.0/0.80 = 5.0 Ω

  2. Qualify the value

    1 mark

    Method

    State that resistance changes elsewhere on the curve.

    Reason

    Filament temperature changes with current.

    Working

    5.0 Ω applies at this point.

Challenge 5

Is it ohmic?

Minimal support

Counterfactual transfer

At 2.0 V a lamp carries 0.50 A. Predict the current at 4.0 V if it were ohmic, then compare with the actual 0.80 A.

Build the constant-resistance prediction first

Hints

Hint 1: make the ohmic prediction
Doubling voltage would double current at constant resistance.
Hint 2: explain the shortfall
The operating filament is hotter at the higher voltage.
View solution step by step

  1. Predict the ohmic current

    Method

    Double 0.50 A to obtain 1.0 A.

    Reason

    An ohmic conductor at constant temperature has I ∝ V.

    Working

    2.0 → 4.0 V predicts 0.50 → 1.0 A.

  2. Interpret the actual current

    Method

    Use 0.80 A < 1.0 A to reject constant resistance.

    Reason

    Heating raises filament resistance, so current grows less than proportionally.

    Working

    Actual response is non-ohmic.

7. Mind Stretchers

Mind stretcher 1: Comparing with an ohmic conductorExtension

Two devices are tested. Device X gives a straight line through the origin on an I–V graph. Device Y gives a curved line that becomes less steep at higher voltages. Identify X and Y.

Show Answer

X is an ohmic conductor (metallic resistor at constant temperature). Y is a filament lamp (non-ohmic due to heating).

Mind stretcher 2: What happens if the filament stays cold?Extension

If the filament could be kept at constant temperature (no heating), what would happen to its I–V graph?

Show Answer

If temperature stays constant, resistance would be constant, so I ∝ V and the I–V graph would be a straight line through the origin (ohmic behaviour).

8. Practice and next step

Use the I–V Characteristics Lab to compare V/I near and far from the origin, then practise the heating explanation in Structured Current Electricity. Continue to the semiconductor diode.

Continue with the next resource in this course.

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