PhysicsICSE Class 9 · Current Electricity

Ohm's Law, Resistance and V–I Graphs: V = IR with Circuit Components (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Ohm's law: at constant temperature the current through a conductor is directly proportional to the potential difference across it, V = IR, where R is the resistance (ohm, Ω = V A⁻¹). The V–I graph of an ohmic conductor is a straight line through the origin whose slope is R. A filament bulb is non-ohmic because its resistance rises with temperature.

On a V–I graph an ohmic conductor gives a straight line whose slope is R (steeper means larger R), while a bulb gives a curve as its resistance rises.

Illustration drawn for QuestionBuddy.

Understand Ohm's Law, Resistance and V–I Graphs

Resistance. The opposition of a conductor to the flow of current. SI unit ohm (Ω): $1 Ω = 1 V A^{- 1}$ .

Ohm’s law. The current through a conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions stay the same:

$V \propto I \Rightarrow \frac{V}{I} = R (constant) \Rightarrow V = I R .$

From $V = I R$ : $I = \frac{V}{R}$ and $R = \frac{V}{I}$ .

V–I graph. Plot $V$ on the y-axis and $I$ on the x-axis.

For an ohmic conductor: a straight line through the origin; slope $= \frac{V}{I} = R$ .
A steeper line means a larger resistance. If $I$ is on the y-axis the line with the smaller slope has the larger resistance.
For a non-ohmic conductor (a filament bulb, a diode): the line is curved because $\frac{V}{I}$ is not constant. In a bulb the filament heats up and its resistance increases.

Circuit components. A circuit has a source (cell, battery or accumulator), a key (switch), connecting wires and a load. A rheostat is a variable resistance that changes the resistance (and so the current) in the circuit; a resistance box supplies chosen values. Symbols: cell, battery, open and closed key, resistor, rheostat, bulb, ammeter (A in a circle), voltmeter (V in a circle), galvanometer (G).

Verifying Ohm’s law. Connect the resistance wire, ammeter (in series), voltmeter (across the wire), rheostat, key and battery. For several rheostat settings record $V$ and $I$ ; the ratio $\frac{V}{I}$ stays the same and the V–I graph is a straight line through the origin. Keep the current small and the key open between readings so that the wire does not heat.

Total resistance of a circuit. For a single circuit with a load and a rheostat in series, the rheostat’s resistance adds to the load’s resistance, and $I = \frac{V}{R _{t o t a l}}$ . Changing the rheostat changes $I$ but not the load’s own resistance.

Cautions. Ohm’s law is not a definition of resistance: $R = \frac{V}{I}$ is always valid at a given instant, but only an ohmic conductor keeps $R$ constant as $V$ changes.

Ohm's Law, Resistance and V–I Graphs: formulas and key facts

Ohm's law: $V = I R$; At constant temperature.
Resistance unit: $1 Ω = 1 V A^{- 1}$; Ohm.
Slope of V–I graph: $slope = \frac{Δ V}{Δ I} = R$; V on y-axis.

How to solve Ohm's Law, Resistance and V–I Graphs questions

Identify the two known quantities among V, I and R.

1Convert mA to A and kΩ to Ω.
2Rearrange $V = I R$ for the unknown.
3For tables compute $V / I$ for each row; a constant value shows an ohmic conductor.
4For graphs, read the slope using two points far apart.
5For a rheostat in series use $R_{t o t a l} = R_{l o a d} + R_{r h eos t a t}$ .

Ohm's Law, Resistance and V–I Graphs in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A dimmer switch

Suppose a 12 V battery drives a lamp of resistance 6 Ω and a rheostat is in series, used to dim it.

The current is 2 A with the rheostat set to zero. Find the rheostat setting for a current of 0.8 A.

Try it first, then show the solutionSolution

$R_{t o t a l} = \frac{12}{0.8} = 15$ Ω.
Rheostat $= 15 - 6 = 9$ Ω.

Answer: 9 Ω.

Memory hook: More resistance, less current.

Tricks to remember Ohm's Law, Resistance and V–I Graphs

Short memory aids. Say each one aloud once, then use it on a question the same day.

VIR triangle
V on top, I and R below: cover the one you need.
Use it when: Rearranging.
Slope is resistance
V up the side, I along the bottom: slope = R.
Use it when: Graph questions.
Ohmic means straight through origin
Curved V–I lines are non-ohmic.
Use it when: Filament bulbs.

Ohm's Law, Resistance and V–I Graphs: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Voltmeter placement · 1 mark · MCQBoard level

A voltmeter is connected to measure the p.d. across a resistor. It must be connected:
(a) in series with the resistor
(b) in parallel with the resistor
(c) in series with the cell only
(d) in parallel with the ammeter only

Try it first, then show the solutionSolution

A voltmeter is connected across the component, in parallel.

Answer: (b)

Example 2: Using V = IR · 2 marks · Short answerBoard level

A heater coil of resistance 22 Ω carries a current of 5 A. Find the p.d. across it. What current would a p.d. of 55 V drive through it?

Try it first, then show the solutionSolution

$V = I R = 5 \times 22 = 110$ V. (1)
$I = \frac{V}{R} = \frac{55}{22} = 2.5$ A. (1)

Answer: 110 V; 2.5 A

Example 3: Ohmic or not? · 3 marks · Short answerHarder board level

Two conductors give these readings. Conductor X: V = 2, 4, 6, 8 V and I = 0.25, 0.50, 0.75, 1.00 A. Conductor Y (a filament bulb): V = 2, 4, 6, 8 V and I = 0.25, 0.45, 0.60, 0.70 A. Find V/I for each reading and say which obeys Ohm’s law.

Try it first, then show the solutionSolution

X: $\frac{V}{I} = 8, 8, 8, 8$ Ω: constant, so it obeys Ohm’s law, with $R = 8$ Ω. (1)
Y: $8, 8.9, 10, 11.4$ Ω: increasing. (1)
Y does not obey Ohm’s law: its filament heats up and the resistance rises with temperature. (1)

Answer: X ohmic (8 Ω); Y non-ohmic

Example 4: Using a rheostat · 3 marks · Short answerHarder board level

A 12 V battery, a lamp and a rheostat are in series. With the rheostat at zero resistance the ammeter reads 2 A. The rheostat is adjusted until the current is 0.8 A. Find the resistance of the lamp and the resistance now offered by the rheostat.

Try it first, then show the solutionSolution

Lamp: $R = \frac{12}{2} = 6$ Ω. (1)
Total resistance now $= \frac{12}{0.8} = 15$ Ω. (1)
Rheostat $= 15 - 6 = 9$ Ω. (1)

Answer: Lamp 6 Ω; rheostat 9 Ω

Example 5: A verification experiment · 4 marks · Case-basedHOTS / competency

In a laboratory a student records these readings for a resistance wire. V (V): 1.0, 2.0, 3.0, 4.0; I (A): 0.21, 0.40, 0.62, 0.80.
(a) Find V/I for each row.
(b) Find the mean resistance.
(c) State the shape of the V–I graph and what it shows.
(d) Which condition must be kept constant, and what controls the current?

Try it first, then show the solutionSolution

(a) $4.76, 5.00, 4.84, 5.00$ Ω. (1)
(b) Mean $= \frac{4.76 + 5.00 + 4.84 + 5.00}{4} \approx 4.9$ Ω. (1)
(c) A straight line through (nearly) the origin: $V \propto I$ , Ohm’s law holds within experimental error. (1)
(d) Temperature of the wire; the current is changed with the rheostat. (1)

Answer: (a) 4.76, 5.00, 4.84, 5.00 Ω (b) 4.9 Ω (c) straight line (d) temperature; rheostat

Where students lose marks in Ohm's Law, Resistance and V–I Graphs

The slips examiners see most often, and what to write instead.

Mistake: Resistance read from the V–I graph as I/V
Slope with V on the y-axis is V/I.
Wrong working
R = 0.5/2
Right working
R = 2/0.5 = 4 Ω
Do this: Check which axis is which.
Usually costs: The answer.
Mistake: Ohm’s law stated without the condition
Constant temperature is part of the law.
Wrong working
Current is proportional to voltage.
Right working
Current is proportional to p.d. provided temperature stays constant.
Do this: Add the condition.
Usually costs: 1 mark.
Mistake: mA used as A
Convert mA to A.
Wrong working
V = 250 × 20 = 5000 V
Right working
V = 0.25 × 20 = 5 V
Do this: Divide mA by 1000.
Usually costs: The answer.

Exam tips

State Ohm’s law with the condition of constant temperature.
Draw the V–I graph with axes labelled and mark the slope as R.
Write ammeter in series and voltmeter in parallel in circuit descriptions.
For a non-ohmic result say "resistance increases with temperature".

Practice questions on Ohm's Law, Resistance and V–I Graphs

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
One ohm is equal to:
(a) 1 A V⁻¹
(b) 1 V A⁻¹
(c) 1 J C⁻¹
(d) 1 C s⁻¹
Hint
V/I.
Answer
(b)
Q2 · 1 mark · Very short answerBoard level
Which instrument is connected in series in a circuit to measure current?
Hint
It measures current.
Answer
The ammeter.
Q3 · 2 marks · Short answerHarder board level
The p.d. across a 40 Ω resistor is 12 V. Find the current.
Hint
I = V/R.
Answer
$I = \frac{12}{40} = 0.3$ A.
Q4 · 3 marks · Short answerHarder board level
A conductor carries 0.12 A when the p.d. is 4.8 V. Find its resistance and the p.d. needed for a current of 0.35 A (assume it is ohmic).
Hint
R is constant.
Answer
$R = \frac{4.8}{0.12} = 40$ Ω; $V = 0.35 \times 40 = 14$ V.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): A filament bulb does not obey Ohm’s law over a wide range of voltage.
Reason (R): The resistance of the filament changes with its temperature.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Temperature.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
The V–I graphs (V on the y-axis) of two wires A and B pass through the origin and through (0.5 A, 2 V) and (0.5 A, 5 V) respectively. Find the resistance of each, say which line is steeper, and find the current each carries at 6 V.
Hint
Slope of each line.
Answer
$R_{A} = \frac{2}{0.5} = 4$ Ω; $R_{B} = \frac{5}{0.5} = 10$ Ω; B is steeper (larger resistance). At 6 V: $I_{A} = 1.5$ A; $I_{B} = 0.6$ A.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Ohm's law formula?Show answer
V = IR.
2.Unit of resistance?Show answer
Ohm (Ω).
3.Shape of V–I graph of an ohmic conductor?Show answer
Straight line through the origin.
4.Function of a rheostat?Show answer
Changes the resistance, and so the current, in a circuit.
5.Is a bulb filament ohmic?Show answer
No; its resistance rises as it heats.

Ohm's Law, Resistance and V–I Graphs: frequently asked questions

What is Ohm's law?

At constant temperature the current through a conductor is directly proportional to the potential difference across it: V = IR.

What does the slope of a V–I graph represent?

The resistance R = V/I (with V on the vertical axis).

Why does a bulb not obey Ohm’s law?

Its filament heats up as current increases, so its resistance rises and V/I is not constant.