ScienceCBSE Class 10 · Electricity

Ohm's Law Class 10: Statement, V-I Graph and Experiment

Updated 14 October 2026 · Fact-checked

Ohm's law says that for a metallic wire kept at a constant temperature, the potential difference across it and the current through it change in the same ratio, so V = IR. The board asks you to state it with the temperature condition, use it in numericals, plot a V-I graph, find resistance from the slope and describe the practical.

Understand Ohm's Law

Imagine connecting a nichrome wire to one cell, then two, three and four cells. Each time you add a cell, the push on the charges (the potential difference V) increases, and the current I through the wire increases too. When you divide V by I for each reading, you get nearly the same number every time. That steady ratio is the heart of Ohm's law.

Ohm's law: for a metallic wire kept at a steady temperature, the potential difference across its ends and the current through it rise and fall in the same ratio. Double V and the current doubles; halve V and the current halves. So $V \propto I$ , which means $\frac{V}{I}$ is a constant. This constant is called the resistance $R$ of the wire, giving $V = I R$ . Georg Simon Ohm found this relationship in 1827.

Why does a wire have resistance at all? The electrons in a metal are not perfectly free to move. They keep colliding with, and being pulled by, the atoms of the metal, which slows their drift along the wire. Resistance is the property of a conductor that opposes the flow of charge. Its SI unit is the ohm ( $Ω$ ). A wire has a resistance of $1 Ω$ when a p.d. of 1 V across it drives a current of 1 A.

If you plot V against I, the points lie on a straight line through the origin. The slope of this line, $\frac{V}{I}$ , is the resistance. A steeper line means a bigger resistance. Rearranging gives $I = \frac{V}{R}$ : for a fixed voltage, current is inversely proportional to resistance. Double the resistance and the current halves. A rheostat (variable resistance) uses this idea to change the current in a circuit without changing the source.

The condition about temperature matters. Resistance changes with temperature, so if the wire heats up, $\frac{V}{I}$ no longer stays constant. NCERT asks you to take the plug key out after each reading; this also stops the wire heating, which would change its resistance.

Ohm's law is the tool you use for the rest of the chapter. It gives the series and parallel formulas, it is used to find current before working out heat with $H = I^{2} R t$ , and it turns $P = V I$ into $P = I^{2} R$ and $P = \frac{V ^{2}}{R}$ .

Ohm's Law: formulas and key facts

Ohm's law: $V = I R$; V in volt, I in ampere, R in ohm. Valid for a metallic conductor at constant temperature.
Current from Ohm's law: $I = \frac{V}{R}$; For a fixed V, current is inversely proportional to R: double R, half I.
Resistance from Ohm's law: $R = \frac{V}{I}$; Also the slope of the V-I graph when V is on the y-axis and I on the x-axis.
Unit of resistance: $1 Ω = \frac{1 V}{1 A}$; Convert mA to A ( $1 m A = 1 0^{- 3} A$ ) before substituting.
Nature of the V-I graph: A straight line through the origin; slope $= \frac{V}{I} = R$; A steeper line means a higher resistance.
Meter connections in the practical: $Ammeter in series with the resistor; voltmeter in parallel across it$; A rheostat or a change in number of cells is used to change the current.

How to solve Ohm's Law questions

Use this method for any Ohm's law numerical, graph or practical question.

1Write down the given quantities with units and mark what is asked.
2Convert to SI units: mA to A, kΩ to Ω.
3Check whether R stays the same (same wire, same temperature). If so, use the ratio idea: V and I change in the same proportion.
4Write the formula in the form you need: $V = I R$ , $I = \frac{V}{R}$ or $R = \frac{V}{I}$ .
5Substitute the values with units and calculate.
6For graph questions, put V on the y-axis and I on the x-axis, draw the best straight line through the origin and take slope $= \frac{Δ V}{Δ I}$ .
7Write the final answer with its unit and check it makes sense (a bigger V on the same wire should give a bigger I).

Ohm's Law 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.

Ohm's law and resistance

Suppose your old torch runs on two cells of 1.5 V each, giving 3 V across its bulb, and the bulb draws 0.3 A. Imagine you add a third 1.5 V cell to make it brighter. For the calculation, treat the filament's resistance as fixed.

(a) Find the resistance of the bulb. (b) Find the new current with three cells. (c) How much more charge passes through the bulb in 10 minutes with three cells than with two? (d) Once the filament gets hot, would Ohm's law still hold exactly? Give a reason.

Try it first, then show the solutionSolution

With two cells, V = 3 V and I = 0.3 A, so $R = \frac{3 V}{0.3 A} = 10 Ω$ .
With three cells, V = 4.5 V, so $I = \frac{4.5 V}{10 Ω} = 0.45 A$ .
t = 10 min = 600 s. Two cells: $Q = I t = 0.3 A \times 600 s = 180 C$ .
Three cells: $Q = 0.45 A \times 600 s = 270 C$ ; extra charge = 270 C − 180 C = 90 C.
Not necessarily: resistance varies with temperature, so a hot filament need not keep the same V/I. Ohm's law holds only when the temperature stays the same.

Answer: (a) 10 Ω; (b) 0.45 A; (c) 90 C more; (d) not exactly, because the filament's temperature changes and resistance varies with temperature, so V/I need not stay constant.

Memory hook: Same bulb, more cells: V up by half, I up by half, as long as it stays at the same temperature.

V-I graph and finding resistance

Imagine your school lab has a resistor whose colour code has worn off. You connect it with an ammeter, a voltmeter, a plug key and a battery, and record: V = 2 V at I = 0.25 A, and V = 4 V at I = 0.5 A. The ammeter can read at most 1 A.

(a) Find the resistance and check that the ratio is constant. (b) What current would you expect at 6 V, and what would the V-I graph look like? (c) Using cells of 2 V each in series, what is the largest number of cells you can use without the current going above 1 A?

Try it first, then show the solutionSolution

$R = \frac{2 V}{0.25 A} = 8 Ω$ .
Check with the second reading: $\frac{4 V}{0.5 A} = 8 Ω$ , so V/I is constant.
At 6 V: $I = \frac{6 V}{8 Ω} = 0.75 A$ .
Since V/I is constant, the V-I graph is a straight line through the origin with slope 8 Ω.
Largest allowed p.d.: $V = I R = 1 A \times 8 Ω = 8 V$ .
Number of 2 V cells = 8 V ÷ 2 V = 4 cells (current exactly 1 A); a fifth cell would give 10 V and 1.25 A, which is too much.

Answer: R = 8 Ω; I = 0.75 A at 6 V; straight line through the origin; at most 4 cells.

Memory hook: No label? Measure V and I, divide, and the resistor tells you its value.

Tricks to remember Ohm's Law

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

The VIR triangle
Write V at the top and I and R side by side below. Cover the one you want: V = I × R, I = V ÷ R, R = V ÷ I.
Use it when: Rearranging Ohm's law quickly in numericals.
Very Important Rule
V-I-R in that order: V = I R. Add 'at the same temperature' every time you state it.
Use it when: Writing the full statement without dropping the condition.
Slope = Steepness = Stubbornness
On a V-I graph, the steeper line belongs to the more 'stubborn' wire, the one with more resistance.
Use it when: Comparing two wires from their graphs.
A is in a line, V stands aside
Ammeter in series (in the line of current), voltmeter stands beside the resistor (parallel).
Use it when: Circuit diagram and practical questions.

Ohm's Law: solved examples

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

Example 1: Finding an unknown resistance with meters · 5 marks · Long answerHarder board level

A 6 V battery, a plug key, an ammeter, a 4 Ω resistor and an unknown resistor X are connected in series. A voltmeter is connected across the 4 Ω resistor and reads 2 V. Treat the ammeter and connecting wires as having no resistance.
(a) What does the ammeter read? (1)
(b) Find the p.d. across X. (1)
(c) Find the resistance of X. (1)
(d) The 4 Ω resistor is removed and X alone is connected across the same battery. Find the new current. (1)
(e) The voltmeter is moved across both resistors together. What will it read, and why? (1)

Try it first, then show the solutionSolution

(a) The same current flows through every part of a series circuit: $I = \frac{V}{R} = \frac{2 V}{4 Ω} = 0.5 A$ . (1)
(b) The battery's p.d. is shared between the resistors: $V_{X} = 6 V - 2 V = 4 V$ . (1)
(c) $R_{X} = \frac{V _{X}}{I} = \frac{4 V}{0.5 A} = 8 Ω$ . Check: $\frac{6 V}{( 4 + 8 ) Ω} = 0.5 A$ . (1)
(d) $I = \frac{6 V}{8 Ω} = 0.75 A$ . (1)
(e) 6 V, because the p.d.s across resistors in series add up: $2 V + 4 V = 6 V$ . (1)

Answer: (a) 0.5 A; (b) 4 V; (c) R of X = 8 Ω; (d) 0.75 A; (e) 6 V.

Example 2: Statement and ratio reasoning · 3 marks · Short answerBoard level

(a) State Ohm's law and the condition under which it holds.
(b) The p.d. across an electric heater is 50 V when it draws 5 A. Find the current when the p.d. is increased to 150 V.
(c) If the p.d. across a resistor is halved, what happens to the current through it?

Try it first, then show the solutionSolution

(a) For a metallic conductor at constant temperature, $V \propto I$ , so $\frac{V}{I}$ is a constant. (1)
(b) $R = \frac{50 V}{5 A} = 10 Ω$ ; $I = \frac{150 V}{10 Ω} = 15 A$ . (1)
(c) R stays constant, so $I = \frac{V}{R}$ also halves. (1)

Answer: (a) V ∝ I at constant temperature; (b) 15 A; (c) the current becomes half.

Example 3: V-I graph from data · 3 marks · Short answerBoard level

A student records these readings for a resistor:
I (A): 0.5, 1.0, 1.5, 2.0, 2.5
V (V): 2.0, 4.0, 6.0, 8.0, 10.0
(a) Describe the V-I graph and how you would draw it.
(b) Find the resistance from the slope.
(c) What current flows when V = 5 V?

Try it first, then show the solutionSolution

(a) Put I on the x-axis and V on the y-axis; plot the points; they give a straight line through the origin. (1)
(b) Slope $= \frac{Δ V}{Δ I} = \frac{10.0 - 2.0}{2.5 - 0.5} = \frac{8.0 V}{2.0 A} = 4 Ω$ . (1)
(c) $I = \frac{5 V}{4 Ω} = 1.25 A$ . (1)

Answer: Straight line through origin; R = 4 Ω; I = 1.25 A.

Example 4: Full practical: verifying Ohm's law · 5 marks · Long answerHarder board level

(a) Describe the circuit for studying how V across a nichrome wire depends on I, listing each component (cells, plug key, ammeter, voltmeter, nichrome wire XY and a rheostat, which is optional) and how it is connected. Give a draw-and-label checklist for the circuit diagram and for the V-I graph.
(b) How are different values of current obtained?
(c) State what is observed and what the graph shows.
(d) Give two precautions.
(e) In one reading, V = 1.5 V and I = 0.5 A. Find R and the current when V = 4.5 V.

Try it first, then show the solutionSolution

(a) Cells, plug key, ammeter and nichrome wire XY all in series; voltmeter connected in parallel across XY. (1)
Draw and label: cell/battery symbol (long line +, short line −), plug key, ammeter A in series, wire XY as a resistor, voltmeter V across X and Y, arrow for current from + terminal round to −. Common slip: putting the voltmeter in series or the ammeter across the wire. (1)
Graph checklist: V on the y-axis and I on the x-axis, units on both axes, a uniform scale, a best-fit straight line through the origin, and a triangle drawn on the line to find the slope.
(b) Use one, two, three and then four cells of 1.5 V (or adjust a rheostat), noting I and V each time. (½)
(c) V/I is nearly the same for every reading; V-I graph is a straight line through the origin, so V ∝ I: Ohm's law. (1)
(d) Take the plug key out after each reading so the wire does not heat; keep connections tight. (½)
(e) $R = \frac{1.5 V}{0.5 A} = 3 Ω$ ; $I = \frac{4.5 V}{3 Ω} = 1.5 A$ . (1)

Answer: Series circuit with voltmeter across the wire; V/I constant, straight-line graph through origin; R = 3 Ω and I = 1.5 A at 4.5 V.

Example 5: Two wires on one graph · 4 marks · Case-basedHarder board level

Read the passage and answer.

Two students test wires P and Q and plot V (y-axis) against I (x-axis). Both get straight lines through the origin. The line for P passes through the point I = 2 A, V = 6 V. The line for Q passes through the point I = 0.5 A, V = 6 V.
(a) Find the resistance of each wire and say which line is steeper. (1)
(b) The same 3 V is applied across each wire separately. Find each current and the ratio of the current in P to the current in Q. (1)
(c) The same current of 0.5 A is passed through each wire separately. Find the p.d. across each and the ratio of the p.d. across P to that across Q. (1)
(d) A third student leaves his wire connected for a long time and finds that his V/I values keep changing. Explain. (1)

Try it first, then show the solutionSolution

(a) $R_{P} = \frac{6 V}{2 A} = 3 Ω$ ; $R_{Q} = \frac{6 V}{0.5 A} = 12 Ω$ . Q's line is steeper because slope = V/I = R. (1)
(b) $I_{P} = \frac{3 V}{3 Ω} = 1 A$ ; $I_{Q} = \frac{3 V}{12 Ω} = 0.25 A$ ; $I_{P} : I_{Q} = 4 : 1$ . (1)
(c) $V_{P} = 0.5 A \times 3 Ω = 1.5 V$ ; $V_{Q} = 0.5 A \times 12 Ω = 6 V$ ; $V_{P} : V_{Q} = 1 : 4$ . (1)
(d) The current heats the wire, and resistance varies with temperature. The temperature is no longer constant, so V/I does not stay fixed and the condition for Ohm's law is not met. (1)

Answer: R_P = 3 Ω, R_Q = 12 Ω (Q steeper); at 3 V: 1 A and 0.25 A, ratio 4 : 1; at 0.5 A: 1.5 V and 6 V, ratio 1 : 4; heating changes the resistance, so the constant-temperature condition of Ohm's law fails.

Example 6: Spotting an odd reading in a practical table · 4 marks · Case-basedHOTS / competency

Read the passage and answer.

In the practical on V and I for a resistor, a student records:
I (A): 0.2, 0.4, 0.6, 0.8, 1.0
V (V): 0.6, 1.2, 2.4, 2.4, 3.0
She finds V/I for every row and reports the mean of all five values as the resistance.
(a) Find V/I for each reading and name the reading that does not fit. (1)
(b) Find the mean resistance using only the readings that fit. Compare it with her mean of all five and say which value should be reported, with a reason. (1)
(c) What should V have been for the odd reading? (1)
(d) Give two possible reasons for the odd reading. (1)

Try it first, then show the solutionSolution

(a) V/I = 3, 3, 4, 3, 3 Ω. The reading at I = 0.6 A (V = 2.4 V) does not fit. (1)
(b) Mean of the four good values = 3 Ω. Her mean of all five = $\frac{3 + 3 + 4 + 3 + 3}{5} = 3.2 Ω$ . Report 3 Ω: for a resistor at constant temperature V/I should be constant, so the odd reading is an error that wrongly pulls the mean up. (1)
(c) $V = I R = 0.6 A \times 3 Ω = 1.8 V$ . (1)
(d) Any two: the voltmeter was misread; a loose connection; the key was left in so the wire got warm and its resistance changed. (1)

Answer: (a) The reading at 0.6 A is odd; (b) R = 3 Ω (not 3.2 Ω), as the faulty reading should be left out; (c) 1.8 V; (d) misreading the meter, a loose connection or the wire heating up.

Where students lose marks in Ohm's Law

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

Mistake: Dropping the temperature condition from the statement of Ohm's law.
Students remember V ∝ I but forget that it holds only at constant temperature.
Wrong working
Ohm's law: V is directly proportional to I.
Right working
Ohm's law: for a metallic wire at constant temperature, the p.d. across it rises and falls in the same ratio as the current through it (V ∝ I).
Do this: Always end the statement with 'at constant temperature'.
Usually costs: ½ to 1 mark in a definition question.
Mistake: Not converting mA to A.
Substituting 2.5 for 2.5 mA gives an answer 1000 times too small.
Wrong working
R = 12 V ÷ 2.5 = 4.8 Ω
Right working
R = 12 V ÷ (2.5 × 10⁻³ A) = 4800 Ω
Do this: Convert every quantity to SI units before substituting.
Usually costs: Usually the final 1 mark of a numerical.
Mistake: Taking the slope as I/V or swapping the axes.
Students plot V on the x-axis and then call the slope resistance.
Wrong working
Slope = ΔI/ΔV = 0.25, so R = 0.25 Ω
Right working
With V on the y-axis and I on the x-axis, slope = ΔV/ΔI = 4 Ω = R
Do this: Put V on the y-axis; if axes are swapped, R = 1/slope.
Usually costs: 1 to 2 marks in a graph question.
Mistake: Connecting the voltmeter in series in the circuit diagram.
Confusing the roles of ammeter and voltmeter.
Wrong working
Battery – key – ammeter – voltmeter – wire, all in one loop.
Right working
Battery – key – ammeter – wire in series; voltmeter across the two ends of the wire.
Do this: Ammeter in the line of current; voltmeter beside the part being measured.
Usually costs: 1 mark in a diagram or practical question.
Mistake: Writing an answer without units.
Rushing the final line.
Wrong working
R = 55
Right working
R = 55 Ω
Do this: Box the final answer with its unit every time.
Usually costs: ½ mark.

Exam tips

When stating Ohm's law, name both quantities, say 'directly proportional' and add the constant-temperature condition.
In graph questions, label both axes with quantity and unit, choose a sensible scale, and show the two points you used for the slope.
For 'what happens if V is doubled/halved' questions, say R stays the same and give the change in I with a one-line reason.
In practical questions, mention series ammeter, parallel voltmeter, changing the number of cells or a rheostat, and taking the key out after each reading.
If V/I comes out slightly different in each reading, say the values are 'approximately' constant; real readings have small errors.

Practice questions on Ohm's Law

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

Q1 · 2 marks · Very short answerBoard level
An electric heater draws 4 A from a 220 V supply. (a) Find its resistance. (b) One evening the supply falls to 200 V. Find the current now, assuming the heater's temperature (and so its resistance) stays the same, and say why this assumption is needed.
Hint
Find R first and keep it fixed.
Answer
(a) $R = \frac{220 V}{4 A} = 55 Ω$ (1 mark). (b) $I = \frac{200 V}{55 Ω} \approx 3.64 A$ . The assumption is needed because Ohm's law ( $V \propto I$ ) holds only at constant temperature; if the heater cools, its resistance changes (1 mark).
Q2 · 3 marks · Short answerHarder board level
A student records the current in milliamperes for a resistor:
I (mA): 100, 200, 300, 400
V (V): 0.5, 1.0, 1.6, 2.0
(a) Convert the currents to amperes and find V/I for each reading. Which reading does not fit? (b) What should V have been for that reading? (c) The ammeter can read at most 500 mA. What is the largest p.d. that can be applied across the resistor?
Hint
100 mA = 0.1 A. A reading that breaks the constant ratio is an error.
Answer
(a) I = 0.1, 0.2, 0.3, 0.4 A; V/I = 5, 5, 5.33, 5 Ω. The reading at 300 mA does not fit; R = 5 Ω (1 mark). (b) $V = 0.3 A \times 5 Ω = 1.5 V$ (1 mark). (c) $V_{ma x} = 0.5 A \times 5 Ω = 2.5 V$ (1 mark).
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When the potential difference across a nichrome wire at constant temperature is doubled, the current through it also doubles.
Reason (R): The resistance of a wire depends on its length, its area of cross-section and the material it is made of.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Both statements are true. Does R tell you why the current doubles?
Answer
(b) Both A and R are true but R is not the correct explanation of A. A follows from Ohm's law: at constant temperature R is fixed, so $I = \frac{V}{R}$ doubles when V doubles. R only lists what decides the value of the resistance.
Q4 · 3 marks · Short answerHarder board level
A resistor carries 60 mA when the p.d. across it is 3 V. The ammeter in the circuit reads at most 200 mA. (a) Find the resistance. (b) Find the largest p.d. that can be used without the ammeter going off-scale, and say whether four 1.5 V cells in series are safe. (c) The resistor is replaced by one of double resistance across the four cells. Find the current. (d) Name the device that could change the current without changing the cells.
Hint
Convert mA to A. Four cells in series give 6 V.
Answer
(a) $R = \frac{3 V}{0.06 A} = 50 Ω$ . (b) $V_{ma x} = 0.2 A \times 50 Ω = 10 V$ ; four cells give 6 V, so $I = \frac{6}{50} = 0.12 A = 120$ mA, which is safe (1 mark each for (a) and (b)). (c) $R = 100 Ω$ , $I = \frac{6}{100} = 0.06 A = 60$ mA. (d) A rheostat (variable resistance) (1 mark for (c) and (d)).
Q5 · 3 marks · Short answerHOTS / competency
Resistors A and B are each connected in turn across the same 12 V supply. A draws three times the current that B draws, and B has a resistance of 24 Ω.
(a) Find the resistance of A.
(b) What p.d. would make B carry the same current that A carries at 12 V?
(c) A student says that on a V-I graph (V on the y-axis), A's line is steeper because A carries more current. Find the error in this reasoning.
Hint
Find the current in B first. Remember slope = V/I = R.
Answer
(a) $I_{B} = \frac{12 V}{24 Ω} = 0.5 A$ , so $I_{A} = 3 \times 0.5 A = 1.5 A$ and $R_{A} = \frac{12 V}{1.5 A} = 8 Ω$ . (b) $V = 1.5 A \times 24 Ω = 36 V$ . (c) The slope of a V-I graph is V/I = R, not the current. B has the larger resistance (24 Ω), so B's line is steeper and A's line is less steep.

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.State Ohm's law in one line.Show answer
V across a metallic wire is directly proportional to I through it, at constant temperature.
2.What is the shape of the V-I graph for a resistor?Show answer
A straight line passing through the origin.
3.What does the slope of a V-I graph (V on y-axis) give?Show answer
The resistance, R = V/I.
4.Define 1 ohm.Show answer
The resistance of a conductor that carries 1 A when the p.d. across it is 1 V.
5.How are the ammeter and voltmeter connected in the Ohm's law practical?Show answer
Ammeter in series with the wire; voltmeter in parallel across it.
6.What does a rheostat do?Show answer
It changes the resistance in a circuit to adjust the current without changing the source.

Ohm's Law: frequently asked questions

Why does Ohm's law need the temperature to stay constant?

Resistance changes with temperature. If the wire heats up, its resistance changes, so V divided by I no longer stays the same. That is why the plug key is taken out after each reading in the practical.

How do I find resistance from a V-I graph?

Plot V on the y-axis and I on the x-axis. Pick two points on the straight line and divide the change in V by the change in I. This slope is the resistance in ohm.

Is V = IR the same as Ohm's law?

V = IR comes from Ohm's law: since V is proportional to I, the ratio V/I is a constant called resistance R. Writing the full statement with the temperature condition is what earns marks for 'state Ohm's law'.

What is the use of a rheostat in the Ohm's law experiment?

A rheostat is a variable resistance. It lets you change the current in the circuit without changing the battery, so you can take several readings of V and I. The NCERT activity instead changes the number of 1.5 V cells.

Does a bulb filament obey Ohm's law?

Not exactly. Ohm's law needs the temperature to stay the same. A filament gets much hotter as the current rises, and resistance changes with temperature, so V/I for a glowing filament does not stay constant.