PhysicsICSE Class 10 · Chapter 14

Ohm's Law and Resistances ICSE Class 10: V = IR, EMF, Internal Resistance and Combinations

Q: Is the derivation of the series and parallel formulas needed?

Yes. The syllabus includes the derivation of the expressions for equivalent resistance.

Q: Why does a filament lamp not obey Ohm's law?

As the current increases, the filament gets much hotter and its resistance rises, so V/I is not constant.

Q: Is emf a force?

No. Despite its name, emf is the energy given per coulomb by the cell, measured in volts.

Q: What is the resistance of an ideal ammeter and voltmeter?

An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance, so neither disturbs the circuit.

Q: Does resistivity change with temperature?

Yes. For metals both resistance and resistivity increase with temperature.

Updated 14 October 2026 · Fact-checked

Ohm's law says the current through a conductor is directly proportional to the potential difference across it at constant temperature: V = IR. In ICSE Class 10 you also use R = ρl/A, emf and internal resistance (V = ε − Ir), and the series and parallel formulas to solve simple networks.

What Ohm's Law and Resistances is about

Electric current is the flow of charge; a potential difference (pd) pushes it through a circuit; resistance opposes it. Ohm's law ties these together and, with the rules for combining resistors, lets you work out the current and voltage anywhere in a simple circuit.

The CISCE 2027 syllabus covers: pd, current, resistance and charge; Ohm's law (statement, $V = I R$ , SI units, experimental verification, V–I graph and resistance from the slope); ohmic and non-ohmic resistors; factors affecting resistance including specific resistance (resistivity); internal resistance; superconductors; electromotive force (emf); combinations of resistors in series and parallel with derivations; and simple numerical problems on simple networks.

The CISCE examiners' comments for 2024 and 2025 list: incomplete statements of Ohm's law (no 'constant temperature'), judging ohmic behaviour without checking the graph passes through the origin, confusing resistivity with resistance, sign errors in $V = ε - I r$ , misreading unusual circuit diagrams, and missing units.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Ohm's Law and Resistances is in the 4. Electricity and Magnetism (i) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Physics (Science Paper 1) syllabus, examination year 2027

Key concepts

Charge, current and potential difference

Current: rate of flow of charge, $I = \frac{Q}{t}$ ; unit ampere (1 A = 1 C s⁻¹).
Potential difference between two points: work done per unit charge in moving charge between them, $V = \frac{W}{Q}$ ; unit volt (1 V = 1 J C⁻¹).
Resistance: opposition to current, $R = \frac{V}{I}$ ; unit ohm (Ω).

An ammeter is connected in series; a voltmeter in parallel across the part being measured.

Worked example

A current of 0.5 A flows through a lamp for 2 minutes across a pd of 12 V. Find the charge that passes and the work done.

Try it first, then show the solutionSolution

$Q = I t = 0.5 \times 120 = 60$ C.
$W = Q V = 60 \times 12 = 720$ J.

Answer: 60 C; 720 J

Ohm's law and the V–I graph

Ohm's law: the current flowing through a conductor is directly proportional to the potential difference across its ends, provided the physical conditions (such as temperature) remain the same:
$V \propto I$ , i.e. $V = I R$ .

Verification: connect a resistor, ammeter, cell, key and rheostat in series, with a voltmeter across the resistor. Change the rheostat to get several readings of V and I. A graph of V against I is a straight line through the origin; its slope equals $R$ .

Worked example

A resistor carries 0.3 A when the pd across it is 6 V. Find its resistance and the current when the pd is 9 V (temperature unchanged).

Try it first, then show the solutionSolution

$R = \frac{6}{0.3} = 20$ Ω.
$I = \frac{9}{20} = 0.45$ A.

Answer: 20 Ω; 0.45 A

Ohmic and non-ohmic conductors

Ohmic conductors obey Ohm's law: V–I graph is a straight line through the origin, so $\frac{V}{I}$ is the same for every reading. Examples: metal wires (nichrome, copper, manganin) at constant temperature.
Non-ohmic conductors do not: the V–I graph is curved, or straight but not through the origin, so $\frac{V}{I}$ changes. Examples: a filament lamp (heats up), a junction diode, an electrolyte, LEDs, LDRs.

Check both the ratio V/I at each reading and whether the line passes through the origin.

Worked example

Readings for a material: V = 2, 4, 6 V; I = 0.5, 0.8, 1.0 A. Is it ohmic?

Try it first, then show the solutionSolution

V/I = 4, 5, 6 Ω: not constant.
So the material does not obey Ohm's law: non-ohmic.

Answer: Non-ohmic

Factors affecting resistance; resistivity

For a wire: $R = ρ \frac{l}{A}$ .

Length $l$ : $R \propto l$ .
Area of cross-section $A$ : $R \propto \frac{1}{A}$ (thicker wire, less resistance).
Material: through the specific resistance (resistivity) $ρ$ , unit Ω m.
Temperature: for metals, resistance rises with temperature.

Resistivity depends only on the material (and its temperature), not on length or thickness. Alloys such as nichrome have high resistivity (heating elements); copper has low resistivity (connecting wires).

Worked example

A wire of resistance 5 Ω is stretched uniformly to double its length. Find its new resistance. Does its resistivity change?

Try it first, then show the solutionSolution

Volume is constant, so doubling $l$ halves $A$ .
$R \propto \frac{l}{A}$ : $R^{'} = 5 \times 2 \times 2 = 20$ Ω.
Resistivity is unchanged: same material, same temperature.

Answer: 20 Ω; resistivity unchanged

EMF, terminal voltage and internal resistance

EMF ( $ε$ ) of a cell: the pd across its terminals when no current is drawn (open circuit); the energy it gives to each coulomb of charge.
Internal resistance ( $r$ ): resistance of the electrolyte and electrodes inside the cell.
Terminal voltage ( $V$ ): pd across the terminals when current flows (closed circuit).

$ε = V + I r$ , so $V = ε - I r$ , and $I = \frac{ε}{R + r}$ . The 'lost volts' $I r$ are used inside the cell. In an open circuit $V = ε$ ; in a closed circuit $V < ε$ .

Worked example

A cell of emf 3 V and internal resistance 0.5 Ω is connected to a 5.5 Ω resistor. Find the current and the terminal voltage.

Try it first, then show the solutionSolution

$I = \frac{3}{5.5 + 0.5} = 0.5$ A.
$V = ε - I r = 3 - 0.25 = 2.75$ V (or $I R = 0.5 \times 5.5$ ).

Answer: 0.5 A; 2.75 V

Resistors in series

In series, the same current flows through each resistor and the pds add: $V = V_{1} + V_{2} + V_{3}$ .

Derivation: $I R = I R_{1} + I R_{2} + I R_{3}$ , so
$R_{s} = R_{1} + R_{2} + R_{3}$ .

The equivalent resistance is greater than the largest. If one component breaks, the whole circuit stops (as in old decorative bulb strings).

Worked example

Resistors of 2 Ω, 3 Ω and 5 Ω are joined in series across 10 V. Find the current and the pd across each.

Try it first, then show the solutionSolution

$R_{s} = 10$ Ω, so $I = 1$ A.
pds: 2 V, 3 V and 5 V (adding to 10 V).

Answer: 1 A; 2 V, 3 V, 5 V

Resistors in parallel and simple networks

In parallel, each resistor has the same pd and the currents add: $I = I_{1} + I_{2} + I_{3}$ .

Derivation: $\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}}$ , so
$\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ .

The equivalent resistance is less than the smallest. For two: $R_{p} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ . Current divides in the inverse ratio of the branch resistances. Household appliances are in parallel.

For a network: combine the parallel groups first, then add series parts, find the total current, then work back.

Worked example

Resistors of 3 Ω and 6 Ω are in parallel across 6 V. Find the equivalent resistance, total current and branch currents.

Try it first, then show the solutionSolution

$R_{p} = \frac{3 \times 6}{9} = 2$ Ω.
$I = \frac{6}{2} = 3$ A.
Branches: $\frac{6}{3} = 2$ A and $\frac{6}{6} = 1$ A.

Answer: 2 Ω; 3 A; 2 A and 1 A

Superconductors

Some materials lose all their resistance when cooled below a certain (critical) temperature; they are superconductors. A current, once started in a superconducting loop, can flow without loss.

Uses (qualitative): powerful electromagnets in MRI scanners and particle accelerators, magnetic levitation trains, and loss-free power cables (being developed).

Worked example

What is a superconductor, and why would superconducting power cables save energy?

Try it first, then show the solutionSolution

A material whose resistance becomes zero below its critical temperature.
With $R = 0$ , no heat ( $I^{2} R t$ ) is produced in the cable, so no energy is wasted.

Answer: Zero resistance below a critical temperature; no heating loss

Physics in real life: Ohm's Law and Resistances

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.

Series and parallel lamps

Suppose four identical 6 Ω bulbs are connected to a 12 V battery, first in series and then in parallel.

Find the current through each bulb in both cases. Which arrangement gives brighter bulbs?

Try it first, then show the solutionSolution

Series: $R = 24$ Ω, $I = 0.5$ A through each (3 V each).
Parallel: each bulb has 12 V, so $I = 2$ A each (8 A in total).
Parallel bulbs carry more current and get more pd: they are brighter.

Answer: 0.5 A each (series); 2 A each (parallel); parallel brighter

Memory hook: Parallel gives every bulb the full voltage.

Lost volts in a torch

Imagine an old torch cell of emf 1.5 V whose internal resistance has grown to 1 Ω, lighting a bulb of 2 Ω.

Find the current and the terminal voltage.

Try it first, then show the solutionSolution

$I = \frac{1.5}{2 + 1} = 0.5$ A.
$V = 1.5 - 0.5 \times 1 = 1$ V.

Answer: 0.5 A; 1 V

Memory hook: Old cells lose volts inside.

Choosing wire for a heater

Suppose you must choose between copper and nichrome wire for a heater element and for the connecting cable.

Which material goes where, and why?

Try it first, then show the solutionSolution

The heater needs high resistance (high resistivity) to produce heat: nichrome.
The cable should waste little energy: low resistivity copper.

Answer: Nichrome for the element; copper for the cable

Memory hook: High ρ to heat, low ρ to carry.

Topics in this chapter

Ohm's Law and Resistances formula sheet

Current and pd: $I = \frac{Q}{t}, V = \frac{W}{Q}$; Units: A = C s⁻¹, V = J C⁻¹.
Ohm's law: $V = I R$; At constant temperature.
Resistance of a wire: $R = ρ \frac{l}{A}$; ρ in Ω m depends on material and temperature.
EMF and internal resistance: $ε = V + I r, I = \frac{ε}{R + r}$; Open circuit: V = ε.
Series: $R_{s} = R_{1} + R_{2} + R_{3}$; Same current.
Parallel: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$; Same pd; for two, R₁R₂/(R₁ + R₂).

Ohm's Law and Resistances: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Circuit to verify Ohm's law

Draw the circuit and describe the experiment.

Draw

A cell (or battery), a plug key and a rheostat in series.
The resistor (wire) under test in series with an ammeter.
A voltmeter connected in parallel across the resistor.
Arrow for conventional current from + to −.

Label

Cell / battery (+ and − terminals)
Key K
Rheostat Rh
Ammeter A (series)
Voltmeter V (parallel)
Resistor R

Watch out: Ammeter always in series; voltmeter always in parallel. Use standard symbols.

V–I graph for ohmic and non-ohmic conductors

Show or interpret V–I graphs.

Draw

Axes: current I (x-axis), pd V (y-axis), with the origin marked.
A straight line through the origin for an ohmic conductor.
A curve (bending upward, e.g. a filament lamp) for a non-ohmic conductor.

Label

Ohmic: straight line through O, slope = R
Non-ohmic: curved (V/I changes)
Axes with units

Watch out: With V on the y-axis, slope = R; with I on the y-axis, slope = 1/R.

How the board asks Ohm's Law and Resistances

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
MCQ and fill-in-the-blanks	What resistivity depends on; emf vs terminal voltage; series or parallel in an odd-looking diagram.	ρ: material only; open circuit V = ε; parallel = same pd across both ends.
Ohmic or non-ohmic from data	A V–I table; decide and justify; state Ohm's law.	V/I constant and graph through origin → ohmic; state the law with 'constant temperature'.
Effect of a change	Remove a bulb from a series string; stretch a wire; long current flow.	Use R = ρl/A, I = V/R and P = I²R; answer with 'increases/decreases/remains same'.
EMF and internal resistance	Find I, V, r or lost volts.	I = ε/(R + r); V = ε − Ir.
Networks (3-4 marks)	Equivalent resistance, total current, branch currents and pds.	Parallel groups first; then series; then work back; units with every answer.

Tricks to remember Ohm's Law and Resistances

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 want.
Use it when: Ohm's law rearrangements.
Series adds, parallel shrinks
Series: total bigger than the biggest. Parallel: total smaller than the smallest.
Use it when: Checking network answers.
ρ is like density
Resistivity, like density, belongs to the material, not the shape.
Use it when: Resistivity MCQs.
Lost volts
V = ε − Ir: some volts are lost inside the cell when current flows.
Use it when: Terminal voltage questions.

Ohm's Law and Resistances: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Resistivity · 1 mark · MCQBoard level

The specific resistance (resistivity) of a wire depends on its:
(a) length
(b) area of cross-section
(c) material
(d) shape

Try it first, then show the solutionSolution

Resistivity is a property of the material (at a given temperature).

Answer: (c)

Example 2: Ohmic graph · 1 mark · Assertion-reasonBoard level

Assertion (A): The V–I graph of a metal wire at constant temperature is a straight line passing through the origin.
Reason (R): For such a conductor, V is directly proportional to I.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Try it first, then show the solutionSolution

Direct proportion gives a straight line through the origin.

Answer: (c)

Example 3: Using Ohm's law · 2 marks · Short answerBoard level

The element of an electric heater has a resistance of 44 Ω. Find (a) the current it draws from a 220 V supply, (b) the current if the supply voltage falls to 200 V (resistance unchanged).

Try it first, then show the solutionSolution

(a) $I = \frac{V}{R} = \frac{220}{44} = 5$ A.
(b) $I = \frac{200}{44} \approx 4.55$ A.

Answer: (a) 5 A (b) about 4.55 A

Example 4: Internal resistance · 3 marks · Short answerHarder board level

A cell of emf 6 V and internal resistance 1 Ω is connected to a 5 Ω resistor. Find (a) the current, (b) the terminal voltage, (c) the 'lost volts'. (d) What is the terminal voltage when the circuit is open?

Try it first, then show the solutionSolution

(a) $I = \frac{ε}{R + r} = \frac{6}{5 + 1} = 1$ A.
(b) $V = I R = 1 \times 5 = 5$ V; (c) lost volts $= I r = 1 \times 1 = 1$ V (and $5 + 1 = 6$ V $= ε$ ).
(d) Open circuit: no current, so $V = ε = 6$ V.

Answer: (a) 1 A (b) 5 V (c) 1 V (d) 6 V

Example 5: Comparing two wires · 3 marks · Short answerHarder board level

Wires A and B are made of the same metal. A is 1 m long with a cross-sectional area of 0.5 mm²; B is 2 m long with a cross-sectional area of 1.5 mm². (a) Find the ratio of their resistances $R_{A} : R_{B}$ . (b) Find the ratio of their resistivities. Explain.

Try it first, then show the solutionSolution

$R = ρ \frac{l}{A}$ with the same $ρ$ : $R_{A} \propto \frac{1}{0.5} = 2$ and $R_{B} \propto \frac{2}{1.5} = \frac{4}{3}$ .
(a) $R_{A} : R_{B} = 2 : \frac{4}{3} = 3 : 2$ .
(b) $1 : 1$ : resistivity depends only on the material (and temperature), not on length or area.

Answer: (a) 3 : 2 (b) 1 : 1

Example 6: Series circuit · 2 marks · Long answerBoard level

Resistors of 4 Ω and 8 Ω are connected in series to a 6 V supply. Find the current and the pd across each resistor.

Try it first, then show the solutionSolution

$R_{s} = 4 + 8 = 12$ Ω, so $I = \frac{6}{12} = 0.5$ A.
pds: $0.5 \times 4 = 2$ V and $0.5 \times 8 = 4$ V (adding to 6 V).

Answer: 0.5 A; 2 V and 4 V

Example 7: Parallel circuit · 3 marks · Long answerHarder board level

A 3 Ω and a 6 Ω resistor are connected in parallel across a 6 V supply. Find (a) the equivalent resistance, (b) the total current, (c) the current in each resistor. (d) Which resistor carries more current, and why?

Try it first, then show the solutionSolution

(a) $\frac{3 \times 6}{3 + 6} = 2$ Ω.
(b) $\frac{6}{2} = 3$ A; (c) 2 A in 3 Ω, 1 A in 6 Ω.
(d) The 3 Ω resistor: with the same pd, current is inversely proportional to resistance.

Answer: (a) 2 Ω (b) 3 A (c) 2 A, 1 A

Example 8: A network with internal resistance · 4 marks · Long answerHOTS / competency

A 4 Ω resistor is in series with a parallel combination of 6 Ω and 3 Ω. The combination is connected to a battery of emf 12 V and internal resistance 2 Ω. Find (a) the total resistance, (b) the current from the battery, (c) the terminal voltage, (d) the current in the 6 Ω resistor.

Try it first, then show the solutionSolution

Parallel part: $\frac{6 \times 3}{9} = 2$ Ω. (a) Total $= 4 + 2 + 2 = 8$ Ω.
(b) $I = \frac{12}{8} = 1.5$ A.
(c) $V = 12 - 1.5 \times 2 = 9$ V.
(d) pd across the parallel part $= 1.5 \times 2 = 3$ V; current in 6 Ω $= \frac{3}{6} = 0.5$ A.

Answer: (a) 8 Ω (b) 1.5 A (c) 9 V (d) 0.5 A

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Decorative lights

Suppose four identical bulbs, each of resistance 6 Ω, are connected to a 12 V battery of negligible internal resistance.

(i) · 3 marks
When they are in series, find the current and the pd across each bulb.
Answer
Answer: 0.5 A; 3 V each
$R = 4 \times 6 = 24$ Ω; $I = \frac{12}{24} = 0.5$ A; pd $= 0.5 \times 6 = 3$ V each.
(ii) · 3 marks
One bulb fuses and is removed; the other three are reconnected in series. State the effect on (a) the total resistance, (b) the current, (c) the brightness of each bulb.
Answer
Answer: Decreases; increases; increases
(a) Resistance decreases (18 Ω). (b) Current increases ( $\frac{12}{18} \approx 0.67$ A). (c) Brightness increases, because each bulb carries more current.
(iii) · 4 marks
If all four are connected in parallel instead, find the current through each bulb and the total current. Why are houses wired in parallel?
Answer
Answer: 2 A each; 8 A
Each bulb has the full 12 V: $I = 2$ A each, total 8 A ( $R = 1.5$ Ω). In parallel, each appliance gets the full mains voltage and can be switched on or off independently; if one fails, the others keep working.

Testing two materials

In a laboratory, two materials were tested. Material X: V = 1, 2, 3, 4 V with I = 0.2, 0.4, 0.6, 0.8 A. Material Y: V = 2, 4, 6 V with I = 0.5, 0.8, 1.0 A.

(i) · 3 marks
Find V/I for each reading of X. Is X ohmic? Find its resistance.
Answer
Answer: 5 Ω each time; ohmic
V/I = 5 Ω for every reading, and the graph passes through the origin: X is ohmic with R = 5 Ω.
(ii) · 3 marks
Is Y ohmic? Justify.
Answer
Answer: No
V/I = 4, 5, 6 Ω: the ratio increases, so the V–I graph is not a straight line through the origin. Y is non-ohmic (e.g. a filament heating up).
(iii) · 4 marks
State Ohm's law fully, and suggest why Y's resistance increases at higher currents.
Answer
Answer: Statement; heating
Ohm's law: the current through a conductor is directly proportional to the pd across it, provided the temperature and other physical conditions remain unchanged. Larger currents heat Y more; for metals, resistance increases with temperature.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A torch bulb draws 0.25 A from a 5 V supply. Find its resistance. (2 marks)

Step	Marks
$R = \frac{V}{I} = \frac{5}{0.25}$	1
$= 20$ Ω	1

Three resistors of 6 Ω each are given. Find the maximum and minimum resistance obtainable and state the combinations. (3 marks)

Step	Marks
Maximum: all in series	1
$= 18$ Ω	1
Minimum: all in parallel, $\frac{6}{3} = 2$ Ω	1

A cell of emf 6 V gives a terminal voltage of 5 V when connected to a 10 Ω resistor. Find the current, the internal resistance, and the terminal voltage with a 4 Ω resistor instead. (4 marks)

Step	Marks
$I = \frac{5}{10} = 0.5$ A	1
$r = \frac{6 - 5}{0.5} = 2$ Ω	1
With 4 Ω: $I = \frac{6}{4 + 2} = 1$ A	1
$V = 1 \times 4 = 4$ V	1

Where students lose marks in Ohm's Law and Resistances

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

Mistake: Incomplete statement of Ohm's law
CISCE examiners' comments, 2024: candidates omitted 'constant temperature', gave only the formula, or wrote I ∝ R or I inversely proportional to V.
Do this: Current ∝ pd, provided temperature and other physical conditions stay the same.
Usually costs: 1 mark.
Mistake: Judging ohmic behaviour by slope alone
CISCE examiners' comments, 2024: from a V–I table, candidates looked only for a constant slope and missed that the line must pass through the origin, or used slope and ratio inconsistently.
Do this: Check V/I for every reading; an ohmic conductor gives the same ratio (line through the origin).
Usually costs: 1 to 2 marks.
Mistake: Resistivity treated like resistance
CISCE examiners' comments, 2024 and 2025: candidates thought specific resistance depends on length or area, and treated resistivity as fixed even when a conductor heats up after a current flows for a long time.
Do this: Resistivity depends on material and temperature only; heating raises both resistance and resistivity of a metal.
Usually costs: 1 mark each.
Mistake: EMF and terminal voltage mixed up
CISCE examiners' comments, 2024 and 2025: candidates missed the word 'closed' and answered e.m.f. for the pd across a cell in a closed circuit, chose V = ε + Ir, and were unclear that terminal voltage equals emf in an open circuit.
Do this: Closed circuit: V = ε − Ir (less than ε). Open circuit: V = ε.
Usually costs: 1 mark.
Mistake: Network slips and missing units
CISCE examiners' comments, 2024: wrong steps such as 1/16 + 1/16 = 1/8 = 8 Ω, missing units, and long calculations where equal branch resistances meant the current simply splits equally.
Do this: Write 1/R = 2/16, so R = 8 Ω, with the unit; look for equal branches before calculating.
Usually costs: 1 to 2 marks.
Mistake: Misreading unusual circuit diagrams
CISCE examiners' comments, 2025: candidates were confused by an unusual orientation and could not tell series from parallel.
Do this: Resistors are in parallel if both their ends join the same two points (same pd).
Usually costs: 1 mark.

Practice questions on Ohm's Law and Resistances

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
The SI unit of resistivity is:
(a) Ω
(b) Ω m
(c) Ω m⁻¹
(d) V A
Hint
ρ = RA/l.
Answer
(b)
Q2 · 1 mark · MCQBoard level
Two 4 Ω resistors are connected in parallel. Their equivalent resistance is:
(a) 8 Ω
(b) 4 Ω
(c) 2 Ω
(d) 1 Ω
Hint
Half of one.
Answer
(c)
Q3 · 1 mark · MCQHarder board level
When no current is drawn from a cell, the pd across its terminals is:
(a) zero
(b) less than its emf
(c) equal to its emf
(d) more than its emf
Hint
No lost volts.
Answer
(c)
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): The resistance of a metal wire increases if a current flows through it for a long time.
Reason (R): The wire heats up, and the resistance of a metal increases with temperature.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Heating effect.
Answer
(c)
Q5 · 2 marks · Short answerHarder board level
A charge of 1.5 C flows through a conductor in 3 s, and 9 J of work is done. Find the current and the pd.
Hint
I = Q/t; V = W/Q.
Answer
$I = 0.5$ A; $V = 6$ V
Q6 · 2 marks · Short answerHarder board level
Resistors of 2 Ω, 3 Ω and 6 Ω are available. Find the maximum and minimum equivalent resistance that can be obtained using all three.
Hint
All in series; all in parallel.
Answer
Maximum $2 + 3 + 6 = 11$ Ω (series); minimum: $\frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ , so $R = 1$ Ω (parallel).
Q7 · 3 marks · Long answerBoard level
(a) State Ohm's law. (b) Draw the V–I graph for an ohmic conductor. (c) How is resistance found from it?
Hint
Straight line through origin.
Answer
(a) Current ∝ pd at constant temperature. (b) Straight line through the origin (V on y-axis, I on x-axis). (c) Resistance = slope = ΔV/ΔI.
Q8 · 3 marks · Long answerHarder board level
A copper wire is 2 m long with cross-sectional area 1 mm². The resistivity of copper is $1.7 \times 1 0^{- 8}$ Ω m. Find its resistance.
Hint
1 mm² = 10⁻⁶ m².
Answer
$R = \frac{1.7 \times 1 0 ^{- 8} \times 2}{1 0 ^{- 6}} = 0.034$ Ω
Q9 · 4 marks · Long answerHOTS / competency
A cell gives a terminal voltage of 1.4 V when connected to a 7 Ω resistor, and 1.2 V when connected to a 3 Ω resistor. Find its emf and internal resistance.
Hint
Find the current in each case, then use ε = V + Ir twice.
Answer
Currents: $\frac{1.4}{7} = 0.2$ A and $\frac{1.2}{3} = 0.4$ A. $ε = 1.4 + 0.2 r = 1.2 + 0.4 r$ , so $0.2 r = 0.2$ , $r = 1$ Ω and $ε = 1.6$ V.
Q10 · 3 marks · Long answerHarder board level
What is a superconductor? Give one use, and explain why superconducting cables would save energy.
Hint
Zero resistance.
Answer
A material whose resistance falls to zero below a critical temperature. Use: strong electromagnets in MRI scanners (or maglev trains). With zero resistance, no heat is produced in the cable, so no energy is lost.
Q11 · 1 mark · MCQHOTS / competency
$n$ identical resistors, each of resistance $R$ , are first joined in series and then in parallel. The ratio of the series resistance to the parallel resistance is:
(a) $n$
(b) $n^{2}$
(c) $\frac{1}{n}$
(d) $\frac{1}{n ^{2}}$
Hint
nR ÷ (R/n).
Answer
(b)
Q12 · 4 marks · Long answerHOTS / competency
Two branches are connected in parallel across a 6 V supply: branch 1 has 2 Ω and 4 Ω in series; branch 2 has 3 Ω and 3 Ω in series. Find (a) the equivalent resistance, (b) the current in each branch, (c) the pd between the mid-point of branch 1 (between 2 Ω and 4 Ω) and the mid-point of branch 2.
Hint
Each branch is 6 Ω.
Answer
(a) 6 Ω ∥ 6 Ω = 3 Ω. (b) 1 A in each branch. (c) Drop across 2 Ω = 2 V, across the first 3 Ω = 3 V, so the mid-points differ by 1 V.

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.Show answer
Current through a conductor is directly proportional to the pd across it, at constant temperature.
2.Unit of resistivity?Show answer
Ω m
3.R = ? (in terms of ρ, l, A)Show answer
R = ρl/A
4.Terminal voltage in a closed circuit?Show answer
V = ε − Ir
5.Series combination formula?Show answer
R = R₁ + R₂ + …
6.Parallel combination formula?Show answer
1/R = 1/R₁ + 1/R₂ + …
7.Give an example of a non-ohmic conductor.Show answer
Filament lamp (or diode).

Best order to study it

Charge, current, pd and resistance.
Ohm's law, its verification and V–I graphs.
Ohmic and non-ohmic conductors.
Factors affecting resistance; resistivity; superconductors.
EMF, terminal voltage and internal resistance.
Series, parallel and simple networks.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you state Ohm's law with its condition?
Can you decide ohmic or non-ohmic from data?
Do you know R = ρl/A and what ρ depends on?
Can you use V = ε − Ir?
Can you derive the series and parallel formulas?
Can you solve a network step by step with units?

Ohm's Law and Resistances: frequently asked questions

Is the derivation of the series and parallel formulas needed?

Yes. The syllabus includes the derivation of the expressions for equivalent resistance.

Why does a filament lamp not obey Ohm's law?

As the current increases, the filament gets much hotter and its resistance rises, so V/I is not constant.

Is emf a force?

No. Despite its name, emf is the energy given per coulomb by the cell, measured in volts.

What is the resistance of an ideal ammeter and voltmeter?

An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance, so neither disturbs the circuit.

Does resistivity change with temperature?

Yes. For metals both resistance and resistivity increase with temperature.

Based on: CISCE ICSE 2027 Physics (Science Paper 1) syllabus; CISCE ICSE 2027 Physics (Science Paper 1) specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Physics (Science Paper 1), 2025; CISCE Analysis of Pupil Performance, ICSE Physics (Science Paper 1), 2024