TOPIC 41 OF 50

Ohm’s Law, Resistance and Resistivity

🎓 Class 10 Science CBSE Theory Ch 11 — Electricity ⏱ ~20 min
🌐 Language: [gtranslate]

This MCQ module is based on: Ohm’s Law, Resistance and Resistivity

[myaischool_lt_science_assessment grade_level="class_10" science_domain="physics" difficulty="intermediate"]

Introduction — The Push and the Push-Back

In Part 1 we saw that a cell supplies a potential difference that pushes charges through a wire. But how much current actually flows? That depends on how strongly the wire resists the flow. In this part we study the simple but powerful link between voltage, current and resistance — discovered by the German physicist Georg Simon Ohm in 1827.

11.5 Ohm's Law

Ohm carried out careful experiments on metallic conductors at constant temperature and found that current through a conductor is directly proportional to the potential difference across its ends.

Ohm's Law: At constant temperature, the potential difference V across the ends of a metallic conductor is directly proportional to the current I flowing through it. \[ V \propto I \qquad \Rightarrow \qquad V = IR \] The constant of proportionality R is called the resistance of the conductor.
Unit of R: ohm (Ω). \(1\ \Omega = 1\ \text{V/A}\).

Rearranging Ohm's law:

\[ I = \dfrac{V}{R}, \qquad R = \dfrac{V}{I} \]

11.5.1 Resistance — What It Means

Every conductor opposes the drift of electrons to some extent. This opposition is called electrical resistance. Inside a wire, free electrons keep colliding with the vibrating atoms of the metal, converting some of the electrical energy into heat.

Resistance (R): The ratio of the potential difference applied across a conductor to the current flowing through it. \[ R = \dfrac{V}{I} \] A conductor has resistance of 1 ohm if a potential difference of 1 volt across it drives a current of 1 ampere through it.

11.5.2 V-I Graph for an Ohmic Conductor

When we plot V (x-axis) against I (y-axis) for a metallic conductor at a fixed temperature, we obtain a straight line passing through the origin. Its slope equals \(1/R\). Conductors that obey this law over a wide range of voltages are called ohmic conductors (e.g., copper, nichrome).

I (A) V (V) Ohmic conductor slope = 1/R I (A) V (V) Non-ohmic (filament bulb) R increases as T rises
Fig 11.4 — V–I graph. Left: metallic conductor at constant temperature gives a straight line. Right: a filament bulb heats up as V rises; its R increases, so the curve bends.

11.5.3 Non-Ohmic Conductors

Many real devices do not give a straight V–I line. Examples: tungsten filament bulbs (resistance rises sharply with temperature), semiconductor diodes, electrolytes, transistors, LEDs. These are called non-ohmic conductors.

Worked Example 1 — Resistance from V and I

Q. A potential difference of 12 V across a resistor drives a current of 0.4 A. Find its resistance.

Solution. \(R = V/I = 12/0.4 = 30\ \Omega\).

Worked Example 2 — Current Through a Resistor

Q. Find the current flowing through a 5 Ω resistor when connected to a 10 V battery.

Solution. \(I = V/R = 10/5 = 2\) A.

11.6 Factors Affecting the Resistance of a Conductor

Careful experiments show that the resistance of a uniform wire depends on four things:

  1. Length (L) of the wire — R ∝ L. Doubling the length doubles the resistance.
  2. Area of cross-section (A) — R ∝ 1/A. A thicker wire offers less resistance.
  3. Nature of the material — silver and copper conduct better than iron or nichrome.
  4. Temperature — for metals, R increases with temperature because lattice vibrations intensify and obstruct electron drift.
Length L → R = R₀ Length 2L (same thickness) → R = 2R₀ Thin wire: small A → larger R Thick wire: large A → smaller R R ∝ L R ∝ 1/A
Fig 11.5 — Doubling the length doubles R; doubling the cross-section halves R.

11.6.1 Resistivity ρ

Combining the first three factors, we write:

R = ρL/A \[ R = \rho\,\dfrac{L}{A} \] Here ρ (rho) is the resistivity or specific resistance of the material. It depends only on the material (and its temperature), not on the size or shape of the sample.
SI unit of ρ: ohm-metre (Ω·m). Numerically, ρ is the resistance of a 1-metre-long wire of 1 m² cross-section of that material.

11.6.2 Table of Resistivities at 20 °C

MaterialTypeResistivity ρ (Ω·m)
SilverConductor1.60 × 10⁻⁸
CopperConductor1.62 × 10⁻⁸
AluminiumConductor2.63 × 10⁻⁸
TungstenConductor (for filaments)5.20 × 10⁻⁸
NickelConductor6.84 × 10⁻⁸
IronConductor10.0 × 10⁻⁸
Manganin (alloy)Alloy (std. resistance wire)44 × 10⁻⁸
Constantan (alloy)Alloy49 × 10⁻⁸
Nichrome (alloy)Alloy (heating element)100 × 10⁻⁸
MercuryLiquid conductor94 × 10⁻⁸
GlassInsulator10¹⁰ – 10¹⁴
Hard rubberInsulator10¹³ – 10¹⁶

Observations: Silver has the lowest resistivity and is therefore the best conductor, but copper (slightly higher ρ, much cheaper) is used for wiring. Alloys like nichrome have resistivities 30–50 times higher than pure metals, yet they do not oxidise readily at high temperatures — that is why they are used in heating elements of electric irons, toasters and heaters.

11.6.3 Conductors, Insulators and Semiconductors

  • Good conductors — resistivity \(\sim 10^{-8}\) Ω·m. E.g., silver, copper, aluminium.
  • Insulators — resistivity \(\sim 10^{12}\) Ω·m or more. E.g., glass, rubber, mica, dry wood.
  • Semiconductors — resistivity between these extremes (\(\sim 10^{-3}\) to \(10^{3}\) Ω·m). E.g., silicon, germanium. Their resistivity decreases with temperature (opposite to metals), which makes them the basis of diodes, transistors and ICs.

11.7 Worked Numericals on Resistance and Resistivity

Numerical 1 — Resistivity from R, L and A

Q. A copper wire of length 2 m and cross-sectional area \(1.0 \times 10^{-7}\) m² has a resistance of 0.32 Ω. Find the resistivity of copper.

Solution. From \(R = \rho L/A\):

\[ \rho = \dfrac{RA}{L} = \dfrac{0.32 \times 1.0 \times 10^{-7}}{2} = 1.6 \times 10^{-8}\ \Omega\cdot\text{m}. \]

Numerical 2 — Length of a Heating Wire

Q. What length of nichrome wire of resistivity \(1.0 \times 10^{-6}\) Ω·m and area of cross-section \(2.0 \times 10^{-7}\) m² is required to make a coil of resistance 50 Ω?

Solution. \(L = \dfrac{RA}{\rho} = \dfrac{50 \times 2.0 \times 10^{-7}}{1.0 \times 10^{-6}} = 10\) m.

Numerical 3 — Doubling the Length of a Wire

Q. A wire has resistance 10 Ω. It is stretched so that its length becomes double (volume remains constant). Find its new resistance.

Solution. When length doubles, area halves (since volume \(L \times A\) = constant). New resistance:

\[ R' = \rho \dfrac{L'}{A'} = \rho \dfrac{2L}{A/2} = 4\,\rho\,\dfrac{L}{A} = 4R = 4 \times 10 = 40\ \Omega. \]

Numerical 4 — Comparing Two Wires

Q. Two wires of the same material have the same length but different diameters in the ratio 1 : 2. Compare their resistances.

Solution. Diameters in ratio 1:2, so areas are in ratio 1:4 (since \(A \propto d^2\)). \(R \propto 1/A\), hence \(R_1 : R_2 = 4 : 1\).

Numerical 5 — Resistance of a Parallel Wire Setup (length-based)

Q. A copper wire has length 1 m and area of cross-section \(1.0 \times 10^{-7}\) m². If ρ = 1.6 × 10⁻⁸ Ω·m, find the resistance. What will the resistance be if the length is tripled?

Solution. \(R = \rho L/A = (1.6 \times 10^{-8}) \times 1 / (1.0 \times 10^{-7}) = 0.16\) Ω. If length triples (same area), \(R' = 3R = 0.48\) Ω.

Numerical 6 — Current from V, Length and Area

Q. A nichrome wire of length 4 m and cross-sectional area \(2 \times 10^{-7}\) m² (ρ = 1.0 × 10⁻⁶ Ω·m) is connected to a 12 V battery. Find the current through it.

Solution. \(R = \rho L/A = (1.0 \times 10^{-6}) \times 4 / (2 \times 10^{-7}) = 20\) Ω. \(I = V/R = 12/20 = 0.6\) A.

Activity 11.2 — Verifying Ohm's LawL3 Apply

Aim: To show that V ∝ I for a metallic resistor at constant temperature.

Materials: Nichrome wire (∼ 5 Ω), four 1.5 V dry cells, an ammeter (0–1 A), a voltmeter (0–5 V), a plug key, connecting wires.

Procedure:

  1. Build the circuit: cells → plug key → ammeter → nichrome resistor. Connect the voltmeter in parallel across the resistor.
  2. Plug in one cell only. Record the ammeter and voltmeter readings.
  3. Repeat with two, three and four cells in series, each time noting V and I.
  4. Tabulate V and I; calculate V/I for each row. Plot V (x-axis) vs I (y-axis).
Predict: What shape will the V–I plot take? What does the value V/I represent? Will it stay constant?

The plotted points lie on a straight line through the origin, confirming \(V \propto I\). The ratio V/I comes out nearly constant — this ratio is the resistance R of the nichrome wire. Tiny variations are due to warming of the wire as more cells are added. Ohm's law is thus verified for a metallic conductor at steady temperature.

Competency-Based Questions

An electrician needs to choose a wire for the heating coil of an electric iron. She has samples of (i) copper, (ii) iron, and (iii) nichrome of identical length and thickness. She tests their resistances at room temperature and finds the nichrome sample has the highest resistance.
Q1. State Ohm's law in words and as an equation. L1 Remember
At constant temperature, the current through a metallic conductor is directly proportional to the potential difference across its ends: V = IR, where R is the resistance.
Q2. (MCQ) In a conductor obeying Ohm's law, if V is doubled, the current will L2 Understand
  • (a) be halved
  • (b) remain the same
  • (c) be doubled
  • (d) become four times
(c) Doubled. Since I = V/R and R is constant, current scales linearly with V.
Q3. Why is nichrome chosen over copper for the heating element? Give TWO reasons. L2 Understand
(i) Nichrome has a high resistivity (~100 × 10⁻⁸ Ω·m vs copper's ~1.6 × 10⁻⁸ Ω·m), so it produces more heat for the same current. (ii) Nichrome has a high melting point and does not oxidise readily at red-hot temperatures, so it lasts long.
Q4. (Numerical) Find the resistance of a copper wire of length 50 m and radius 0.5 mm. ρ = 1.6 × 10⁻⁸ Ω·m. L3 Apply
Area \(A = \pi r^2 = 3.14 \times (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}\) m². \(R = \rho L/A = (1.6 \times 10^{-8}) \times 50 / (7.85 \times 10^{-7}) ≈ 1.02\) Ω.
Q5. (HOT) Why are copper wires used for household wiring even though silver has slightly lower resistivity? L4 Analyse
Silver is much more expensive and softer than copper. Copper has only marginally higher resistivity (1.62 vs 1.60 × 10⁻⁸ Ω·m), excellent ductility, is cheap and durable — so it is the practical choice for wiring.

Assertion–Reason Questions

Options: (A) Both A & R true, R correctly explains A. (B) Both A & R true, R does NOT explain A. (C) A true, R false. (D) A false, R true.

Assertion (A): A thick wire has lower resistance than a thin wire of the same material and length.
Reason (R): Resistance is inversely proportional to the cross-sectional area of the conductor.
(A) — Both true and R correctly explains A. From R = ρL/A.
Assertion (A): Alloys like nichrome are used in heating appliances.
Reason (R): Alloys have low resistivity and are easily melted.
(C) — A is true but R is false. Alloys have high resistivity and high melting points, which is why they are preferred for heating elements.
Assertion (A): The V–I graph of a filament bulb is not a straight line.
Reason (R): Resistance of the tungsten filament increases with temperature as the current heats it.
(A) — Both true; R correctly explains A. Since R is not constant, the V–I plot curves.

Frequently Asked Questions — Ohm's Law, Resistance & Resistivity

What is ohm's law, resistance & resistivity in Class 10 Science (CBSE board)?

Ohm's Law, Resistance & Resistivity is a key topic in NCERT Class 10 Science Chapter 11 — Electricity. It explains ohm's law v = ir, factors affecting resistance and the concept of resistivity. Core ideas covered include Ohm's law, resistance, ohm, resistivity. Mastering this subtopic is essential for scoring well in the CBSE Class 10 Science board exam because board papers repeatedly test these concepts through MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.

Why is Ohm's law important in NCERT Class 10 Science?

Ohm's law is important in NCERT Class 10 Science because it forms the foundation for understanding ohm's law, resistance & resistivity in Chapter 11 — Electricity. Without a clear idea of Ohm's law, students cannot answer higher-order CBSE board questions involving resistance, ohm, resistivity. Board papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link Ohm's law to real-life situations. Building clarity here pays off directly in board marks.

How is ohm's law, resistance & resistivity tested in the Class 10 Science CBSE board exam?

The CBSE Class 10 Science board exam tests ohm's law, resistance & resistivity through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or balanced equations) and 4-mark competency-based questions. Expect direct questions on Ohm's law, resistance, ohm and application-based questions drawn from NCERT activities. Students who follow NCERT thoroughly and practice this chapter's questions consistently score in the 90%+ range.

What are the key terms to remember for ohm's law, resistance & resistivity in Class 10 Science?

The key terms to remember for ohm's law, resistance & resistivity in NCERT Class 10 Science Chapter 11 are: Ohm's law, resistance, ohm, resistivity, factors affecting resistance, length. Each of these concepts carries exam weightage and regularly appears in the CBSE board paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 10 Science board exam.

Is Ohm's Law, Resistance & Resistivity included in the Class 10 Science syllabus for 2025–26 CBSE board exam?

Yes, Ohm's Law, Resistance & Resistivity is a part of the NCERT Class 10 Science syllabus (2025–26) prescribed by CBSE. It falls under Chapter 11 — Electricity — and is examined in the annual board paper. The current syllabus retains the full treatment of Ohm's law, resistance, ohm as per the NCERT textbook. Because CBSE bases every board question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.

How should I prepare ohm's law, resistance & resistivity for the CBSE Class 10 Science board exam?

Prepare ohm's law, resistance & resistivity for the CBSE Class 10 Science board exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of Ohm's law, resistance, ohm. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practice competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the board exam.

AI Tutor
Science Class 10 — NCERT (2024-25)
Ready
Hi! 👋 I'm Gaura, your AI Tutor for Ohm’s Law, Resistance and Resistivity. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.