Ohm's Law, Resistance, and Factors Affecting Resistance for CBSE Class 10

Why does an electric toaster glow red-hot and toast bread, while the copper cord connecting it to the wall outlet remains completely cool to the touch? Both elements carry the exact same electrical current from the household power grid. The secret behind this stark difference in thermal behavior lies in an intrinsic electrical property possessed by all material conductors: electrical resistance.

In CBSE Class 10 Science, Chapter 11 (Electricity) introduces the single most important law in electrical physics: Ohm's Law. Understanding the linear relationship between voltage and current, the nature of resistance, and the material property called resistivity ($\rho$) is essential for mastering circuit analysis.

What You Will Learn

• Statement and experimental verification of Ohm's Law: $V = IR$
• The linear $V-I$ characteristic graph and calculating resistance from its slope
• Definition and SI unit of Resistance ($R$): The Ohm ($\Omega$)
• The four physical factors governing electrical resistance
• The master equation of resistance: $R = \rho \frac{l}{A}$
• Definition, formula, and units of Resistivity (Specific Resistance, $\rho$)
• Why heating appliances use alloys (Nichrome) rather than pure metals
• Wire stretching problems: What happens when length is doubled?
• Solved CBSE board examination numerical problems and common traps

1. Ohm's Law: The Voltage-Current Relationship

In 1827, German physicist Georg Simon Ohm performed a classic laboratory experiment measuring the current flowing through a metallic wire as the potential difference across its ends was systematically varied:

                            V-I Characteristic Graph
                     Potential
                     Difference (V)
                          ^              / Slope = ΔV / ΔI = Resistance R
                          |             /
                          |            /
                          |           /
                          |          /
                          |         /
                          +--------+----------------> Current (I)
                          O (Origin)

Statement of Ohm's Law

The electric current ($I$) flowing through a metallic conductor is directly proportional to the potential difference ($V$) applied across its terminals, provided its temperature and other physical conditions remain constant.

$\mathbf{V \propto I} \quad \Longleftrightarrow \quad \mathbf{V = I \times R}$ where $R$ is a constant of proportionality called the resistance of the conductor.

2. What is Electrical Resistance?

Physical Definition

Resistance is the intrinsic property of a conductor by virtue of which it opposes or resists the flow of electric charges (electrons) through it.

• Microscopic Cause of Resistance: As free electrons drift through a metallic crystal lattice under an electric field, they constantly collide with positive metal ions. These collisions impede the forward drift of electrons, generating resistance and dissipating kinetic energy as heat.

Mathematical Formula and SI Unit:

$R = \frac{V}{I}$

• The SI unit of resistance is the Ohm, represented by the Greek letter $\mathbf{\Omega}$.
• Definition of $1\text{ Ohm}$:

  <u>One Ohm ($1\ \Omega$) is the resistance of a conductor such that when a potential difference of 1 Volt is applied across its ends, a current of 1 Ampere flows through it ($1\ \Omega = 1 ext{ V/A}$).</u>

Slope of the $V-I$ Graph:

• In a $V-I$ graph (with $V$ on the $y$-axis and $I$ on the $x$-axis), the line is a straight line passing through the origin.
• Slope of $V-I$ graph = Resistance ($R$): $\mathbf{\text{Slope} = \frac{\Delta V}{\Delta I} = R}$
• A steeper line indicates a higher resistance.

3. Factors on Which Resistance of a Conductor Depends

Careful laboratory experiments show that the electrical resistance of a uniform conductor depends on four distinct parameters:

1. Length of the Conductor ($l$): Resistance is directly proportional to length: $\mathbf{R \propto l}$ Doubling the length of a wire doubles its resistance.
2. Area of Cross-Section ($A$): Resistance is inversely proportional to cross-sectional area (thickness): $\mathbf{R \propto \frac{1}{A}}$ A thick wire has less resistance than a thin wire of the same material and length because a wider cross-section provides more pathways for electrons to flow.
3. Nature of the Material: Resistance depends on the atomic structure of the material (captured by resistivity $\rho$).
4. Temperature: For pure metallic conductors, resistance increases as temperature increases (higher thermal vibrations cause more frequent electron-lattice collisions).

4. The Resistivity Equation

Combining the length and cross-sectional area dependencies: $R \propto \frac{l}{A} \quad \implies \quad \mathbf{R = \rho \frac{l}{A}}$ where $\rho$ (rho) is a constant of proportionality called the electrical resistivity (or specific resistance) of the material.

$\mathbf{\rho = \frac{R \times A}{l}}$

Definition and SI Unit of Resistivity:

Resistivity is defined as the resistance offered by a cylindrical conductor of the material having unit length ($1\text{ m}$) and unit cross-sectional area ($1\text{ m}^2$).

• SI Unit of Resistivity: $\text{Unit of } \rho = \frac{\Omega \times \text{m}^2}{\text{m}} = \mathbf{\Omega \cdot \text{m} \text{ (Ohm-metre)}}$

Important: <u>Resistivity is an INTRINSIC material property! Resistivity depends ONLY on the nature of the material and temperature. Resistivity DOES NOT change when the length, thickness, or shape of the wire is altered!</u>

5. Conductors vs. Alloys vs. Insulators

Why Are Alloys Used in Electric Heating Devices? (CBSE High-Frequency Question)

1. Alloys (such as Nichrome—an alloy of nickel, chromium, manganese, and iron) have a much higher resistivity than their constituent pure metals, producing abundant Joule heat ($H = I^2Rt$).
2. <u>Alloys do not oxidize (burn) easily even at high red-hot temperatures ($800^\circ ext{C}$), preventing the heating element from melting or deteriorating!</u>

6. Solved CBSE Board Examination Problems

Solved Example 1: Wire Stretching / Doubling Problem (Board Classic)

Problem: A $4\ \Omega$ resistance wire is doubled on it. Calculate the new resistance of the wire.

Solution:

1. Understand "Doubled on it": Folding a wire in half ("doubled on it") halves its length and doubles its cross-sectional area!
   • Initial state: Length $l$, Area $A$. $R_1 = \rho \frac{l}{A} = 4\ \Omega$
   • New state: New length $l' = \frac{l}{2}$, New cross-sectional area $A' = 2A$.
2. Calculate New Resistance ($R_2$): $R_2 = \rho \frac{l'}{A'} = \rho \frac{l/2}{2A} = \rho \frac{l}{4A} = \frac{1}{4} \left(\rho \frac{l}{A}\right)$
3. Substitute $R_1 = 4\ \Omega$: $R_2 = \frac{1}{4} \times 4 = \mathbf{1\ \Omega}$
4. Therefore, <u>the new resistance of the wire is $1\ \Omega$</u>.

Solved Example 2: Wire Stretched to Double Its Length

Problem: A wire of resistance $R$ is stretched so that its length is doubled. What will be its new resistance and new resistivity?

Solution:

1. Analyze Volume Conservation: When a wire is stretched, its total physical volume remains constant: $\text{Volume} = A \times l = A' \times l'$ Since new length $l' = 2l$: $A \times l = A' \times (2l) \implies A' = \frac{A}{2}$
2. Compute New Resistance ($R'$): $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = \rho \frac{4l}{A} = 4 \left(\rho \frac{l}{A}\right) = \mathbf{4R}$
3. New Resistivity: Since resistivity is an intrinsic material property independent of dimensions, <u>the new resistivity remains completely UNCHANGED ($ho$)</u>!
4. Therefore, <u>the new resistance is $4R$, while its resistivity remains identical</u>.

7. Summary and Examination Tips

Exam Tip: In questions asking: "If a wire is stretched to double its length, what happens to its resistivity?", the answer is NO CHANGE! Many students incorrectly calculate $4\rho$; remember that resistance quadruples ($4R$), but resistivity remains constant!

Common Mistake: Forgetting that when a wire's radius is halved ($r' = r/2$), its area becomes one-fourth ($A' = \pi (r/2)^2 = A/4$). Since $R \propto 1/A$, resistance increases by $16$ times ($2 \times 2 \times 4$) if stretched to maintain volume!