What four physical factors determine the electrical resistance of a conductor?

Resistance R = \rho \frac{l}{A}, where: 1. **Length (l):** R \propto l (doubling length doubles resistance). 2. **Cross-sectional Area (A):** R \propto \frac{1}{A} (thicker wire has lower resistance). 3. **Nature of material (Resistivity \rho):** Characteristic property of the substance. 4. **Temperature:** Resistance of pure metals increases with rising temperature.

What are the equivalent resistance formulas for Resistors in Series versus Resistors in Parallel?

- **Series Combination:** R_s = R_1 + R_2 + R_3 (same current I flows through each; total resistance is greater than the largest individual resistor). - **Parallel Combination:** \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} (same potential difference V across each; total equivalent resistance is smaller than the smallest individual resistor).

What does Joule’s Law of Heating state and what is its mathematical formula?

Joule’s Law states that heat produced in a resistor is directly proportional to: (1) Square of current (I^2), (2) Resistance (R), and (3) Time (t) for which current flows: H = I^2Rt = VIt = \frac{V^2}{R}t\text{ (in Joules)}

Why are heating elements in toasters, geysers, and electric irons made of Nichrome alloy rather than pure copper?

1. Nichrome (an alloy of \text{Ni, Cr, Mn, Fe}) has very **high electrical resistivity** compared to pure metals, generating high heat (H = I^2Rt). 2. It has a **high melting point** and does not undergo oxidation (burn) even at glowing red-hot temperatures (800-1000^\circ\text{C}).

What is the commercial unit of electrical energy and how many Joules does $1\text{ kWh}$ contain?

The commercial unit is **Kilowatt-hour (kWh)**, commonly called a "unit" of electricity. 1\text{ kWh} = 1\text{ kW} \times 1\text{ h} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ Joules (J)}

Why are domestic electrical appliances always connected in parallel and never in series?

1. Each appliance receives the full mains supply voltage (220\text{ V} in India). 2. Each appliance operates independently with its own on/off switch; if one appliance fails or burns out, other appliances continue working uninterrupted. 3. The total equivalent resistance is low, allowing sufficient current draw for power-hungry devices.

All Revision Notes

ScienceCh-11 15 min comprehensive revision

NCERT Class 10 Science — Chapter 12

Electricity

Electric current ( $I = Q/t$ ), potential difference ( $V = W/Q$ ), Ohm's law ( $V = IR$ ), resistivity factors ( $\rho, l, A$ ), series & parallel resistor combinations, Joule's law of heating ( $H = I^2Rt$ ), electric power ( $P = VI = I^2R = V^2/R$ ), and commercial energy units (kWh).

Quick Key Takeaways:

Current & Potential Difference:

I = \frac{Q}{t} \text{ (Amperes, A)}, \quad V = \frac{W}{Q} \text{ (Volts, V)}

Ohm's Law & Resistance:

V = I R \implies R = \frac{V}{I}, \quad R = \rho \frac{l}{A}

(Resistivity

\rho

depends strictly on the material and temperature, measured in

\Omega \cdot \text{m}

Series vs Parallel Resistor Combinations:
- Series: Current

I

is identical through all resistors;

R_s = R_1 + R_2 + R_3 + \dots

- Parallel: Voltage

V

is identical across all branches;

\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots

Joule's Law of Heating & Electric Power:

H = I^2 R t = V I t = \frac{V^2}{R} t, \quad P = V I = I^2 R = \frac{V^2}{R} \text{ (Watts, W)}

Commercial Energy Unit:

1\text{ kWh (Unit)} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ Joules}

Physics Numerical CalculatorCompute equivalent resistance, Joule heating, and electric power

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Jump to Concept:1. 1. Current, Potential, Ohm's Law & Resistivity Factors 2. 2. Series/Parallel Derivations, Joule's Heating & Electrical Energy 3. 3. High-Yield Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. Practical Laboratory & Competency-Based Case Drill: Stretching a Cylindrical Wire: Resistance & Resistivity (NCERT Problem)5. 5. CBSE Examiner Marking Scheme, Scientific Notation & Deduction Traps

1. Current, Potential, Ohm's Law & Resistivity Factors

Fundamental Principles

Core scientific laws, chemical equations, anatomical structures, and visual model for Electricity.

• Ohm's Law & Factors Affecting Electrical Resistance

•

Statement of Ohm's Law: The electric current flowing through a conductor is directly proportional to the potential difference across its ends, provided temperature and other physical conditions remain constant:

V \propto I \implies V = IR

•

Factors Affecting Resistance ( $R = \rho \frac{l}{A}$ ):
1. Length ( $l$ ):

R \propto l

(doubling length doubles resistance).
2. Cross-Sectional Area ( $A = \pi r^2$ ):

R \propto \frac{1}{A}

(thick wire has lower resistance than thin wire).
3. Material Nature ( $\rho$ ): Conductors (Copper, Aluminium) have very low

\rho

(

10^{-8}\,\Omega\cdot\text{m}

); Alloys (Nichrome, Manganin) have high

\rho

and do not oxidize easily at high temperatures (used in electric irons, heaters, toasters); Insulators have extremely high

\rho

(

10^{12}-10^{17}\,\Omega\cdot\text{m}

📊 Electricity: Ohm's Law Circuit & Series-Parallel NetworksVisual Model

Visual schematic mapping the ammeter-voltmeter circuit diagram, linear V-I Ohm slope, series vs parallel resistor networks, and Joule heating power equations.

2. Series/Parallel Derivations, Joule's Heating & Electrical Energy

Mechanisms & Experiments

Step-by-step chemical reaction mechanisms, experimental activities, and physiological pathways for Electricity.

• Derivations of Equivalent Resistance

•

Series Derivation: Total voltage

V = V_1 + V_2 + V_3

. Since current

I

is constant:

I R_s = I R_1 + I R_2 + I R_3 \implies \mathbf{R_s = R_1 + R_2 + R_3}

•

Parallel Derivation: Total current

I = I_1 + I_2 + I_3

. Since voltage

V

is constant:

\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} \implies \mathbf{\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}

•

Why Domestic Circuits are Connected in Parallel (Not Series):
1. In parallel, each appliance gets the full rated line voltage (220 V).
2. Each appliance operates independently with its own on/off switch.
3. If one appliance breaks down or burns out, all other appliances continue functioning.
4. Total effective resistance decreases, allowing sufficient current flow for heavy load appliances.

3. High-Yield Solved Board Examination Questions (3-Mark & 5-Mark)

Solved Board Questions

Standard CBSE board exam questions with complete scientific justifications and marking scheme step protocols.

• 3-Mark Standard Board Question: An electric lamp of resistance 20 Ω and a conductor of 4 Ω resistance are connected in series to a 6 V battery.
(a) Calculate the total resistance of the circuit.
(b) Calculate the current flowing through the circuit.
(c) Calculate the potential difference across the electric lamp and the conductor.

•

(a) Total Resistance ( $R_s$ ):
In series:

R_s = R_{\text{lamp}} + R_{\text{conductor}} = 20\,\Omega + 4\,\Omega = \mathbf{24\,\Omega}

•

(b) Total Current ( $I$ ):

I = \frac{V}{R_s} = \frac{6\text{ V}}{24\,\Omega} = \mathbf{0.25\text{ A}}

•

(c) Potential Difference Across Each Component:
- Across Conductor (

4\,\Omega

V_1 = I R_1 = 0.25\text{ A} \times 4\,\Omega = \mathbf{1.0\text{ V}}

.
- Across Lamp (

20\,\Omega

V_2 = I R_2 = 0.25\text{ A} \times 20\,\Omega = \mathbf{5.0\text{ V}}

.
(Check: $V_1 + V_2 = 1.0 + 5.0 = 6.0\text{ V} = V_{\text{total}}$ ).

• 5-Mark Comprehensive Question / Numerical: An electric refrigerator rated 400 W operates 8 hours/day and an electric television rated 100 W operates 6 hours/day. What is the cost of energy to operate them for 30 days at Rs 3.00 per kWh?

•

Step 1 (Daily Energy Consumed by Refrigerator):

E_1 = P_1 \times t_1 = 400\text{ W} \times 8\text{ h} = 3200\text{ Wh} = 3.2\text{ kWh/day}

•

Step 2 (Daily Energy Consumed by TV):

E_2 = P_2 \times t_2 = 100\text{ W} \times 6\text{ h} = 600\text{ Wh} = 0.6\text{ kWh/day}

•

Step 3 (Total Daily Energy Consumption):

E_{\text{daily}} = 3.2 + 0.6 = 3.8\text{ kWh/day}

•

Step 4 (Total Energy Consumed in 30 Days):

E_{\text{total}} = 3.8\text{ kWh/day} \times 30\text{ days} = \mathbf{114\text{ kWh (Units)}}

•

Step 5 (Total Electricity Bill Cost):

\text{Cost} = 114\text{ kWh} \times \text{Rs } 3.00 = \mathbf{\text{Rs } 342.00}

•

Final Boxed Answer:

\mathbf{\text{Total Cost of Energy} = \text{Rs } 342}

4. Practical Laboratory & Competency-Based Case Drill: Stretching a Cylindrical Wire: Resistance & Resistivity (NCERT Problem)

Practical & Case Drill

Experimental observation analysis, chemical gas tests, and assertion-reason drills.

• Laboratory Activity Context: Stretching a Cylindrical Wire: Resistance & Resistivity (NCERT Problem)

A cylindrical metal wire of resistance $R$ , length $l$ , and area of cross-section $A$ is stretched to double its original length ( $l' = 2l$ ) without changing its mass.

•

Q1: How does the cross-sectional area change upon stretching?

\rightarrow

Since the volume of the wire remains constant (

V = A \cdot l = A' \cdot l'

A' \cdot (2l) = A \cdot l \implies A' = \mathbf{A/2}

(area is halved).

•

Q2: Calculate the new resistance ( $R'$ ) in terms of original resistance ( $R$ ).

\rightarrow

R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4\left(\rho \frac{l}{A}\right) = \mathbf{4R}

. (Resistance increases by a factor of 4).

•

Q3: How does the resistivity ( $\rho$ ) of the material change?

\rightarrow

Resistivity remains completely unchanged because resistivity is an intrinsic material property that depends only on the nature of substance and temperature, not on dimensions.

5. CBSE Examiner Marking Scheme, Scientific Notation & Deduction Traps

Important Solved Board Questions

Examiner step-marking allocations, mandatory scientific terminology, and common error avoidance.

• Step-by-Step Marking Rubric & Key Terminology

•

1 Mark: Statement and mathematical formulation of Ohm's Law.

•

2 Marks: Step-by-step derivation of series (

R_s = R_1+R_2

) or parallel (

1/R_p = 1/R_1+1/R_2

) formulas.

•

1 Mark: Accurate unit conversion from Watts/hours to commercial kilowatt-hours (

1\text{ kWh} = 3.6\times 10^6\text{ J}

•

1 Mark: Boxed numerical answers with correct electrical units (A, V,

\Omega

, W, kWh).

• Common Error Deduction Traps

•

Trap 1: Assuming that stretching a wire leaves area unchanged (stretching doubles length AND halves cross-sectional area, making resistance

4\times

•

Trap 2: Stating that resistivity changes when wire is cut or stretched (resistivity is constant for a material).

•

Trap 3: Calculating commercial energy using seconds instead of hours (kWh requires power in kW and time in hours).

Authentic Board Question (3 Marks)Topic: Electricity Laws of Physics & Numerical Problem Solving

State the governing physical law, write the standard formula with Cartesian sign conventions, and solve the numerical/diagram application for Electricity.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Law / Formula & Sign Convention Setup: State the formal governing physical law (Ohm's Law, Joule's Heating, Mirror/Lens formula) with Cartesian sign conventions (

u, v, f

1 Mark

Step 2: Step-by-Step Algebraic Substitution & Calculation: Substitute given values systematically showing all intermediate algebraic simplification steps.

1 Mark

Step 3: Boxed Final Answer with Proper SI Units & Direction: State the final numerical result clearly boxed with mandatory SI units (

\Omega, \text{V}, \text{A}, \text{W}, \text{J}, \text{cm}, \text{D}

) and ray/field directional arrows.

1 Mark

Model Student Answer (Target: Full 3/3 Marks):

To score full 3 marks on Electricity in CBSE Science (Physics):

1. Formula & Conventions: Write the governing formula (e.g.,

V = IR

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

) and assign correct signs to given quantities.
2. Calculation Steps: Substitute values clearly and show each arithmetic reduction line.
3. Final Result: Box the final numerical value with mandatory SI units (e.g.

R = 10\,\Omega

P = 100\,\text{W}

f = -15\,\text{cm}

Examiner Mark Deduction Traps:

•Always assign Cartesian sign conventions before substituting into mirror/lens formulas.

•Never write a pure number without its mandatory SI unit—examiners deduct ½ mark for missing units.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.

Ohm’s Law states: The electric current flowing through a metallic conductor is directly proportional to the potential difference across its ends, provided temperature and physical conditions remain constant:

V = IR

. A linear

V-I

graph passing through the origin verifies Ohm’s Law, where the slope

\frac{V}{I}

equals resistance

R

Electricity

1. Current, Potential, Ohm's Law & Resistivity Factors

2. Series/Parallel Derivations, Joule's Heating & Electrical Energy

3. High-Yield Solved Board Examination Questions (3-Mark & 5-Mark)

4. Practical Laboratory & Competency-Based Case Drill: Stretching a Cylindrical Wire: Resistance & Resistivity (NCERT Problem)

5. CBSE Examiner Marking Scheme, Scientific Notation & Deduction Traps

Official CBSE Step-by-Step Marking Breakdown:

High-Frequency Conceptual Doubts & FAQs

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