Electricity

Theory/Concept

• Resistance of a wire is directly proportional to its length: \[ R \propto L \]
• When a wire is cut into equal parts, resistance reduces in the same ratio.
• For resistors in parallel: \[ \frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} + \dots \]
• If \( n \) identical resistors each of resistance \( r \) are connected in parallel: \[ R' = \frac{r}{n} \]

Solution Roadmap

1. Find resistance of each piece after cutting.
2. Apply parallel combination formula.
3. Calculate equivalent resistance.
4. Compute ratio \( \frac{R}{R'} \).

Step-by-Step Solution

Step 1: Original resistance

\[ R \]

Step 2: After cutting into 5 equal parts

\[ R_n = \frac{R}{5} \]

Step 3: Apply parallel formula

\[ \frac{1}{R'} = \frac{1}{R_n} + \frac{1}{R_n} + \frac{1}{R_n} + \frac{1}{R_n} + \frac{1}{R_n} \]

\[ \frac{1}{R'} = 5 \times \frac{1}{R_n} \]

Step 4: Substitute \( R_n \)

\[ \frac{1}{R'} = 5 \times \frac{1}{\frac{R}{5}} \]

\[ \frac{1}{R'} = 5 \times \frac{5}{R} \]

\[ \frac{1}{R'} = \frac{25}{R} \]

Step 5: Take reciprocal

\[ R' = \frac{R}{25} \]

Step 6: Required ratio

\[ \frac{R}{R'} = \frac{R}{\frac{R}{25}} = 25 \]

Final Answer: (d) 25

Concept Visualization

Exam Significance

• Tests understanding of proportionality \( R \propto L \).
• Very common in CBSE board MCQs and case-based questions.
• Frequently asked in NTSE and Olympiads.
• Key shortcut: \[ R' = \frac{R}{n^2} \] when a wire is cut into \( n \) equal parts and connected in parallel.

Theory/Concept

• Electrical power is defined as the rate of consumption of electrical energy.
• Basic formula: \[ P = VI \]
• Using Ohm’s Law: \[ V = IR \quad \text{and} \quad I = \frac{V}{R} \]
• Thus, power can also be written as: \[ P = I^2 R \quad \text{and} \quad P = \frac{V^2}{R} \]

Solution Roadmap

1. Start with standard power formula.
2. Derive equivalent expressions using Ohm’s Law.
3. Compare each option with valid expressions.
4. Identify the incorrect one.

Step-by-Step Solution

Step 1: Standard formula of power

\[ P = VI \]

Step 2: Using Ohm’s Law

\[ V = IR \quad \Rightarrow \quad P = I \times (IR) \]

\[ P = I^2 R \]

Step 3: Another form using \( I = \frac{V}{R} \)

\[ P = V \times \frac{V}{R} \]

\[ P = \frac{V^2}{R} \]

Step 4: Compare with options

• \( I^2 R \) ✔ Valid expression of power
• \( VI \) ✔ Valid expression of power
• \( \frac{V^2}{R} \) ✔ Valid expression of power
• \( IR^2 \) ✘ Not derived from any valid formula of power

Final Answer: (b) \( IR^2 \)

Exam Significance

• Very frequently asked conceptual MCQ in CBSE board exams.
• Tests clarity of formula manipulation using Ohm’s Law.
• Common trap: confusing \( I^2R \) with \( IR^2 \).
• Important for NTSE, Olympiads, and engineering entrance basics.
• Quick tip: Always verify expressions by dimensional correctness.

Theory/Concept

• Power of an electrical device: \[ P = \frac{V^2}{R} \]
• For a given device (like a bulb), resistance remains constant.
• If voltage changes, power changes according to: \[ P \propto V^2 \]

Solution Roadmap

1. Use rated values to calculate resistance of the bulb.
2. Apply the same resistance at new voltage.
3. Compute new power.

Step-by-Step Solution

Step 1: Given values

\[ V = 220 \text{ V}, \quad P = 100 \text{ W} \]

Step 2: Find resistance of bulb

\[ P = \frac{V^2}{R} \quad \Rightarrow \quad R = \frac{V^2}{P} \]

\[ R = \frac{(220)^2}{100} \]

\[ R = \frac{48400}{100} \]

\[ R = 484 \, \Omega \]

Step 3: New voltage applied

\[ V' = 110 \text{ V} \]

Step 4: Find new power

\[ P' = \frac{V'^2}{R} \]

\[ P' = \frac{(110)^2}{484} \]

\[ P' = \frac{12100}{484} \]

\[ P' = 25 \text{ W} \]

Final Answer: (d) 25 W

Concept Visualization

Exam Significance

• Very important for CBSE board MCQs and case-based questions.
• Tests understanding that resistance remains constant for a given device.
• Key concept: Power varies as square of voltage.
• Frequently asked in NTSE, Olympiads, and basic engineering exams.
• Shortcut: \[ \frac{P'}{P} = \left(\frac{V'}{V}\right)^2 \Rightarrow \frac{P'}{100} = \left(\frac{110}{220}\right)^2 = \frac{1}{4} \Rightarrow P' = 25 \text{ W} \]

Theory/Concept

• Resistance depends on material, length, and area. Since wires are identical: \[ R_1 = R_2 = R \]
• Equivalent resistance:
  • Series: \( R_s = R_1 + R_2 \)
  • Parallel: \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \)
• Heat produced (Joule’s Law): \[ H = I^2 R t \]
• Also, using \( I = \frac{V}{R} \): \[ H = \frac{V^2}{R} \, t \]

Solution Roadmap

1. Assign resistance \( R \) to each wire.
2. Find equivalent resistance in series and parallel.
3. Use heat formula \( H = \frac{V^2}{R} t \).
4. Compute ratio.

Step-by-Step Solution

Step 1: Let resistance of each wire

\[ R \]

Step 2: Series combination

\[ R_s = R + R = 2R \]

Heat in series

\[ H_s = \frac{V^2}{R_s} \times t \]

\[ H_s = \frac{V^2}{2R} \, t \]

Step 3: Parallel combination

\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \]

\[ R_p = \frac{R}{2} \]

Heat in parallel

\[ H_p = \frac{V^2}{R_p} \times t \]

\[ H_p = \frac{V^2}{R/2} \, t \]

\[ H_p = \frac{2V^2}{R} \, t \]

Step 4: Ratio

\[ \frac{H_s}{H_p} = \frac{\frac{V^2}{2R}}{\frac{2V^2}{R}} \]

\[ \frac{H_s}{H_p} = \frac{V^2}{2R} \times \frac{R}{2V^2} \]

\[ \frac{H_s}{H_p} = \frac{1}{4} \]

Final Answer: (c) 1 : 4

Exam Significance

• Very important conceptual problem combining resistance and heating effect.
• Frequently appears in CBSE board MCQs and case-based questions.
• Tests understanding of how heat depends on resistance at constant voltage.
• Common in NTSE and Olympiad exams.
• Key insight: Lower resistance in parallel → higher heat produced.

Theory/Concept

• Potential difference (voltage) is the work done per unit charge between two points.
• A voltmeter measures the potential difference across a component.
• A voltmeter has very high resistance so that it draws negligible current.
• To measure voltage correctly, the device must be connected across (parallel to) the component.

Solution Roadmap

1. Understand what voltage measurement requires.
2. Analyze why parallel connection is necessary.
3. Explain why series connection is incorrect.

Step-by-Step Explanation

Step 1: Purpose of a voltmeter

A voltmeter measures the potential difference between two points in a circuit.

Step 2: Correct method of connection

To measure the voltage across a component, the voltmeter must be connected in parallel with that component.

Reason:

• In parallel connection, the voltmeter experiences the same potential difference as the component.
• Because of its high resistance, it draws negligible current.
• This ensures that the circuit behavior remains unchanged.

Step 3: Why not series connection?

• If connected in series, voltmeter adds large resistance to the circuit.
• This reduces current significantly.
• It leads to incorrect readings or may even stop current flow.

Final Answer: A voltmeter is connected in parallel across the two points where potential difference is to be measured.

Exam Significance

• Very common theory question in CBSE board exams.
• Often asked in short-answer or reasoning-based questions.
• Concept is foundational for practical circuits and experiments.
• Also important for NTSE and basic electronics questions.
• Key takeaway: Voltmeter → Parallel, Ammeter → Series.

Theory/Concept

• Resistance of a wire: \[ R = \rho \frac{l}{A} \]
• Cross-sectional area of wire: \[ A = \pi r^2 \]
• Resistance is inversely proportional to area: \[ R \propto \frac{1}{A} \]

Solution Roadmap

1. Convert diameter to radius and SI units.
2. Calculate cross-sectional area.
3. Find length using resistance formula.
4. Analyze effect of doubling diameter.

Step-by-Step Solution

Step 1: Given values

\[ d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \]

\[ r = \frac{d}{2} = 0.25 \times 10^{-3} \, \text{m} \]

\[ \rho = 1.6 \times 10^{-8} \, \Omega \text{m}, \quad R = 10 \, \Omega \]

Step 2: Cross-sectional area

\[ A = \pi r^2 \]

\[ A = \pi (0.25 \times 10^{-3})^2 \]

\[ A = \pi \times 6.25 \times 10^{-8} \]

\[ A = 1.9635 \times 10^{-7} \, \text{m}^2 \]

Step 3: Length of wire

\[ \begin{aligned} R &= \rho \frac{l}{A} \\ \Rightarrow l &= \frac{R A}{\rho} \end{aligned} \]

\[ l = \frac{10 \times 1.9635 \times 10^{-7}}{1.6 \times 10^{-8}} \]

\[ l = 122.7 \, \text{m} \]

Step 4: When diameter is doubled

New diameter \( d' = 1.0 \, \text{mm} \Rightarrow r' = 0.5 \times 10^{-3} \, \text{m} \)

\[ A' = \pi (0.5 \times 10^{-3})^2 \]

\[ A' = \pi \times 25 \times 10^{-8} \]

\[ A' = 7.854 \times 10^{-7} \, \text{m}^2 \]

Step 5: New resistance

\[ R' = \rho \frac{l}{A'} \]

Using ratio method:

\[ R' = \frac{R \times A}{A'} \]

\[ R' = \frac{10 \times 1.9635 \times 10^{-7}}{7.854 \times 10^{-7}} \]

\[ R' = 2.5 \, \Omega \]

Final Result:

• Length of wire = \( 122.7 \, \text{m} \)
• New resistance = \( 2.5 \, \Omega \)

Key Insight:

\[ \text{If diameter doubles} \Rightarrow \text{Area becomes 4 times} \Rightarrow R \text{ becomes } \frac{1}{4} \]

Exam Significance

• Very important numerical for CBSE board exams.
• Tests unit conversion, substitution, and proportional reasoning.
• Frequently asked in NTSE and Olympiad exams.
• Common mistake: forgetting to convert mm to meters.
• Key takeaway: Resistance varies inversely with square of diameter.

Theory/Concept

• Ohm’s Law: \[ V = IR \]
• For an ohmic conductor, graph of \( V \) vs \( I \) is a straight line passing through origin.
• Slope of \( V-I \) graph gives resistance: \[ R = \frac{V}{I} \]

Solution Roadmap

1. Plot \( I \) on X-axis and \( V \) on Y-axis.
2. Mark given points.
3. Draw best-fit straight line.
4. Find slope \( \frac{V}{I} \).

Step-by-Step Solution

Step 1: Observe data

The ratio \( \frac{V}{I} \) for different points:

\[ \frac{1.6}{0.5} = 3.2, \quad \frac{3.4}{1.0} = 3.4, \quad \frac{6.7}{2.0} = 3.35 \]

\[ \frac{10.2}{3.0} = 3.4, \quad \frac{13.2}{4.0} = 3.3 \]

Step 2: Average resistance

\[ R \approx 3.3 \, \Omega \]

Step 3: Graph interpretation

• Graph is a straight line → conductor obeys Ohm’s law.
• Slope of line gives resistance.

Final Answer: Resistance \( \approx 3.3 \, \Omega \)

Exam Significance

• Very important graph-based question in CBSE board exams.
• Tests understanding of Ohm’s law and slope interpretation.
• Frequently asked in practical exams and viva.
• Also important for NTSE and foundational physics exams.
• Key idea: Straight line → constant resistance.

Theory/Concept

• Ohm’s Law: \[ R = \frac{V}{I} \]
• SI unit of current is ampere (A), so milliampere must be converted: \[ 1 \, \text{mA} = 10^{-3} \, \text{A} \]

Solution Roadmap

1. Convert current into amperes.
2. Apply Ohm’s law.
3. Compute resistance.

Step-by-Step Solution

Step 1: Given values

\[ V = 12 \, \text{V}, \quad I = 2.5 \, \text{mA} \]

Step 2: Convert current to SI unit

\[ I = 2.5 \times 10^{-3} \, \text{A} \]

Step 3: Apply Ohm’s Law

\[ R = \frac{V}{I} \]

\[ R = \frac{12}{2.5 \times 10^{-3}} \]

Step 4: Simplify calculation

\[ R = \frac{12}{0.0025} \]

\[ R = 4800 \, \Omega \]

Step 5: Express in kilo-ohms

\[ R = 4.8 \, \text{k}\Omega \]

Final Answer: \( 4.8 \, \text{k}\Omega \)

Exam Significance

• Very common direct numerical in CBSE board exams.
• Tests unit conversion and correct use of Ohm’s law.
• Frequently asked in NTSE and basic physics exams.
• Common mistake: not converting mA to A.
• Key takeaway: Always convert to SI units before calculation.

Theory/Concept

• In series combination, resistances add: \[ R_{\text{total}} = R_1 + R_2 + R_3 + \dots \]
• Current is same through all components in series.
• Ohm’s Law: \[ I = \frac{V}{R} \]

Solution Roadmap

1. Find total resistance of the circuit.
2. Apply Ohm’s law to find current.
3. Use property of series circuit to conclude current in 12 Ω resistor.

Step-by-Step Solution

Step 1: Given values

\[ V = 9 \, \text{V} \]

Resistances:

\[ 0.2 \, \Omega,\; 0.3 \, \Omega,\; 0.4 \, \Omega,\; 0.5 \, \Omega,\; 12 \, \Omega \]

Step 2: Total resistance

\[ R_{\text{total}} = 0.2 + 0.3 + 0.4 + 0.5 + 12 \]

\[ R_{\text{total}} = 13.4 \, \Omega \]

Step 3: Apply Ohm’s Law

\[ I = \frac{V}{R_{\text{total}}} \]

\[ I = \frac{9}{13.4} \]

\[ I \approx 0.67 \, \text{A} \]

Step 4: Current through 12 Ω resistor

Since all resistors are in series, same current flows through each resistor.

\[ I_{12\Omega} = 0.67 \, \text{A} \]

Final Answer: \( 0.67 \, \text{A} \)

Exam Significance

• Very common numerical based on series circuits.
• Tests understanding that current is same in series.
• Important for CBSE board exams and NTSE.
• Common mistake: calculating current separately for each resistor.
• Key idea: One current → entire series circuit.

Theory/Concept

• Ohm’s Law: \[ R = \frac{V}{I} \]
• For identical resistors in parallel: \[ R_{\text{eq}} = \frac{R}{n} \]
• Current increases when resistors are connected in parallel due to decrease in equivalent resistance.

Solution Roadmap

1. Find total equivalent resistance required.
2. Use parallel formula for identical resistors.
3. Calculate number of resistors.

Step-by-Step Solution

Step 1: Given values

\[ V = 220 \, \text{V}, \quad I = 5 \, \text{A} \]

Step 2: Find equivalent resistance

\[ R_{\text{eq}} = \frac{V}{I} \]

\[ R_{\text{eq}} = \frac{220}{5} \]

\[ R_{\text{eq}} = 44 \, \Omega \]

Step 3: Use parallel resistor relation

\[ R_{\text{eq}} = \frac{R}{n} \]

\[ 44 = \frac{176}{n} \]

Step 4: Solve for \( n \)

\[ n = \frac{176}{44} \]

\[ n = 4 \]

Final Answer: 4 resistors

Concept Visualization

Exam Significance

• Very important application-based numerical for CBSE boards.
• Tests understanding of parallel combination and Ohm’s law.
• Common in NTSE and basic electrical problems.
• Key shortcut: \( n = \frac{R}{R_{\text{eq}}} \).
• Conceptual takeaway: Parallel connection reduces resistance and increases current capacity.

Theory/Concept

• Series combination: \[ R_s = R_1 + R_2 + \dots \]
• Parallel combination: \[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots \]
• Parallel reduces resistance, series increases resistance.

Solution Roadmap

1. Try combinations of series and parallel.
2. First combine two resistors.
3. Then combine result with third resistor.

Step-by-Step Solution

Each resistor:

\[ R = 6 \, \Omega \]

(i) Required resistance = 9 Ω

Step 1: Take two resistors in parallel

\[ \frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} \]

\[ \frac{1}{R_p} = \frac{2}{6} \]

\[ R_p = 3 \, \Omega \]

Step 2: Add third resistor in series

\[ R_{\text{total}} = R_p + 6 \]

\[ R_{\text{total}} = 3 + 6 = 9 \, \Omega \]

Conclusion: Two in parallel, then one in series.

(ii) Required resistance = 4 Ω

Try all possibilities:

• All in series: \[ R = 6 + 6 + 6 = 18 \, \Omega \]
• All in parallel: \[ \frac{1}{R} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} \Rightarrow R = 2 \, \Omega \]
• Two in series, then parallel with third: \[ R_s = 6 + 6 = 12 \, \Omega \] \[ \frac{1}{R} = \frac{1}{12} + \frac{1}{6} = \frac{1 + 2}{12} = \frac{3}{12} \Rightarrow R = 4 \, \Omega \]

Conclusion: Two resistors in series, then this combination in parallel with third gives 4 Ω.

Final Answer:

• (i) 9 Ω → Two parallel + one series
• (ii) 4 Ω → Two series + one parallel

Concept Visualization

Exam Significance

• Very important conceptual question in CBSE exams.
• Tests ability to design resistor networks.
• Frequently asked in NTSE and Olympiads.
• Key skill: combining series and parallel logically.
• Common mistake: assuming only one arrangement works.

Theory/Concept

• Electrical power: \[ P = VI \]
• In parallel connection:
  • Voltage across each bulb is same (220 V).
  • Total current is sum of individual currents.
• Total power consumed: \[ P_{\text{total}} = n \times P_{\text{bulb}} \]

Solution Roadmap

1. Find maximum total power allowed.
2. Divide by power of one bulb.
3. Get number of bulbs.

Step-by-Step Solution

Step 1: Given values

\[ V = 220 \, \text{V}, \quad I_{\text{max}} = 5 \, \text{A} \]

Step 2: Maximum total power

\[ P_{\text{max}} = V \times I \]

\[ P_{\text{max}} = 220 \times 5 \]

\[ P_{\text{max}} = 1100 \, \text{W} \]

Step 3: Power of one bulb

\[ P_{\text{bulb}} = 10 \, \text{W} \]

Step 4: Number of bulbs

\[ n = \frac{P_{\text{max}}}{P_{\text{bulb}}} \]

\[ n = \frac{1100}{10} \]

\[ n = 110 \]

Final Answer: 110 bulbs

Concept Visualization

Exam Significance

• Very important application-based question for CBSE boards.
• Tests understanding of power, current, and parallel circuits.
• Common in NTSE and Olympiad exams.
• Key idea: Total current limit determines number of appliances.
• Shortcut: \( n = \frac{VI}{P_{\text{bulb}}} \).

Theory/Concept

• Ohm’s Law: \[ I = \frac{V}{R} \]
• Series combination: \[ R_s = R_1 + R_2 \]
• Parallel combination: \[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \]
• Current increases when resistance decreases and vice versa.

Solution Roadmap

1. Calculate current for one coil.
2. Find equivalent resistance in series and calculate current.
3. Find equivalent resistance in parallel and calculate current.

Step-by-Step Solution

Given:

\[ V = 220 \, \text{V}, \quad R_1 = R_2 = 24 \, \Omega \]

(i) When used separately

\[ I = \frac{V}{R} \]

\[ I = \frac{220}{24} \]

\[ I \approx 9.17 \, \text{A} \]

(ii) When connected in series

Step 1: Equivalent resistance

\[ R_s = 24 + 24 = 48 \, \Omega \]

Step 2: Current

\[ I = \frac{220}{48} \]

\[ I \approx 4.58 \, \text{A} \]

(iii) When connected in parallel

Step 1: Equivalent resistance

\[ \frac{1}{R_p} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} \]

\[ R_p = 12 \, \Omega \]

Step 2: Current

\[ I = \frac{220}{12} \]

\[ I \approx 18.33 \, \text{A} \]

Final Answers:

• Separate: \( 9.17 \, \text{A} \)
• Series: \( 4.58 \, \text{A} \)
• Parallel: \( 18.33 \, \text{A} \)

Concept Visualization

Exam Significance

• Very important multi-case numerical for CBSE exams.
• Tests understanding of series vs parallel behavior.
• Frequently asked in NTSE and Olympiads.
• Key concept: Resistance controls current.
• Common mistake: forgetting that same voltage is applied in all cases.

Theory/Concept

• Ohm’s Law: \[ I = \frac{V}{R} \]
• Power formulas: \[ P = I^2 R \quad \text{and} \quad P = \frac{V^2}{R} \]
• In series:
  • Current is same through all resistors.
• In parallel:
  • Voltage across each resistor is same.

Solution Roadmap

1. Case (i): Find current, then power using \( I^2 R \).
2. Case (ii): Use direct voltage across resistor.
3. Compare both results.

Step-by-Step Solution

(i) Series circuit

Step 1: Total resistance

\[ R_{\text{total}} = 1 + 2 = 3 \, \Omega \]

Step 2: Current

\[ I = \frac{6}{3} = 2 \, \text{A} \]

Step 3: Power in 2 Ω resistor

\[ P = I^2 R \]

\[ P = (2)^2 \times 2 \]

\[ P = 4 \times 2 = 8 \, \text{W} \]

(ii) Parallel circuit

Step 1: Voltage across 2 Ω resistor

\[ V = 4 \, \text{V} \]

Step 2: Power

\[ P = \frac{V^2}{R} \]

\[ P = \frac{4^2}{2} \]

\[ P = \frac{16}{2} = 8 \, \text{W} \]

Final Comparison:

\[ P_{\text{series}} = 8 \, \text{W}, \quad P_{\text{parallel}} = 8 \, \text{W} \]

Conclusion: Power used in the 2 Ω resistor is same in both cases.

Exam Significance

• Very important conceptual comparison problem.
• Tests understanding of power formulas in different circuits.
• Frequently asked in CBSE board case-based questions.
• Common in NTSE and Olympiad exams.
• Key insight: Same power can arise from different circuit conditions.

Theory/Concept

• Electrical power: \[ P = VI \]
• In parallel connection:
  • Voltage across each device is same.
  • Total power is sum of individual powers.
• Total current drawn: \[ I_{\text{total}} = \frac{P_{\text{total}}}{V} \]

Solution Roadmap

1. Add powers of both lamps.
2. Use \( P = VI \) to find total current.

Step-by-Step Solution

Step 1: Given values

\[ P_1 = 100 \, \text{W}, \quad P_2 = 60 \, \text{W}, \quad V = 220 \, \text{V} \]

Step 2: Total power

\[ P_{\text{total}} = P_1 + P_2 \]

\[ P_{\text{total}} = 100 + 60 = 160 \, \text{W} \]

Step 3: Total current

\[ I = \frac{P_{\text{total}}}{V} \]

\[ I = \frac{160}{220} \]

\[ I \approx 0.73 \, \text{A} \]

Final Answer: \( 0.73 \, \text{A} \)

Concept Visualization

Exam Significance

• Very common CBSE board numerical.
• Tests understanding of power addition in parallel circuits.
• Important for NTSE and basic electrical calculations.
• Common mistake: adding currents directly instead of powers.
• Key idea: Same voltage → add powers, then find current.

Theory/Concept

• Electrical energy: \[ E = P \times t \]
• Power \(P\) is in watts (W), time \(t\) must be in hours.
• Energy is measured in watt-hour (Wh) or kilowatt-hour (kWh).

Solution Roadmap

1. Convert time into hours.
2. Calculate energy for both devices.
3. Compare values.

Step-by-Step Solution

Case 1: TV

\[ P = 250 \, \text{W}, \quad t = 1 \, \text{h} \]

\[ E_{\text{TV}} = P \times t \]

\[ E_{\text{TV}} = 250 \times 1 = 250 \, \text{Wh} \]

Case 2: Toaster

Step 1: Convert time

\[ t = 10 \, \text{minutes} = \frac{10}{60} = \frac{1}{6} \, \text{h} \]

Step 2: Calculate energy

\[ E_{\text{toaster}} = 1200 \times \frac{1}{6} \]

\[ E_{\text{toaster}} = 200 \, \text{Wh} \]

Final Comparison:

\[ E_{\text{TV}} = 250 \, \text{Wh}, \quad E_{\text{toaster}} = 200 \, \text{Wh} \]

Conclusion: TV uses more energy.

Exam Significance

• Very common conceptual numerical in CBSE exams.
• Tests correct use of \( E = Pt \) and unit conversion.
• Common mistake: not converting minutes to hours.
• Frequently asked in NTSE and Olympiads.
• Key insight: Higher power does not always mean more energy; time matters.

Theory/Concept

• Heating effect of current (Joule’s Law): \[ H = I^2 R t \]
• Rate of heat production (Power): \[ P = \frac{H}{t} = I^2 R \]
• Power is measured in watts (W), where: \[ 1 \, \text{W} = 1 \, \text{J/s} \]

Solution Roadmap

1. Identify required quantity (rate of heat → power).
2. Apply formula \( P = I^2 R \).
3. Substitute values and calculate.

Step-by-Step Solution

Step 1: Given values

\[ I = 5 \, \text{A}, \quad R = 44 \, \Omega \]

Step 2: Formula for rate of heat production

\[ P = I^2 R \]

Step 3: Substitute values

\[ P = (5)^2 \times 44 \]

\[ P = 25 \times 44 \]

\[ P = 1100 \, \text{W} \]

Final Answer: \( 1100 \, \text{W} \)

Interpretation: The heater produces 1100 joules of heat energy every second.

Exam Significance

• Very important direct application of Joule’s law.
• Frequently asked in CBSE board exams.
• Tests understanding of difference between heat and rate of heat.
• Common mistake: multiplying by time unnecessarily.
• Key idea: "rate of heat" = power = \( I^2 R \).

Theory/Concept

• Resistance of a conductor: \[ R = \rho \frac{l}{A} \]
• Heating effect of current depends on resistance and material properties.
• Material choice depends on conductivity, melting point, and durability.

Solution Roadmap

1. Use material properties (melting point, resistivity, strength).
2. Apply circuit principles (series vs parallel).
3. Use resistance relation where required.

Step-by-Step Explanation

(a) Why is tungsten used for filaments?

• Very high melting point (~3422°C) → does not melt at high temperature.
• High resistivity → produces sufficient heat and light.
• Low rate of evaporation → longer life.
• Can be drawn into thin wires (good ductility).

(b) Why are alloys used in heating devices?

• Alloys have higher resistivity → produce more heat.
• Do not oxidize easily at high temperatures.
• Have high melting point and strength.
• Resistance changes less with temperature → stable performance.

(c) Why is series arrangement not used in domestic circuits?

• Same current flows → appliances cannot operate independently.
• If one device fails, entire circuit stops.
• Voltage divides → appliances do not get required voltage.
• Parallel connection ensures full voltage and independent operation.

(d) Variation of resistance with area

\[ R = \rho \frac{l}{A} \]

• \( R \propto \frac{1}{A} \)
• Increasing area → resistance decreases.
• Doubling area → resistance becomes half.

(e) Why copper and aluminium are used for transmission?

• Low resistivity → high conductivity.
• Less power loss during transmission.
• Copper: high strength and excellent conductivity.
• Aluminium: lightweight and economical.
• Both resist corrosion → long life.

Exam Significance

• Very important theory-based question for CBSE board exams.
• Frequently appears as 3–5 mark descriptive question.
• Tests conceptual clarity of materials and circuits.
• Important for NTSE and foundational physics exams.
• Key tip: Always justify answers using properties (melting point, resistivity, etc.).