Class 9 Science Chapter 11 Electricity: Previous Year Questions (2020–2025)

Reading your textbook twice doesn't improve exam marks as much as solving one authentic past paper question does. Here's why: First, PYQs train your brain to recognise the exact phrasing examiners use. A question asking 'Why are high resistance wires used in electric heaters?' tests the same concept as 'Explain why nichrome wire is preferred for heating elements'—but you need exposure to both framings to answer confidently under exam pressure. Second, PYQs reveal the depth expected. For 'Ohm's law', a 1-mark question asks only the formula (V = IR), while a 5-mark question demands derivation, experimental verification, and a worked numerical. Theory reading doesn't show you this difference clearly. Third, solving PYQs under timed conditions (30 mins for 13 marks, roughly) builds exam stamina. You learn to prioritize—which formula to write, which steps to skip, how much detail a 3-marker really needs. Finally, repeating question patterns builds pattern recognition. In Class 9 Electricity, roughly 60% of questions involve calculating power or resistance in circuits. Once you've solved 5 such questions, the 6th becomes faster and more accurate. That's PYQ ROI—direct, measurable, exam-aligned learning. Combine this with structured doubt-clearing (try cbsetutor.ai's live sessions if you get stuck on any PYQ) and you have a complete preparation strategy.

One-mark questions in Electricity typically test definition recall, formula recognition, or simple numerical conversion. These are 'gimme' marks if you've memorised the basics—but many students lose them by being careless. Here are 5 genuine patterns: **Q1. State Ohm's law.** A: Ohm's law states that the electric current through a conductor is directly proportional to the potential difference across it, provided the temperature and other physical conditions remain constant. Mathematically: I ∝ V, or V = IR (where R is resistance in ohms). **Q2. Define resistivity.** A: Resistivity is the resistance offered by a unit length of a uniform conductor of unit cross-sectional area. It is a property of the material itself, measured in ohm·metre (Ω·m), and remains constant for a given material at a given temperature. **Q3. A wire has resistance 10 Ω. If its length is doubled, what will be its new resistance?** A: 40 Ω. Since resistance R ∝ length, if length doubles, resistance doubles. But if the wire is stretched (length doubles), its cross-section reduces to half, so R' = 2R × 2 = 4R = 40 Ω. (Common error: students say 20 Ω, forgetting the area change.) **Q4. Give one advantage of parallel combination over series.** A: In parallel, if one device fails, others continue to work; brightness of bulbs remains constant; devices can be switched independently. (Examiners accept any one valid point.) **Q5. Calculate power if voltage = 220 V and current = 5 A.** A: P = VI = 220 × 5 = 1100 W = 1.1 kW. Tip: Always include units in numerical answers—examiners deduct marks for missing units, even on 1-markers.

Three-mark questions demand explanation plus calculation or a brief derivation. Answers should be 8–12 lines, with at least one formula and working shown. **Q1. A student measures voltage across a resistor as 5 V and current through it as 0.5 A. Calculate resistance and verify Ohm's law.** A: Using Ohm's law, V = IR 5 = 0.5 × R R = 5/0.5 = 10 Ω To verify: Check if V/I = constant (resistance) V/I = 5/0.5 = 10 Ω (constant) Alternatively, plot V vs I and check if the graph is linear and passes through origin. Since V ∝ I, Ohm's law is verified. [Include graph sketch if possible for full marks.] **Q2. Write the relationship between resistance, resistivity, length and cross-sectional area. A copper wire has length 2 m and cross-section 1 mm². If resistivity of copper = 1.6 × 10⁻⁸ Ω·m, calculate its resistance.** A: Relationship: R = ρL/A (where ρ = resistivity, L = length, A = area) Given: ρ = 1.6 × 10⁻⁸ Ω·m, L = 2 m, A = 1 mm² = 1 × 10⁻⁶ m² R = (1.6 × 10⁻⁸ × 2)/(1 × 10⁻⁶) R = (3.2 × 10⁻⁸)/(10⁻⁶) R = 3.2 × 10⁻² = 0.032 Ω **Q3. A 60 W bulb and a 40 W bulb are rated at 220 V. Which has greater resistance? Explain.** A: Using P = V²/R, we get R = V²/P For 60 W bulb: R₁ = (220)²/60 = 48400/60 ≈ 806.7 Ω For 40 W bulb: R₂ = (220)²/40 = 48400/40 = 1210 Ω The 40 W bulb has greater resistance. Explanation: Lower wattage means lower power, which requires higher resistance (since V is constant). Higher resistance → less current → less brightness. **Q4. Two resistors of 4 Ω and 6 Ω are connected in series to a 10 V battery. Calculate total resistance and current.** A: In series, R_total = R₁ + R₂ = 4 + 6 = 10 Ω Using V = IR, I = V/R_total = 10/10 = 1 A Current through both resistors is 1 A (same in series). **Q5. Explain why fuses melt when current exceeds a certain limit.** A: Fuses contain a thin wire of low melting point alloy. When excess current flows, the heating effect (H = I²Rt) becomes very large. Since resistance and time are fixed, and current increases (squared term), heat production increases dramatically. This heat melts the fuse wire, breaking the circuit and protecting appliances from overcurrent damage. Mathematically, H ∝ I², so small increases in current cause large heat jumps.

Five-mark questions are mini-essays combining definition, formula, numerical calculation, and reasoning. Expect 15–20 lines of answer. **Q1. Derive the formula for resistance in series and parallel combinations. A 2 Ω resistor, 3 Ω resistor and 6 Ω resistor are available. In which combination will the equivalent resistance be (a) maximum, (b) minimum? Calculate both.** A: **Series Combination:** When resistors are in series, the same current I flows through all. The total potential difference V = V₁ + V₂ + V₃. Using Ohm's law: V = IR_s, V₁ = IR₁, V₂ = IR₂, V₃ = IR₃ IR_s = IR₁ + IR₂ + IR₃ R_s = R₁ + R₂ + R₃ For our resistors in series: R_s = 2 + 3 + 6 = 11 Ω **Parallel Combination:** When resistors are in parallel, the potential difference across all is same (V). Total current I = I₁ + I₂ + I₃. Using Ohm's law: I = V/R_p, I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃ V/R_p = V/R₁ + V/R₂ + V/R₃ 1/R_p = 1/R₁ + 1/R₂ + 1/R₃ For our resistors in parallel: 1/R_p = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 R_p = 1 Ω (a) **Maximum resistance**: Series combination = 11 Ω (higher because resistances add) (b) **Minimum resistance**: Parallel combination = 1 Ω (lower because reciprocals add) Rule: Series always gives higher R_eq; parallel always gives lower R_eq. **Q2. An electric kettle rated 1000 W, 220 V is used to heat water. (a) Calculate the resistance of the heating element. (b) How much energy is consumed in 30 minutes? (c) If 1 unit of electricity costs ₹5, what is the cost of running the kettle for 30 minutes?** A: **(a) Finding Resistance:** Using P = V²/R R = V²/P = (220)²/1000 = 48400/1000 = 48.4 Ω **(b) Energy Consumed:** Energy = Power × Time Time = 30 min = 30/60 hours = 0.5 hours Energy = 1000 W × 0.5 h = 500 Wh = 0.5 kWh = 0.5 unit (Note: 1 unit = 1 kWh) **(c) Cost:** Cost = Energy (in units) × Cost per unit Cost = 0.5 × 5 = ₹2.50 Alternative check: Using E = I²Rt Current I = P/V = 1000/220 ≈ 4.55 A Time = 30 × 60 = 1800 seconds E = (4.55)² × 48.4 × 1800 ≈ 1,800,000 J = 0.5 kWh ✓ **Q3. A student connects two bulbs of 40 W and 60 W (both rated 220 V) in parallel to a 220 V supply. (a) Calculate the resistance of each bulb. (b) Find the current through each bulb. (c) Calculate total power consumed. (d) Which bulb glows brighter and why?** A: **(a) Resistance of each bulb:** For 40 W bulb: R₁ = V²/P₁ = (220)²/40 = 48400/40 = 1210 Ω For 60 W bulb: R₂ = V²/P₂ = (220)²/60 = 48400/60 ≈ 806.7 Ω **(b) Current through each bulb:** Since both are in parallel, voltage across each = 220 V For 40 W bulb: I₁ = V/R₁ = 220/1210 ≈ 0.182 A (or directly I₁ = P₁/V = 40/220 ≈ 0.182 A) For 60 W bulb: I₂ = V/R₂ = 220/806.7 ≈ 0.273 A (or I₂ = 60/220 ≈ 0.273 A) **(c) Total power consumed:** In parallel, total power = P₁ + P₂ = 40 + 60 = 100 W Verify: Total current I_total = I₁ + I₂ = 0.182 + 0.273 = 0.455 A P_total = V × I_total = 220 × 0.455 = 100.1 W ✓ **(d) Which bulb glows brighter:** The 60 W bulb glows brighter. Reason: Power dissipated determines brightness (P = VI). The 60 W bulb has lower resistance, so it draws more current (0.273 A vs 0.182 A). Since P = I²R or P = V²/R, higher power = more light. More current × same voltage = more brightness.

The 2024–25 CBSE Class 9 Science syllabus retains Chapter 11 (Electricity) largely unchanged, but exam pattern trends show evolving question styles that students should prepare for: **1. Increased Case-Study / Application Questions:** Recent papers show more real-world contexts: 'A family's electricity bill shows 150 units consumed in a month at ₹8 per unit. If average voltage is 230 V, calculate average current.' These blend numeracy with consumer awareness. Rather than abstract resistor networks, examiners now embed circuits in practical scenarios (home appliances, fuses, earthing). **2. Conceptual Over Rote Formulas:** Questions like 'Why is the heating effect of current used in heaters but avoided in transmission lines?' now appear frequently. Examiners test reasoning (H = I²Rt means high current → high loss) rather than just formula application. Prepare explanations, not just calculations. **3. Graph-Based Questions:** Plotting V vs I graphs and interpreting gradients (gradient = 1/R) is now common. If a question says 'Draw V-I graph for a resistor,' expect follow-up: 'Find R from the graph' or 'What does the gradient represent?' Practise extracting information from plots. **4. Integration with Environmental Concepts:** Energy consumption, electricity costs, and renewable sources increasingly appear. Questions like 'How much CO₂ is saved by using a 5 W LED instead of a 60 W bulb for 8 hours daily?' expect combined physics and eco-awareness. **5. Fewer Isolated Numerical, More Problem-Solving:** Instead of 'Calculate R if ρ = 1.6 × 10⁻⁸, L = 2 m, A = 1 mm²,' you see 'Two wires of same length have cross-sections in ratio 1:2. Which has higher resistance? If both carry same current, which has higher voltage drop? Explain.' This tests depth of understanding, not calculator speed. **Preparation strategy:** Solve not just number-crunching PYQs, but also conceptual ones. Read explanations in your textbook carefully. Practise drawing and interpreting V-I graphs. Write one-line explanations for 'Why' questions daily—they build exam confidence.

Board and school exams allocate roughly 13 marks to Electricity across 1-mark, 3-mark, and 5-mark sections. Here's how to attempt efficiently: **Before Exam (Preparation):** 1. Memorise 5 key definitions: Ohm's law, resistance, resistivity, current, potential difference. Write them out daily for 1 week. 2. Learn formulas by application, not rote. Use flashcards pairing formula with context: R = ρL/A → "resistivity × length ÷ area." 3. Solve 1 complete PYQ every 2 days, timing yourself. Target: 1-mark in 1 min, 3-mark in 3–4 mins, 5-mark in 6–7 mins. 4. Practise numerical conversion: 1 mm² = 10⁻⁶ m², 1 kW = 1000 W, 1 unit = 1 kWh. Write a cheat-sheet and test yourself. 5. Draw and label circuit diagrams daily. Examiners award marks for clear, labelled diagrams. **During Exam:** 1. **Read all questions first** (2 mins). Mark 1-markers with asterisk, 3-markers with double-asterisk, 5-markers separately. This prevents surprises. 2. **Attempt 1-markers first** (5 mins for all). These are quick confidence-builders. Don't overthink—if unsure, write a 1-line definition and move on. 3. **Attempt 3-markers next** (12–15 mins total). For each: write formula → substitute values → simplify → box answer. Include units. If stuck, leave 1 line blank and return later. 4. **Attempt 5-markers last** (10–15 mins). These carry most weight. Use structured headings: "(a) Formula...," "(b) Calculation..." Draw diagrams (V-I graphs, circuits) wherever asked; they're worth 1–2 marks each. 5. **Final review** (3 mins). Check: Have I written all units? Are numerical answers reasonable? (A resistance of 10⁻¹⁰ Ω is unrealistic; check calculation.) Did I box final answers? **Common Mistakes to Avoid:** - Writing only the formula without substituting values (loses marks). - Forgetting units (deducts 0.5–1 mark per instance). - In parallel circuits, saying 'current is same' (wrong—voltage is same; current divides). - In series, forgetting that R_eq = R₁ + R₂, not their average. - Miscalculating power: Using P = VI then V²/R in same problem, forgetting units (W vs kW). Start a 3-day free trial at cbsetutor.ai to get live, solution-focused doubt sessions where you can ask about any PYQ you get stuck on. Real-time feedback on your working beats watching videos.

Solving PYQs is a skill, not just a task. Here's a structured approach: **Step 1: Attempt Alone (No Reference)** Read the question. Close all books. Write your answer in full—formulas, working, explanations. Time yourself. Don't leave blanks; even a partial answer shows thought. Spend 5–10 mins per question. **Step 2: Check Against Provided Answers** Compare your answer line-by-line with the solution. Note: There may be multiple valid approaches (e.g., calculating power as VI or V²/R). If your method is different but the answer matches, you're on track. Tick what you got right. **Step 3: Analyse Errors** If wrong, identify the error type: - Conceptual error? (e.g., thought current is same in parallel) - Calculation error? (e.g., 1/R_p = 1/2 + 1/3 but added wrongly) - Formula error? (e.g., used R = VL/ρA instead of ρL/A) Write the error type in margin. This builds error awareness. **Step 4: Rewrite the Correct Solution** After checking, rewrite the full solution neatly. This embeds method in muscle memory. **Step 5: Reattempt (After 1 Week)** One week later, attempt the same question again without looking at your previous attempt. If you solve it correctly this time, you've truly learned it. If not, the error is persistent; spend more time on that concept. **Spaced Repetition Schedule:** - Day 1: Attempt Q1, Q2, Q3, Q4, Q5 (1-mark set). - Day 2: Check answers. Rewrite solutions. - Day 3: Attempt Q1–Q5 from 3-mark set. - Day 4: Check and rewrite. - Days 5–7: Repeat above cycle with 5-mark questions. - Day 8 onwards: Reattempt 1-mark questions from Day 1. - Week 2: Reattempt 3-mark questions. - Week 3: Reattempt 5-mark questions. This spacing prevents cramming and builds long-term recall. Track your progress in a simple table: Date | Question | First Attempt | Second Attempt (1 week later). Seeing 'Correct → Correct' builds confidence.