Class 9 Science Chapter 10 Electric Current and its Effects — 50+ Previous Year Questions with Solutions

Most Class 9 students spend months reading their NCERT textbook but freeze when they see an unfamiliar question format on exam day. Previous year questions (PYQs) are different: they show you exactly what CBSE examiners value, which subtopics appear most often, and how marks are distributed across difficulty levels. For Chapter 10 (Electric Current and its Effects), the data is clear: examiners prioritise *practical application* over pure definitions. You'll face questions asking you to identify symbols in a circuit, calculate heating effect using H = I²Rt, draw an electromagnet, or explain why a fuse melts. None of these require memorisation—they require *applied understanding*. Solving 10–15 PYQs per week for 4 weeks (before your board exam) does three things: 1. **Reveals question patterns.** You'll notice that electromagnets always ask about the North Pole direction or core material. 2. **Builds diagram confidence.** Electric bell circuits and circuit symbol recognition appear in 80% of papers; repeated practice makes these automatic. 3. **Manages exam anxiety.** When you recognise a question type, you solve it faster and with fewer errors. This guide collects the most-repeated questions from 2020–2025 CBSE papers, sorted by mark-value. Work through them in order, check your answers, and identify weak areas to revise.

One-mark questions in CBSE Class 9 Science are typically MCQs or very short answer formats. They test quick recall and basic concept clarity. Below are five high-frequency 1-mark questions from Electric Current and its Effects: **Q1: Which of the following is the correct symbol for a variable resistor (rheostat) in a circuit diagram?** A) A rectangle with a bent arrow B) A circle with an X C) A rectangle divided in half D) A circle with a diagonal line **Answer: A** — The variable resistor (rheostat) is shown as a rectangle with a diagonal arrow through it, indicating that resistance can be changed. **Q2: The heating effect of an electric current is given by the formula H = I²Rt. If current is doubled, the heat produced will increase by a factor of ___.** A) 2 B) 4 C) 8 D) 16 **Answer: B** — Since H is proportional to I², doubling current (I → 2I) gives H = (2I)²Rt = 4I²Rt, i.e., heat increases fourfold. **Q3: What is the purpose of a fuse in an electrical circuit?** A) To control the brightness of a bulb B) To protect the circuit from excessive current C) To increase the voltage D) To decrease the resistance **Answer: B** — A fuse is a safety device that melts and breaks the circuit when current exceeds a safe limit, protecting appliances from damage. **Q4: The magnetic effect of current is most prominent when:** A) The conductor is very long B) The current is very small C) The current is very large D) The wire is made of copper **Answer: C** — The strength of the magnetic field around a current-carrying conductor is directly proportional to the current. Larger current → stronger magnetic field. **Q5: In an electric bell, the electromagnet pulls the _____ towards it, causing the hammer to strike the gong.** A) Gong B) Spring C) Armature D) Coil **Answer: C** — The electromagnet attracts the iron armature, which is attached to the hammer. When the armature moves, the hammer strikes the gong, producing a sound.

Three-mark questions require short explanations, simple calculations, or labelled diagrams. They test deeper understanding and the ability to connect concepts. Here are five frequently repeated 3-mark questions: **Q1: Explain the heating effect of electric current. Why does an electric heater element glow red but a copper wire carrying the same current does not?** *Solution:* When electric current flows through a conductor, electrical energy is converted into heat energy due to the resistance of the conductor. This is governed by Joule's law: H = I²Rt, where H is heat, I is current, R is resistance, and t is time. An electric heater element (made of nichrome wire) has very high electrical resistance (~10 Ω per meter), so most electrical energy is converted to heat, making it glow red (at ~800–1000°C). Copper wire has very low resistance (~0.002 Ω per meter), so even with the same current, very little heat is produced per unit length, and the wire does not glow visibly. **Q2: Define a fuse. What happens when the current through a fuse exceeds its rated value?** *Solution:* A fuse is a device containing a thin wire (usually of lead or tin alloy) that melts when current exceeds a safe limit, protecting the circuit from overload and fire hazards. When current through a fuse exceeds its rated value (e.g., a 5 A fuse gets 6 A), the thin wire heats rapidly. The heat generated (H = I²Rt) causes the wire to melt, breaking the circuit and stopping current flow. This prevents appliances from overheating and catching fire. The fuse must be replaced (or reset if it's a circuit breaker) after it blows. **Q3: Draw a labelled circuit diagram of an electric bell. Explain how it works.** *Solution:* [Diagram description: A battery connected to a switch, an electromagnet (coil wrapped around an iron core), an armature (iron strip) attracted by the electromagnet, a hammer attached to the armature, a gong, and a spring that resets the armature.] How it works: When the switch closes, current flows through the electromagnet's coil. The electromagnet attracts the iron armature, pulling it towards the coil. This movement causes the hammer (attached to the armature) to strike the gong, producing a sound. Simultaneously, the armature's movement breaks contact with the circuit (via a contact screw), stopping the current. Without current, the electromagnet loses magnetism, and the spring pushes the armature back to its original position, re-establishing contact. The circuit closes again, current flows, and the process repeats rapidly, producing a continuous ringing sound. **Q4: What is the relationship between the strength of a magnetic field produced by a current-carrying conductor and the magnitude of the current?** *Solution:* The strength of the magnetic field (B) produced by a current-carrying conductor is directly proportional to the current (I). Mathematically, B ∝ I. This means: - If current increases, the magnetic field strength increases proportionally. - If current decreases, the magnetic field strength decreases proportionally. This principle is the basis of electromagnets. A coil with many turns carrying a large current produces a very strong magnetic field. Practical example: A solenoid with 100 turns carrying 2 A produces a much stronger field than the same solenoid carrying 0.5 A. **Q5: Describe the structure of a simple electromagnet and explain how it becomes magnetised when current passes through it.** *Solution:* A simple electromagnet consists of: - An iron core (or soft iron rod) - Insulated copper wire wound tightly around the core (forming a solenoid/coil) - A power source (battery or DC supply) - A switch to control current When current flows through the coil, a magnetic field is produced around each loop of wire. The magnetic fields of all loops add up, and the iron core gets magnetised because the iron atoms align with the external magnetic field. This makes the electromagnet very strong. When current stops, the field collapses and the iron core loses magnetism (unlike a permanent magnet). Key advantage: The strength can be controlled by adjusting current or changing the number of coil turns.

Five-mark questions require detailed explanations, multi-step calculations, or comprehensive diagrams with description. They test synthesis and problem-solving ability. Here are three high-frequency 5-mark questions: **Q1: A 100 W electric heater is operated at 220 V. Calculate (a) the current flowing through the heater, (b) the resistance of the heater element, and (c) the heat produced in 2 minutes. [Given: Power P = VI; R = V/I; H = Pt = I²Rt]** *Full Solution:* Given: P = 100 W, V = 220 V, t = 2 minutes = 120 seconds (a) Current: P = VI → I = P/V = 100/220 = 10/22 = 5/11 A ≈ 0.45 A (b) Resistance: R = V/I = 220/(5/11) = 220 × 11/5 = 2420/5 = 484 Ω (c) Heat produced: H = Pt = 100 × 120 = 12,000 J = 12 kJ Alternative check: H = I²Rt = (5/11)² × 484 × 120 = (25/121) × 484 × 120 = (25 × 484 × 120)/121 = 12,000 J ✓ **Q2: A student wants to study the magnetic effect of current using a solenoid. She winds 200 turns of insulated copper wire around a cardboard tube and connects it to a battery via a switch. (a) Explain what happens when the switch closes. (b) She then changes the number of turns to 400. How will this affect the strength of the magnetic field? (c) If she replaces the cardboard tube with an iron core, what additional effect will this have?** *Full Solution:* (a) When the switch closes, current flows through the coil. The magnetic field produced by each turn adds up, creating a strong magnetic field inside and around the solenoid. The field direction is given by the right-hand rule: if fingers curl in the direction of current in the coil, the thumb points to the North Pole. (b) The magnetic field strength is proportional to the number of turns. If turns increase from 200 to 400 (doubled), the field strength approximately doubles (assuming the same current). This is because more loops mean more magnetic field lines are superimposed. (c) Replacing the cardboard tube with an iron core dramatically increases the field strength—often by a factor of 100 or more. The iron core gets magnetised by the external field from the coil, and the alignment of iron atoms amplifies the total magnetic field. This is why electromagnets use iron cores; they are far more powerful than air-core solenoids. **Q3: Draw a labelled circuit diagram of an electric bell. Explain the operation, and state two practical applications of electromagnets other than electric bells.** *Full Solution:* [Labelled diagram showing: Battery → Switch → Electromagnet (coil + iron core) → Armature (iron strip) with hammer attached → Gong → Spring → Contact screw that controls the circuit] **Operation:** When the switch closes, current flows through the electromagnet's coil. The magnetic field attracts the iron armature, pulling it and the hammer towards the coil. The hammer strikes the gong. As the armature moves, it loses contact with the circuit (at the contact point), breaking the circuit. No current flows, so the electromagnet loses its magnetic field. The spring immediately pushes the armature back to its original position, re-establishing contact. Current flows again, the electromagnet activates, and the cycle repeats 20–30 times per second, creating a continuous ringing sound. **Two practical applications of electromagnets:** 1. **Electric door locks and magnetic latches:** Electromagnets hold doors securely closed until a signal activates them. Commonly used in banks, offices, and laboratories. 2. **Telephone earpieces and loudspeakers:** A varying electric signal controls the electromagnet, which vibrates a diaphragm or speaker cone to produce sound. This allows conversion of electrical signals back into audible sound.

The CBSE has signalled shifts in assessment patterns starting from the 2026–27 academic year, particularly an increased focus on application-based and competency-based questions. For Chapter 10 (Electric Current and its Effects), expect these changes: **1. More "case study" or "scenario-based" questions:** Instead of asking "Define a fuse," papers will ask: "A household circuit has lights, a heater, and a fan. A fuse rated 15 A is installed. When the heater (10 A) and fan (6 A) are switched on together, the fuse blows. Explain why and suggest a solution." This tests real-world problem-solving, not rote definition. **2. Increased focus on energy calculations:** Expect more numerical problems involving H = I²Rt, P = VI, and cost of electricity. The 2024–25 papers show a trend of 2–3 calculation-based questions per paper. Budget time accordingly. **3. Diagram labelling with reasoning:** Simply drawing a circuit is no longer enough. Questions now ask: "Draw the circuit and explain why the electromagnet in an electric bell must use an iron core, not copper." This bridges diagram skills with conceptual understanding. **4. Reduced "state the formula" questions:** Direct formula questions (1-mark) are declining. Instead, you'll see: "A wire carries 2 A and has a resistance of 5 Ω. Calculate the heat produced in 10 seconds." The formula is implicit; the focus is on calculation and unit awareness. **5. Integration with other chapters:** Expect hybrid questions linking Chapter 10 (current and effects) with Chapter 11 (household electricity). Example: "In a household circuit, a 1000 W heater and 500 W fan are connected in parallel to a 220 V supply. Calculate the total current and suggest an appropriate fuse rating." This tests both magnetic/heating effects AND understanding of parallel circuits. **Recommendation:** As you revise, always ask yourself: "Why does this concept matter in real life?" rather than just memorising definitions. Your future exam questions will expect you to apply knowledge to unfamiliar situations.

Time management during the exam is critical. Here's a tested strategy for tackling Chapter 10 questions: **Step 1: Read all questions first (2 minutes).** Skim the entire paper and identify all Chapter 10 questions. This helps you mentally organise your approach and avoid missing a question. **Step 2: Classify by confidence level.** Sort questions into three buckets: - Green flag: Questions you're very confident about (1-mark MCQs, standard definitions). - Yellow flag: Moderately confident (3-mark concept questions with some calculation). - Red flag: Least confident (5-mark questions requiring multi-step solutions or detailed diagrams). **Step 3: Tackle in order (Green → Yellow → Red).** This strategy boosts morale early and ensures you secure marks in your strongest areas before time pressure builds. **Step 4: For 1-mark questions (30 seconds max per question):** If it's an MCQ, pick the most direct answer. Avoid overthinking. If it's "state" type, write 1–2 lines max. Remember: examiners give 1 mark for correct answer, not for lengthy explanations. **Step 5: For 3-mark questions (2–3 minutes each):** Use this format: - Sentence 1: Define or state the concept (e.g., "A fuse is a safety device...") - Sentence 2–3: Explain the mechanism or provide example (e.g., "When current exceeds rated value, the thin wire melts...") - Sentence 4 (if space): State consequence or application (e.g., "This breaks the circuit and protects appliances.") Draw a simple diagram if asked; label it clearly. **Step 6: For 5-mark questions (4–5 minutes each):** - Write the formula and given values clearly at the top. - Show each calculation step, even if it seems obvious. - Include units in every answer (e.g., 12,000 J, not 12,000). - For diagrams, label at least 5 components and explain the process in 3–4 sentences below the diagram. - Double-check your arithmetic before moving on. **Step 7: Diagram best practices (applies to all questions):** - Use a ruler for circuit diagrams. Wobbly lines lose marks even if conceptually correct. - Draw symbols accurately: electromagnet ≠ simple coil. Check your NCERT figure. - Label every component. Examiners expect: Battery, Switch, Electromagnet, Armature, Spring, Hammer, Gong, etc. - If asked to draw a circuit symbol (rheostat, fuse, etc.), compare with your textbook before finalising. **Step 8: Common mistakes to avoid:** - Confusing "magnetic field direction" with "current direction." Use right-hand rule consistently. - Writing H = I²Rt without showing substitution of values. Examiners want to see the working. - Forgetting units in final answers. "12" without "Joules" or "Amperes" loses 0.5 marks per question. - Drawing incomplete electric bell circuits (missing spring or contact point). This is a common oversight. **Step 9: Time buffer.** If you're on track with time, invest 1 minute per question to proofread for: - Spelling of technical terms ("electromagnet," not "electro-magnet"). - Correct formula substitution. - Logical flow of explanation. This strategy has been refined by coaching students across 50+ schools. Practise it with past papers at home before your exam. Start a 3-day free trial at cbsetutor.ai to access solved past papers with video explanations and step-by-step solving strategies tailored to this chapter.

Before your exam, bookmark these five critical points: **1. Heating Effect of Electric Current:** H = I²Rt (or H = Pt, or H = V²t/R) - H = heat in Joules, I = current in Amperes, R = resistance in Ohms, t = time in seconds, P = power in Watts, V = voltage in Volts. - Practical: A 5 A current produces 25 times more heat than a 1 A current (since heat ∝ I²). **2. Magnetic Field and Current:** B ∝ I (magnetic field strength is directly proportional to current). - Right-hand rule: Curl fingers in the direction of current, thumb points to North Pole. - Solenoid (electromagnet) field strength increases with more turns and higher current. - Iron core amplifies field by 50–100 times compared to air-core solenoid. **3. Fuse Rating:** Always choose a fuse rated slightly above the normal operating current. - Example: A 1000 W heater on 220 V uses I = P/V = 1000/220 ≈ 4.5 A. Use a 5 A fuse, not a 4 A. - Fuse rating determines at what current the wire melts and breaks the circuit. **4. Electric Bell Mechanism:** Electromagnet → Armature attracts → Hammer strikes → Contact breaks → Spring resets → Cycle repeats. - The key is understanding the feedback loop: current flows → electromagnet attracts → armature moves → contact breaks → electromagnet deactivates → spring pushes back → contact restores. **5. Circuit Symbols to memorise:** - Resistor: Rectangle - Variable resistor (rheostat): Rectangle with diagonal arrow - Electromagnet: Coil symbol (spiral) or coil around iron core - Fuse: Short line with bulge (or special symbol per textbook) - Electric bell: Bell symbol or circle with "B" Practise drawing these 20 times until they're automatic. Diagram accuracy in exams directly correlates with marks scored on 3-mark and 5-mark questions.