Class 9 Science Chapter 10: Electric Current and its Effects — Important Questions & Answers

Chapter 10 sits at the intersection of theory and practical demonstration. The CBSE Class 9 board exam emphasizes application-based learning, meaning examiners move beyond 'define heating effect' and ask 'why does a fuse wire melt?' or 'design an electromagnet for a specific purpose.' The 2024–25 rationalized syllabus removed heavy derivations but strengthened conceptual depth and safety awareness (e.g., role of fuses in household circuits, why electromagnets are preferred over permanent magnets in appliances). Multi-step numerical questions on current, resistance, and power remain board staples. Additionally, case-study questions increasingly feature real-world scenarios like circuit safety or design of electric bells. Practising these 18 questions ensures you master the exact question patterns, vocabulary, and depth examiners expect—boosting both confidence and marks.

**Question 1:** Which of the following is the correct symbol for a cell in a circuit diagram? (a) ⊗ (b) —||— (c) ≈≈≈ (d) ⚙ **Answer:** (b) —||— (Two parallel lines, one longer and one shorter, represent a cell. The longer line is the positive terminal.) **Question 2:** The heating effect of electric current is utilised in which of the following? (a) Electric motor (b) Electric heater (c) Galvanometer (d) Electromagnet **Answer:** (b) Electric heater. When current flows through a high-resistance wire (nichrome), electrical energy converts to heat. Motors use magnetic effect; electromagnets and galvanometers use magnetic effect; heaters exploit heating effect (H = I²Rt). **Question 3:** A fuse wire melts and breaks the circuit if the current exceeds a safe limit. This happens because: (a) Fuse has infinite resistance (b) Fuse generates excessive heat (c) Fuse repels electrons (d) Fuse absorbs all energy **Answer:** (b) Fuse generates excessive heat. A fuse wire is made of an alloy with a specific melting point. When overcurrent flows, heat (H = I²Rt) builds up rapidly, melting the wire and breaking the circuit before damage occurs. **Question 4:** The magnetic effect of electric current is used in: (a) Electric kettle (b) Electric iron (c) Electromagnet (d) Both (a) and (b) **Answer:** (c) Electromagnet. Electromagnets rely on the magnetic field produced when current flows through a coil. Kettles and irons use heating effect. **Question 5:** In an electric bell, the electromagnetic force is used to: (a) Generate light (b) Attract the armature, creating a sound (c) Heat the filament (d) Repel the hammer **Answer:** (b) Attract the armature, creating a sound. When current flows through the electromagnet, it attracts the armature (soft iron), which moves and strikes the gong. The motion and contact breaking create the ringing sound.

**Question 1:** Explain why a fuse is placed in series in a household circuit, not in parallel. **Answer:** A fuse must be in series because it needs to control the same current flowing through the entire circuit. If placed in parallel, the fuse carries only a fraction of the total current and may not melt even if the main circuit is overloaded, failing to protect the circuit. In series, all current passes through the fuse wire; any excess melts the wire, breaking the entire circuit safely. **Question 2:** A nichrome wire has a resistance of 5 Ω. When a current of 2 A flows for 10 seconds, calculate the heat generated. (H = I²Rt) **Answer:** H = I²Rt = (2)² × 5 × 10 = 4 × 5 × 10 = 200 J. The heat generated is 200 joules, which is why nichrome wires are used in heating appliances—they produce significant heat at safe currents. **Question 3:** Distinguish between the heating effect and magnetic effect of electric current with one example each. **Answer:** **Heating effect:** Electrical energy converts to heat due to resistance. Example: Electric heater. Current flows through nichrome wire, which has high resistance, producing heat (H = I²Rt). **Magnetic effect:** Current-carrying conductor creates a magnetic field. Example: Electromagnet. Current in a coil produces a magnetic field strong enough to attract iron, used in door locks and electric bells. **Question 4:** Why is copper preferred over nichrome wire for main circuit wiring, but nichrome is used in heating elements? **Answer:** Copper has low resistance (~1.7 × 10⁻⁸ Ω·m), so minimal heat is wasted in main wiring. Nichrome has high resistance (~100 × 10⁻⁸ Ω·m), generating useful heat. Using nichrome in mains would cause excessive heat loss and fire risk; using copper in heaters would not produce enough heat. **Question 5:** Explain why an electric bell continues to ring as long as current flows, not just once. **Answer:** In an electric bell circuit, the electromagnet attracts the armature, which strikes the gong. But when the armature moves forward, it breaks the contact (opening the circuit), demagnetizing the electromagnet. A spring then pulls the armature back, re-establishing contact and re-energizing the electromagnet. This cycle repeats continuously, creating continuous rings.

**Question 1:** Draw a circuit diagram showing a cell, switch, ammeter, and a resistor (nichrome wire) arranged in series. Label the symbols correctly and explain why the ammeter is placed in series. **Answer:** [Description of circuit: Cell (—||—) connected to Switch (⚙), then to Ammeter (circle with 'A'), then to Resistor (zigzag), back to cell in a closed loop.] The ammeter measures the current flowing through the circuit. It must be in series because current is the same at all points in a series circuit. If placed in parallel, it would provide an alternate low-resistance path, bypassing the resistor and giving incorrect readings. **Question 2:** A copper wire and a nichrome wire, both 1 meter long and 1 mm² cross-section, are heated by a current of 5 A for 30 seconds. Which wire gets hotter? Why? (Given: ρ_copper ≈ 1.7 × 10⁻⁸ Ω·m, ρ_nichrome ≈ 100 × 10⁻⁸ Ω·m) **Answer:** Using R = ρL/A: R_copper = (1.7 × 10⁻⁸ × 1) / 1 × 10⁻⁶ = 0.017 Ω. R_nichrome = (100 × 10⁻⁸ × 1) / 1 × 10⁻⁶ = 1 Ω. Heat H = I²Rt: H_copper = 25 × 0.017 × 30 = 12.75 J. H_nichrome = 25 × 1 × 30 = 750 J. Nichrome wire gets much hotter because it has much higher resistance, dissipating far more heat at the same current. **Question 3:** Explain the construction and working of an electromagnet. What advantage does an electromagnet have over a permanent magnet? **Answer:** **Construction:** A soft iron core is wrapped with an insulated copper wire, forming a coil. The ends of the coil are connected to a power source. **Working:** When current flows through the coil, each loop generates a magnetic field that adds up, magnetizing the soft iron core. The entire assembly acts as a strong magnet. **Advantages:** (1) Magnetic strength can be controlled by varying current; (2) Magnetism can be switched on/off instantly; (3) Can be made stronger by increasing loops or core material; (4) Permanent magnet strength is fixed and cannot be controlled. Electromagnets are thus used in devices like electric bells, door locks, and cranes. **Question 4:** A classroom has ten 40 W bulbs and five 2 kW heaters connected to a 5 A fuse. Will the fuse blow? Justify your answer using calculations. (Assume supply voltage = 220 V) **Answer:** Total power = (10 × 40) + (5 × 2000) = 400 + 10,000 = 10,400 W. Using P = VI: I = P/V = 10,400 / 220 ≈ 47.3 A. The safe current is 5 A, but required current is 47.3 A. The fuse will blow immediately because current far exceeds the safe limit, causing excessive heat in the fuse wire (H = I²Rt), which melts and breaks the circuit to prevent fire.

**Question 1:** Explain the principle of the heating effect of electric current (Joule's law). Derive the relationship H = I²Rt and discuss its applications in daily life with at least two examples. **Answer:** **Principle:** When electric current flows through a conductor with resistance, the kinetic energy of electrons increases. These electrons collide with atoms, transferring energy as heat. The amount of heat depends on current intensity, resistance, and time. **Derivation:** Electrical energy supplied: E = VIt. By Ohm's law, V = IR, so E = IRIt = I²Rt. This energy converts to heat in the conductor, hence H = I²Rt (joules). **Applications:** (1) **Electric heater:** A nichrome wire with high resistance produces heat efficiently. For example, a 1000 W heater at 220 V draws I = P/V = 1000/220 ≈ 4.5 A. Over 1 hour, H = (4.5)² × R × 3600 produces sufficient heat to warm water. (2) **Electric kettle:** Current passes through a heating element; H = I²Rt causes rapid heating of water inside. (3) **Fuse protection:** A fuse wire is intentionally made from low-melting-point alloy so that excessive current generates enough heat (H = I²Rt) to melt the wire, breaking the circuit before damage occurs. Higher current = higher heat production = fuse melts faster. **Question 2:** Describe the complete structure and working mechanism of an electric bell circuit. Include a circuit diagram description, roles of each component, and explain why the bell rings continuously when switched on. **Answer:** **Structure:** (1) Power source (cell or battery), (2) Switch, (3) Electromagnet (coil around soft iron), (4) Armature (soft iron attached to spring), (5) Hammer/clapper, (6) Gong (bell), (7) Contact screw. **Circuit Diagram:** Cell → Switch → Contact screw → Electromagnet coil → back to cell. The armature is mechanically linked to the hammer and spring. **Working:** (1) When switch is ON, current flows through electromagnet, magnetizing the soft iron core. (2) Magnetic field attracts the armature, pulling it forward. The hammer strikes the gong, producing sound. (3) As armature moves forward, it breaks contact with the contact screw, cutting current flow. (4) Without current, electromagnet loses magnetism instantly. (5) Spring pulls armature back to original position, re-establishing contact. (6) Current flows again, electromagnet re-magnetizes, and cycle repeats continuously. **Why continuous ringing:** The circuit is self-breaking and self-making; the mechanical action of the armature controls the on-off cycle, causing repeated collisions and continuous ringing as long as switch remains ON. **Question 3:** Explain the function and importance of fuses in household electrical circuits. Discuss the principle behind fuse operation, calculate the correct fuse rating for a given load, and explain what happens if a thicker-than-required fuse wire is used. **Answer:** **Function:** A fuse is a safety device that automatically breaks an electrical circuit when current exceeds a safe limit, preventing fire, electric shock, and equipment damage. **Principle of Operation:** A fuse wire is made from a low-melting-point alloy (typically tin, lead, copper, zinc). When overcurrent flows, heat generated (H = I²Rt) rapidly raises the wire's temperature above its melting point. The wire melts, creating a break in the circuit. The higher the current, the faster the melting. **Calculation Example:** A household circuit has: (1) Three 100 W bulbs, (2) One 2 kW electric heater. Total power = 300 + 2000 = 2300 W. At 230 V supply: I = P/V = 2300/230 = 10 A. A fuse rated 10–15 A is appropriate. **Using thicker fuse wire:** If a 20 A fuse is used instead, it would not melt even at 15 A (unsafe current). Excessive current would heat the main wiring, melting insulation and causing fire risk. A fuse must be rated at or slightly above (10–20%) the normal operating current so it melts only during actual faults, providing effective protection.

**Case-Study Question:** A school laboratory is setting up a new circuit board with five 100 W lights, two 500 W fans, and one 2 kW heater. The main supply is 220 V, and the existing wiring and fuse are rated 15 A. The lab technician wants to add a circuit with an electromagnet (100 Ω resistance) that operates at 5 A for controlling access doors. (1) Will the current total exceed the fuse rating if all devices are switched ON simultaneously? (2) What is the maximum safe power that can be supplied with the current 15 A fuse? (3) Design a solution to safely integrate the electromagnet without tripping the fuse. (4) Explain why the electromagnet circuit should have a separate fuse. **Solution:** **Step 1: Calculate current for existing load.** Lights: P₁ = 5 × 100 = 500 W Fans: P₂ = 2 × 500 = 1000 W Heater: P₃ = 2000 W Total existing: P_total = 500 + 1000 + 2000 = 3500 W Current: I₁ = P_total / V = 3500 / 220 ≈ 15.9 A **Step 2: Add electromagnet current.** Electromagnet: I₂ = 5 A (given) Total current if all ON: I_total = 15.9 + 5 = 20.9 A **Answer (1):** YES, total current (20.9 A) exceeds fuse rating (15 A). The fuse will blow. **Step 3: Calculate maximum safe power.** With 15 A fuse: P_max = VI = 220 × 15 = 3300 W **Answer (2):** Maximum safe power = 3300 W. **Step 4: Design solution.** Remove the 2 kW heater from the main circuit temporarily, or use it on a separate circuit. With only lights and fans: P = 500 + 1000 = 1500 W, I = 1500/220 ≈ 6.8 A. Adding electromagnet (5 A): Total I = 6.8 + 5 = 11.8 A < 15 A. The circuit now safely operates without tripping the main fuse. The heater can run on a dedicated 20 A circuit. **Answer (3):** Separate heavy-load devices (heater) onto different circuits with appropriately rated fuses. **Step 5: Explain separate fuse for electromagnet.** The electromagnet circuit carries 5 A continuously. If a fault occurs (short circuit, insulation breakdown), current could spike to 20–30 A instantly. A dedicated fuse (rated 5–8 A) on this circuit would melt within milliseconds, protecting only the electromagnet winding and control circuit. The main 15 A fuse protects the entire building. Layered protection (separate fuses) is a best practice: each circuit has its own fuse matched to its load, preventing cascading failures. **Answer (4):** A separate fuse ensures fault in one circuit doesn't trip the main supply, and it provides precise protection matched to the electromagnet's 5 A rating.

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