ICSE Class 10 Physics: Current Electricity — Important Questions with Answers 2026

Short Answer Questions (2-3 Marks)

• Q: Define electric current. State its SI unit.
  Ans: Electric current is the rate of flow of electric charge through a conductor. I = Q/t. SI unit: Ampere (A). One ampere is the current flowing when one coulomb of charge passes through a cross-section of a conductor in one second.
• Q: State Ohm's Law. What is the condition for it to be valid?
  Ans: Ohm's Law states: The current flowing through a conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions remain constant. V = IR, where R is the resistance. It is valid only at constant temperature.
• Q: Define resistance. State the factors affecting resistance of a conductor.
  Ans: Resistance is the opposition offered by a conductor to the flow of current. R = V/I. SI unit: Ohm (Ω). Factors: (1) Length — R ∝ l (directly proportional). (2) Area of cross-section — R ∝ 1/A (inversely proportional). (3) Material — depends on resistivity (ρ). (4) Temperature — increases with temperature for metals. R = ρl/A.
• Q: Derive the formula for resistors in series.
  Ans: In series, same current flows through each resistor. Total voltage: V = V₁ + V₂ + V₃. By Ohm's law: IR = IR₁ + IR₂ + IR₃. Dividing by I: R_total = R₁ + R₂ + R₃. Total resistance is the sum of individual resistances.
• Q: Derive the formula for resistors in parallel.
  Ans: In parallel, same voltage across each resistor. Total current: I = I₁ + I₂ + I₃. By Ohm's law: V/R = V/R₁ + V/R₂ + V/R₃. Dividing by V: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃. Equivalent resistance is always less than the smallest individual resistance.

Long Answer / Application Questions (4-6 Marks)

• Q: What is electrical energy? Derive the expression E = VIt.
  Ans: Electrical energy is the work done by electric current. Work W = charge × potential difference = Q × V. Since Q = I × t, we get: E = VIt. Using V = IR: E = I²Rt. Using I = V/R: E = V²t/R. SI unit: Joule. Commercial unit: kWh (1 kWh = 3.6 × 10⁶ J).
• Q: State Joule's law of heating. Give two applications.
  Ans: Joule's law states: The heat produced in a conductor is (1) directly proportional to the square of current (H ∝ I²), (2) directly proportional to resistance (H ∝ R), (3) directly proportional to time (H ∝ t). H = I²Rt. Applications: (1) Electric heater/iron — nichrome wire has high resistance. (2) Electric fuse — melts when current exceeds safe limit.
• Q: Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel. Find the equivalent resistance.
  Ans: 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. Therefore R = 1 Ω. The equivalent resistance (1 Ω) is less than the smallest resistor (2 Ω).
• Q: An electric heater of resistance 50 Ω draws 4 A current. Find (i) voltage, (ii) power, (iii) energy consumed in 2 hours.
  Ans: (i) V = IR = 4 × 50 = 200 V. (ii) P = VI = 200 × 4 = 800 W = 0.8 kW. (iii) E = Pt = 0.8 × 2 = 1.6 kWh = 1.6 units.
• Q: What is the function of a fuse? Why is it connected in series?
  Ans: A fuse is a safety device that protects electrical appliances from damage due to overloading or short circuit. It is a thin wire of low melting point (tin-lead alloy) that melts and breaks the circuit when current exceeds the rated value. It is connected in series with the live wire so that when it melts, it breaks the circuit completely, stopping current flow to the appliance.

Exam Tips for This Chapter

• Revise all definitions and laws from Current Electricity — they are commonly asked as 1-2 mark questions
• Practice diagrams related to Current Electricity — neat labelled diagrams carry 2-3 marks
• For numericals, always show formula → substitution → answer with correct units
• Previous year analysis shows Current Electricity carries 8-12 marks in the board exam