CBSE Class 12 Physics: Current Electricity — Important Questions with Answers 2026

Short Answer Questions (2-3 Marks)

• Q: State and explain Kirchhoff's laws.
  Ans: First Law (Junction Rule): The algebraic sum of currents at any junction is zero. ΣI = 0 (based on conservation of charge). Currents entering are positive, leaving are negative. Second Law (Loop Rule): The algebraic sum of potential differences around any closed loop is zero. ΣV = 0 (based on conservation of energy). Apply sign convention: EMF positive if traversed from - to +, IR drop negative in direction of current.
• Q: Derive the condition for balance in a Wheatstone bridge.
  Ans: Four resistors P, Q, R, S arranged in a diamond. Galvanometer between junctions B and D. At balance, no current through galvanometer (Ig = 0). Applying Kirchhoff's loop rule: Loop ABDA: I₁P - I₂R = 0 → I₁P = I₂R ... (i). Loop BCDB: I₁Q - I₂S = 0 → I₁Q = I₂S ... (ii). Dividing (i) by (ii): P/Q = R/S. This is the balance condition.
• Q: Explain the principle of a potentiometer. How is it used to compare EMFs of two cells?
  Ans: Principle: A potentiometer is a uniform wire of length L with constant current. Potential drop across any length l is proportional to l: V = (E/L)×l = kl. To compare EMFs: Connect cell 1 → find balance length l₁ (E₁ = kl₁). Connect cell 2 → find balance length l₂ (E₂ = kl₂). Ratio: E₁/E₂ = l₁/l₂. Advantage: Potentiometer draws no current from the cell at balance (null method).
• Q: Define drift velocity. Derive the relation between current and drift velocity.
  Ans: Drift velocity (vd) is the average velocity of free electrons in a conductor when an electric field is applied. Derivation: In a conductor of length l, cross-section A, with n free electrons per unit volume. Volume = Al. Number of electrons = nAl. Charge = nAle. Time to cross length l: t = l/vd. Current I = charge/time = nAle/(l/vd) = nAevd. Therefore I = nAevd.

Long Answer / Application Questions (4-6 Marks)

• Q: A cell of EMF 1.5 V and internal resistance 0.5 Ω is connected to a 2.5 Ω resistor. Find current, terminal voltage, and power dissipated.
  Ans: Current I = E/(R+r) = 1.5/(2.5+0.5) = 1.5/3 = 0.5 A. Terminal voltage V = E - Ir = 1.5 - 0.5(0.5) = 1.25 V. Power dissipated in R: P = I²R = (0.5)²(2.5) = 0.625 W. Power dissipated internally: P_int = I²r = (0.5)²(0.5) = 0.125 W.

Exam Tips for This Chapter

• Revise all definitions and laws from Current Electricity — 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 significant marks in the board exam