Current Electricity - Previous Year Questions Solved Examples (Class 12 Physics)
Current electricity extends Class 10 concepts to advanced topics including EMF, internal resistance, and circuit analysis. These questions require both con
TL;DR: Current electricity extends Class 10 concepts to advanced topics including EMF, internal resistance, and circuit analysis. These questions require bot…
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Current electricity extends Class 10 concepts to advanced topics including EMF, internal resistance, and circuit analysis. These questions require both con
Current Electricity - Previous Year Questions — Solved Numerical Examples (Step by Step)
Example 1: Define EMF and internal resistance. Explain why internal resistance of a cell is important. [3 marks]
Solution: EMF (Electromotive Force) is the energy per unit charge provided by a cell, equal to the potential difference when no current flows. Internal resistance (r) is the resistance of the cell material itself. When current flows, terminal voltage V = EMF - Ir. Internal resistance is important because: (1) It causes voltage drop across the cell, (2) It limits maximum current that can flow, (3) It affects efficiency and power output, (4) It increases with age of battery, (5) It determines heating effect inside the cell.
Example 2: A cell of EMF 2V and internal resistance 1 ohm is connected to an external resistance of 9 ohm. Find the current flowing and terminal voltage. [3 marks]
Solution: Using Ohm's law for complete circuit: EMF = I(R + r). 2 = I(9 + 1). 2 = 10I. I = 0.2 A. Terminal voltage V = EMF - Ir = 2 - 0.2(1) = 1.8 V. Or V = IR = 0.2 × 9 = 1.8 V.
Example 3: Derive the condition for maximum power transfer to the external load. [4 marks]
Solution: Power in external circuit: P = I²R = (EMF/(R+r))² × R = EMF² × R / (R+r)². To find maximum, differentiate with respect to R: dP/dR = EMF² × [(R+r)² - R × 2(R+r)] / (R+r)⁴ = 0. This gives (R+r)² = 2R(R+r). R+r = 2R. R = r. Maximum power is transferred when external resistance equals internal resistance.
Example 4: Explain the working of a Wheatstone bridge. Derive the null point condition. [4 marks]
Solution: Wheatstone bridge consists of four resistors P, Q, R, S arranged in bridge form with galvanometer connecting the middle points. Current distribution creates potential difference across galvanometer. At null point (balanced condition), galvanometer shows zero current, meaning: V1 = V2 (potential at middle points is equal). Using Ohm's law: I1 × P = I2 × R and I1 × Q = I2 × S. Dividing: P/Q = R/S. This is the balancing condition. At null point: P × S = Q × R.
Example 5: Two cells of EMFs E1 and E2 with internal resistances r1 and r2 are connected in series. Find equivalent EMF and internal resistance. [3 marks]
Solution: When cells are in series (same terminals), EMFs add: Equivalent EMF = E1 + E2. Internal resistances add: Equivalent internal resistance = r1 + r2. Total current I = (E1 + E2) / (R + r1 + r2), where R is external resistance. This arrangement increases both voltage and internal resistance.
Example 6: State Kirchhoff's laws and apply them to analyze a circuit. [4 marks]
Solution: Kirchhoff's First Law (Junction Rule): Algebraic sum of currents at a junction is zero: ΣI = 0. This follows from conservation of charge. Kirchhoff's Second Law (Loop Rule): Algebraic sum of potential differences around a closed loop is zero: ΣV = 0. This follows from conservation of energy. Application: In multi-loop circuits, apply these laws to set up simultaneous equations and solve for unknown currents.
Example 7: A carbon resistor has a resistance of 100 ohm at 0°C. If the temperature coefficient is 0.0005/°C, find resistance at 50°C. [2 marks]
Solution: Using R = R0(1 + αΔT). R0 = 100 ohm, α = 0.0005/°C, ΔT = 50°C. R = 100(1 + 0.0005 × 50) = 100(1 + 0.025) = 100 × 1.025 = 102.5 ohm.
Tips
- Always distinguish between EMF and terminal voltage; they are equal only when no current flows.
- In circuit problems, use both Ohm's law and Kirchhoff's laws systematically.
- For series and parallel combinations, remember rules: V same in parallel, I same in series.
- Temperature effects on resistance are important; use R = R0(1 + αΔT).
Frequently Asked Questions
Why does terminal voltage drop when more current is drawn from a cell?
As current increases, voltage drop across internal resistance (Ir) increases, so terminal voltage V = EMF - Ir decreases.
What is the difference between a voltmeter and an ammeter?
Voltmeter measures potential difference and is connected in parallel. Ammeter measures current and is connected in series.
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