Electricity - Previous Year Questions Solved Examples (Class 10 Science)
Electricity questions test understanding of circuits, Ohm's law, and power calculations. These board-style problems require both conceptual clarity and num
TL;DR: Electricity questions test understanding of circuits, Ohm's law, and power calculations. These board-style problems require both conceptual clarity an…
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Electricity questions test understanding of circuits, Ohm's law, and power calculations. These board-style problems require both conceptual clarity and num
Electricity - Previous Year Questions — Solved Numerical Examples (Step by Step)
Example 1: Three resistors of 2 ohm, 3 ohm, and 6 ohm are connected in parallel. Find the equivalent resistance. [2 marks]
Solution: For parallel combination: 1/Rp = 1/R1 + 1/R2 + 1/R3. 1/Rp = 1/2 + 1/3 + 1/6. Finding LCM: 1/Rp = 3/6 + 2/6 + 1/6 = 6/6 = 1. Therefore Rp = 1 ohm.
Example 2: A bulb rated 60W, 220V and another 40W, 220V are connected in series across a 220V supply. Which bulb glows brighter and why? [3 marks]
Solution: Resistance of bulbs: R1 = V²/P1 = (220)²/60 = 806.67 ohm. R2 = (220)²/40 = 1210 ohm. In series, current is same through both: I = V/(R1 + R2) = 220/2016.67 = 0.109 A. Power in each: P = I²R. P1 = (0.109)² × 806.67 = 9.6 W. P2 = (0.109)² × 1210 = 14.4 W. The 40W bulb (higher resistance) dissipates more power and glows brighter.
Example 3: Calculate the heat produced in a 5 ohm resistor when 2A current flows through it for 5 minutes. [2 marks]
Solution: Using H = I²Rt. I = 2A, R = 5 ohm, t = 5 min = 300 s. H = (2)² × 5 × 300 = 4 × 5 × 300 = 6000 J = 6 kJ.
Example 4: A wire of resistance 10 ohm is bent into a square. What is the equivalent resistance when current is passed between two adjacent corners? [3 marks]
Solution: When wire is bent into square, each side has resistance = 10/4 = 2.5 ohm. When current enters at one corner and exits at an adjacent corner, one path has 1 side (2.5 ohm) and parallel path has 3 sides (7.5 ohm). Using 1/Req = 1/2.5 + 1/7.5 = 3/7.5 + 1/7.5 = 4/7.5. Req = 7.5/4 = 1.875 ohm.
Example 5: State Ohm's law and verify it using a circuit diagram. [3 marks]
Solution: Ohm's law: V = IR, where V is potential difference, I is current, and R is resistance. The law states that at constant temperature, the current through a conductor is directly proportional to the potential difference applied. Verification: Set up circuit with variable power supply, ammeter in series, voltmeter in parallel. For different voltages, measure corresponding currents and calculate V/I = constant, proving Ohm's law.
Example 6: A fuse wire melts at 5A. What is the maximum power that can be safely drawn from a 220V supply through this fuse? [2 marks]
Solution: Using P = VI. Maximum current through fuse = 5A, Voltage = 220V. P = 220 × 5 = 1100 W.
Example 7: Two wires made of the same material have the same resistance but different lengths and diameters. If one wire is twice as long as the other, what is the ratio of their diameters? [3 marks]
Solution: Resistance R = rho × L/A, where rho is resistivity, L is length, A is cross-sectional area. For equal resistance: R1 = R2. rho × L1/A1 = rho × L2/A2. L1/A1 = L2/A2. Given L1 = 2L2, we get 2L2/A1 = L2/A2. So A1 = 2A2. Since A = pi × (d/2)², we have d1²/d2² = 2, giving d1/d2 = √2.
Tips
- Remember sign conventions in circuits: current flows from positive terminal, use proper ammeter and voltmeter connections.
- In series circuits, same current flows through all components; in parallel, same voltage appears across all components.
- Power rating of devices (e.g., 60W, 220V) tells you maximum safe operating conditions.
- Use dimensional analysis to check answers: resistance in ohms, current in amperes, voltage in volts.
Frequently Asked Questions
Why does a thicker wire have lower resistance?
Resistance R = rho × L/A increases with length and decreases with cross-sectional area. Thicker wire has larger area, allowing more charge carriers to flow, reducing resistance.
What is the purpose of a fuse in a circuit?
A fuse melts and breaks the circuit when current exceeds a safe limit, protecting the circuit and appliances from damage due to overload or short circuit.
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