Answer: Charge on one electron ( e = 1.6 \times 10^{-19} , \text{C} ) Number of electrons in 1 C = ( \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} ) In-Text Questions (Page 202) Q1. Name a device that helps to maintain a potential difference across a conductor. Answer: A cell, battery, or power supply.
Answer: ( R = V/I = 12 / (2.5 \times 10^{-3}) = 4800 , \Omega )
Answer: Power ( P = V \times I = 220 \times 5 = 1100 , \text{W} ) Energy = ( P \times t = 1100 \times 2 = 2200 , \text{Wh} = 2.2 , \text{kWh} ) Chapter Exercise Solutions (Page 221–222) Q1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is: (a) 1/25 (b) 1/5 (c) 5 (d) 25 Answer: Each piece resistance = R/5. Parallel: 1/R' = 5/(R/5) → R' = R/25 → R/R' = 25 → (d) class 10 electricity ncert solutions
Answer: ( H = V \times I \times t = V \times Q = 50 \times 96000 = 4.8 \times 10^6 , \text{J} )
Answer: From Ohm’s law ( V = IR ), if ( R ) constant and ( V ) becomes half, ( I ) also becomes half. Answer: Charge on one electron ( e = 1
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Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below: I (amperes) : 0.5, 1.0, 2.0, 3.0, 4.0 V (volts) : 1.6, 3.4, 6.7, 10.2, 13.2 Plot V–I graph and calculate resistance. Slope ( \Delta V / \Delta I ) ≈ 3.3 Ω. Answer: ( R = V/I = 12 / (2
(a) 100 W (b) 75 W (c) 50 W (d) 25 W Answer: Resistance ( R = V^2 / P = 220^2 / 100 = 484 , \Omega ) At 110 V: ( P = V^2 / R = 110^2 / 484 = 25 , \text{W} ) → (d)