Class 10 Electricity Ncert Solutions Link

Answer: In parallel across the two points.

Answer: ( R = V/I = 12 / (2.5 \times 10^{-3}) = 4800 , \Omega )

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. class 10 electricity ncert solutions

Answer: Each bulb current ( I = P/V = 10/220 = 1/22 , \text{A} ) Number = total current / bulb current = ( 5 / (1/22) = 110 ) bulbs

Answer: Required ( R_p = V/I = 220/5 = 44 , \Omega ) ( n ) resistors in parallel: ( R_p = 176 / n ) → ( n = 176/44 = 4 ) Answer: In parallel across the two points

Answer: (i) 9 Ω = two in parallel (3 Ω) + one in series (6 Ω) (ii) 4 Ω = two in series (12 Ω) in parallel with one (6 Ω): ( 1/R = 1/12 + 1/6 = 3/12 ) → R=4 Ω

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1 Answer: Each R, series ( R_s = 2R ), parallel ( R_p = R/2 ) Heat ( H = V^2 t / R ) → ( H_s / H_p = R_p / R_s = (R/2) / (2R) = 1/4 ) → (c) Answer: A cell, battery, or power supply

Answer: ( R = \rho l / A ) → ( l = R A / \rho ) ( A = \pi (0.25 \times 10^{-3})^2 = \pi \times 6.25 \times 10^{-8} ) ( l = (10 \times \pi \times 6.25 \times 10^{-8}) / (1.6 \times 10^{-8}) ) ( l \approx 122.7 , \text{m} ) If diameter doubled, area 4× → resistance becomes 1/4 → 2.5 Ω.