Class 10 Electricity Ncert Solutions ~repack~ May 2026
Answer: Required ( R_p = V/I = 220/5 = 44 , \Omega ) ( n ) resistors in parallel: ( R_p = 176 / n ) → ( n = 176/44 = 4 )
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
I’ll provide answers to the and the exercise questions at the end of the chapter. In-Text Questions (Page 200) Q1. What does an electric circuit mean? Answer: An electric circuit is a continuous and closed path for the flow of electric current, consisting of a source of electricity (like a cell or battery), connecting wires, a switch, and a device (like a bulb). Answer: Required ( R_p = V/I = 220/5
(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) I’ll provide answers to the and the exercise
Answer: ( H = I^2 R t = 5^2 \times 20 \times 30 = 25 \times 20 \times 30 = 15000 , \text{J} ) In-Text Questions (Page 220) Q1. What determines the rate at which energy is delivered by a current? Answer: Electric power ( P = V \times I = I^2 R = V^2 / R )
Answer: In parallel across the two points.
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.
