Class 10 Electricity Ncert Solutions (2027)
Answer: The SI unit of electric current is ampere (A) . 1 ampere = 1 coulomb per second (1 A = 1 C/s).
(a) I²R (b) IR² (c) VI (d) V²/R Answer: (b) IR² class 10 electricity ncert solutions
(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: The SI unit of electric current is ampere (A)
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 ) What determines the rate at which energy is
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 = \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.
(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)
