Class 10 Electricity Ncert Solutions -

Answer: From Ohm’s law ( V = IR ), if ( R ) constant and ( V ) becomes half, ( I ) also becomes half.

Answer: Required ( R_p = V/I = 220/5 = 44 , \Omega ) ( n ) resistors in parallel: ( R_p = 176 / n ) → ( n = 176/44 = 4 ) class 10 electricity ncert solutions

Answer: ( R = V/I = 12 / (2.5 \times 10^{-3}) = 4800 , \Omega ) Answer: From Ohm’s law ( V = IR

(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: Electric power ( P = V \times

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: ( H = V \times I \times t = V \times Q = 50 \times 96000 = 4.8 \times 10^6 , \text{J} )