(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)
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 class 10 electricity ncert solutions
Answer: Required ( R_p = V/I = 220/5 = 44 , \Omega ) ( n ) resistors in parallel: ( R_p = 176 / n ) → ( n = 176/44 = 4 ) (a) 100 W (b) 75 W (c) 50
Answer: From Ohm’s law ( V = IR ), if ( R ) constant and ( V ) becomes half, ( I ) also becomes half. \text{m} ) If diameter doubled
Answer: Total R = 0.2+0.3+0.4+0.5+12 = 13.4 Ω ( I = 9 / 13.4 \approx 0.67 , \text{A} ) (same in series)
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.
Here are the (based on the latest NCERT textbook).