Answer: Each bulb current ( I = P/V = 10/220 = 1/22 , \text{A} ) Number = total current / bulb current = ( 5 / (1/22) = 110 ) bulbs
Answer: Alloys have higher resistivity and do not oxidize easily at high temperatures. class 10 electricity ncert solutions
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: Each bulb current ( I = P/V
Answer: The SI unit of electric current is ampere (A) . 1 ampere = 1 coulomb per second (1 A = 1 C/s). Answer: Electric power ( P = V \times
Answer: Required ( R_p = V/I = 220/5 = 44 , \Omega ) ( n ) resistors in parallel: ( R_p = 176 / n ) → ( n = 176/44 = 4 )
Answer: Each bulb current ( I = P/V = 10/220 = 1/22 , \text{A} ) Number = total current / bulb current = ( 5 / (1/22) = 110 ) bulbs
Answer: Alloys have higher resistivity and do not oxidize easily at high temperatures.
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: The SI unit of electric current is ampere (A) . 1 ampere = 1 coulomb per second (1 A = 1 C/s).
Answer: Required ( R_p = V/I = 220/5 = 44 , \Omega ) ( n ) resistors in parallel: ( R_p = 176 / n ) → ( n = 176/44 = 4 )
