Answer: Each bulb current ( I = P/V = 10/220 = 1/22 , \text{A} ) Number = total current / bulb current = ( 5 / (1/22) = 110 ) bulbs
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: From Ohm’s law ( V = IR ), if ( R ) constant and ( V ) becomes half, ( I ) also becomes half. Answer: Each bulb current ( I = P/V
Answer: Alloys have higher resistivity and do not oxidize easily at high temperatures. Answer: The SI unit of electric current is ampere (A)
Here are the (based on the latest NCERT textbook).
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: The SI unit of electric current is ampere (A) . 1 ampere = 1 coulomb per second (1 A = 1 C/s).