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# ​four bulbs B1,B2,B3,B4of 100w each are connected to220v main.the reading in an ideal ammeter will be

Arun
25763 Points
one year ago
We know that P = V2/R Here V across all the bulbs is 220 V and P = 100 W. Then the resistance of the bulb is R = V2/P = (220)2/100 = 48400/100 = 484 Ω The circuit when reduce gives 4 resistors each of resistance R = 484 Ω in parallel to each other. Hence, the equivalent resistance is R/4 = 484/4 Ω = 121 Ω Total current in the circuit is V/R = 220/121 = 20/11 and this is equally divided among the bulbs. Current flowing into B2, B3 and B4 will be reading of the ammeter = 20/11 x 3/4 = 15/11 A = 1.36 A
Vikas TU
14149 Points
one year ago
Dear student
Current through each bulb  =100/220=5/11A
Therefore current through ideal ammeter  = 3×5/11=1.35A
Good Luck