​four bulbs B_1,B₂,B₃,B_{4of 100w each are connected to220v main.the reading in an ideal ammeter will be}

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

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