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+2 votes
39.6k views
in Current electricity by (25 points)
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Four bulbs B1, B2, B3 and B4  of 100 W each are connected to 220 V main as shown in the figure. The reading in an ideal ammeter will be:

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1 Answer

+1 vote
by (323k points)

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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