Question

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Answer 1

(V*V)/R

1st lamp-

R1 =(V*V)/P

R1 = (220*220)/200 = 242 ohms

2nd lamp-

R2= (220*220)/100 = 484 ohms

Now, they are connected in series, so net resistance = 242+ 484 = 726 ohms

V= 220V

I= 220/726 = 0.303A

Power dissipated in first lamp, P1= (I*I)*R1 = (0.303*0.303)*242 = 22.222W

Power dissipated in 2nd lamp, P2= (I*I)*R2 = (0.303*0.303)*484 = 44.44 W

Power ratio, P1/P2 = 22.22/44.44 = 0.5

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Answer 2

Hey buddy here is ur answer !!!!

Solution:

Both the bulbs are connected in parallel.

Therefore, potential difference across each of them will be same that is 220 V, Current drawn by the bulb of rating 100 W is given by

I = P/V

=100/220

=0.4545A

Similarly, current drawn by the bulb of rating 60 W is given by

I = P/V

=60/220

= 0.2727A

∴ the total current drawn from the line

= 0.4545+0.2727 = 0.727A