We know that power = voltage^2 ÷resistance
Therfore r1 = 220^2÷100= 484ohm
And r2 =220^2÷60=806.7ohm .so 2nd one s greater resistance.
Hope it helps
Two bulbs have rating 100W, 220V and 60W, 220V.
To Find :-
Which one has the greatest resistance.
Formula to be used :-
P = V²/R
Putting all the values, we get
For Bulb A
⇒ 100 = (220×220)/R₁
⇒ R₁ = 484 ohm
For Bulb B
⇒ 60 = (220 × 220)/R₂
⇒ R₂ = 806.6 ohm
Hence, bulb B having 60W, 220V will have greater resistance.
Terms to be known :-
P is power in Watt (W) .
R is resistance in ohm .
V is voltage in volts (V) .
Resistance = voltage^2/ power
R= 220^2/100 = 48400/100 = 484 ohms
R= 220^2/60 = 48400/60 = 806 ohms
The bulb with rating 60 W and 220 V has greater resistance.
You can do it directly by recognising that power is inversely proportional to resistance as voltage is constant. Hence, the bulb which requires more power will have less resistance and the one which requires less power will have greatest resistance
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