an electric bulb is rated as 60 W -220 V. what is the resistance and safe limit of current through it
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P = VI
60 = 220 x I
I = 3/11 A = 0.2727...A
R = V/I
= 220 / 3/11
= 220 x 11 /3
= 806.667 ohms
therefore safe limit of current is approx 0.3 A
60 = 220 x I
I = 3/11 A = 0.2727...A
R = V/I
= 220 / 3/11
= 220 x 11 /3
= 806.667 ohms
therefore safe limit of current is approx 0.3 A
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