An electric iron is rated 2kW at 220 V. Calculate the capicity of the fuse that should be used for the electric iron.
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as we know,
power (P) = V.I
= V.V/R
= ( V^2 )/R
according to question,
2000 = 220×220/R
R = 220×220/2000
R = 24.2 ohm
now the capacity or the possible current passes from the fuse will be,
I = V/R
I = 220/24.2
I = 9.09 amp
power (P) = V.I
= V.V/R
= ( V^2 )/R
according to question,
2000 = 220×220/R
R = 220×220/2000
R = 24.2 ohm
now the capacity or the possible current passes from the fuse will be,
I = V/R
I = 220/24.2
I = 9.09 amp
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