A household circuit has a fuse of 5A rating.Calculate the max. number of bulbs of rating 60W,220V each which can be connected in the circuit.
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26
Maximum current that can flow out from the mains 220V : 5 A
let there be N bulbs connected in parallel.
current through each bulb = P/V = 60/220 A
max N = 5 / (60/220) = 55/3 = 18 (truncate the decimals)
let there be N bulbs connected in parallel.
current through each bulb = P/V = 60/220 A
max N = 5 / (60/220) = 55/3 = 18 (truncate the decimals)
Answered by
6
Answer: let maximum no. Of bulb connected in parallel be X
Current drawn by one bulb
P=VI
I = P/V =60/220 =3/11 Ampere
No. Of bulbs that can be safely used with 5 A fuse
X = 5/3/11 = 55/3 = 18.33
Hence 18 bulbs can be safely used with 5 A fuse
Explanation:
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