Question

A 110 V power line is protected by a 15 A fuse. What is the maximum number of 400 W lamps that can be simultaneously operated in parallel on this line without "blowing" the fuse because of an excess of current

Answer 1

Total number of lamps will be 4

Step-by-step explanation:

We have given power of the lamp W = 400 watt

Potential difference across the lamp V=110 volt

We know that power is equal to
`P=VI`

So
`400=110* I`

`I=3.636A`

Total current is given 15 A

As it is given that lamps are connected in parallel so total current is the sum of current through each lamp

So number of lamp will be
`n=(15)/(3.636)=4.125`

As the lamp can not be in negative

So total number of lamps will be 4