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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

Respuesta :

Answer:

Total number of lamps will be 4            

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 [tex]P=VI[/tex]

So [tex]400=110\times I[/tex]

[tex]I=3.636A[/tex]

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 [tex]n=\frac{15}{3.636}=4.125[/tex]

As the lamp can not be in negative

So total number of lamps will be 4