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rjkz [21]
2 years ago
15

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
Physics
1 answer:
Nitella [24]2 years ago
5 0

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 P=VI

So 400=110\times 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=\frac{15}{3.636}=4.125

As the lamp can not be in negative

So total number of lamps will be 4

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