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PhysicsCurrent ElectricityElectrical PowerEasy2 minPYQ_2020
PhysicsEasysingle choice

In a building there are15bulbs of45W,15bulbs of100W,15small fans of10Wand2heaters of1kW. The voltage of electric main supply is220V. The minimum fuse capacity (rated value) of the building will be:

Options:

Answer:
D
Solution:

Total power is 15×45+15×100+15×10+2×1000
=4325W
So current =4325220=19.66 A 20 A

so we have to use a fuse which can tolerate atleast 20 A current.

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical Power
2mℹ️ Source: PYQ_2020

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