Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical PowerEasy2 minPYQ_2020
PhysicsEasysingle choice

In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. If the voltage of the electric main is 220 V, then the minimum fuse capacity (in A) of the building should be

Options:

Answer:
A
Solution:

Total power is


(15×45)+(15×100)+(15×10)+(2×1000)=4325 W
So current is =4325220=19.66 A

Ans is 20 A

 

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Electrical Power
⏱ 2mℹ️ Source: PYQ_2020

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