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QJEEM14Phc0a6: JEE Main 2014 Physics - Class 12 JEEM14Phc0a6 Solution: In a large building, three are bulbs of , bulbs of , fans of…

In a large building, three are 1515 bulbs of 4040 WW , 55 bulbs of 100100 WW , 55 fans of 8080 WW and 11 heater of 11 kW.kW. The voltage of electric mains is 220220 V.V. The minimum capacity of the main fuse of the building will be:

Multiple Choice Options

  • A 88 AA
  • B 1010 AA
  • C 1212 AA
  • D 1414 AA

Step-by-Step Verified Solution

Total power consumed by electrical appliances in the building, Ptotal=2500W{P_{total}} = 2500W

Watt == Volt ×\times ampere

⇒2500=V×I\Rightarrow 2500 = V \times {\rm I}

⇒2500=220\Rightarrow 2500 = 220 I{\rm I}

⇒I=2500220=11.36≈12A\Rightarrow I = {{2500} \over {220}} = 11.36 \approx 12A

(Minimum capacity of main fuse)

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