In a building there are 15 bulbs of 45 W , 15 bulbs of 100 W , 15 small fans of 10 W and 2 heaters of 1 k W . The voltage of electric main is 220 V . The minimum fuse capacity (rated value) of the building will be

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>10</mn> <mtext> </mtext> <mi mathvariant="normal">A</mi> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>20</mn> <mtext> </mtext> <mi mathvariant="normal">A</mi> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>25</mn> <mtext> </mtext> <mi mathvariant="normal">A</mi> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mn>15</mn> <mtext> </mtext> <mi mathvariant="normal">A</mi> </math> </span></p>
3 Views|Posted 6 months ago
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6 months ago
Correct Option - 4
Detailed Solution:

Total power is ( 15 * 45 ) + ( 15 * 100 ) + ( 15 * 10 ) + ( 2 * 1000 ) = 4325 W .

So, current is 4325 220 = 19.66 A

Answer is 20 A m p .

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Physics Current Electricity 2025

Physics Current Electricity 2025

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