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In a large building there are 15 bulbs of 40W, 5 bulbs of 100 W. 5 fans of 80 W and banner
voltage of the clectric mains is 220 V. The minimum capacity of the mom fuse of the bo
WEE Mai 2014
(1) 8A
(2) 10 A
(3) 12 A
In the circuit shown the current in the 10 resistor is
JEE Mai 2015
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Answer: AMPS = Watt/Volt
Explanation: To find the number of amps, you divide the watts by the volts.
15@40= 600
5@100 = 500
5@80 = 400
Total watts = 1500
1500/220= 6.8A. So the minimum would be 8A fuse.
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