In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The maximum capacity of the main fuse of the building will be
A
12 A
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
14 A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
8 A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10 A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 12 A Here, total power, P=(15×40)+(5×100)+(5×80)+(1×1000)=2500W Now, using relation, P = VI or I=PV I=2500220 = 11.3 A Thus, the minimum capacity should be 12A.