Current Electricity 5 Question 8
8. 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 minimum capacity of the main fuse of the building will be
(2014 Main)
(a) $8 A$
(b) $10 A$
(c) $12 A$
(d) $14 A$
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Solution:
- Total power $(P)$ consumed
$=(15 \times 40)+(5 \times 100)+(5 \times 80)+(1 \times 1000)=2500 W$
As we know,
Power, $P=V I \Rightarrow I=\frac{2500}{220} A=\frac{125}{11}=11.3 A$
Minimum capacity should be $12 A$.
