Current Electricity Ques 119
- 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 $ $A$
Show Answer
Answer:
Correct Answer: 119.(c)
Solution:
Formula:
- 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 $12 A$.
BETA
