Question

In a large building, there are $15$ bulbs of $40\text{W}$, $5$ bulbs of $100\text{W}$, $5$ fans of $80\text{W}$ and $1$ heater of $1\text{kW}$. The voltage of the electric mains is $220\text{V}$. The minimum capacity of the main fuse of the building will be:

• A. $14\text{A}$
• B. $8\text{A}$
• C. $10\text{A}$
• D. $12\text{A}$

Answer 1

The correct option is D $12\text{A}$

Given: $15$ bulbs $\to 40W$
$5$ bulbs $\to 100W$
$5$ fans $\to 80W$
$1$ heater $\to 1kW$
$V=220V$

Total power,

$P=(15\times 40)+(5\times 100)+(5\times 80)+(1\times 1000)P=2500\text{W}$

Current capacity, $i=\frac{P}{V}$

$\Rightarrow I=\frac{2500}{220}$

$I=\frac{125}{11}=11.3\text{A}$

Minimum capacity of main fuse should be $12\text{A}.$

Hence, option $\left(C\right)$ is correct.