Question

In a building there are $15$ bulbs of $45\text{W},15$ bulbs of $100\text{W},15$ small fans of $10\text{W}$ and $2$ heaters of $1\text{kW}$. The voltage of electric main is $220\text{V}$. The minimum fuse capacity (rated value) of the building will be:

• A. $10\text{A}$
• B. $25\text{A}$
• C. $15\text{A}$
• D. $20\text{A}$

Answer 1

The correct option is D $20\text{A}$

$\text{Net power}$

$P=(15\times 45)+(15\times 100)+(15\times 10)+(2\times 1000)$
$P=(15\times 155)+2000=4325\text{W}$
$\text{Power}P=VI$
$\Rightarrow I=\frac{P}{V}$
$\therefore I_{min}=\frac{4325}{220}=19.66\text{A}\approx 20\text{A}$

Hence, option $\left(\text{D}\right)$ is correct.