# In a building there are 15 bulbs of 45 W

Question:

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

1. (1) $10 \mathrm{~A}$

2. (2) $25 \mathrm{~A}$

3. (3) $15 \mathrm{~A}$

4. (4) $20 \mathrm{~A}$

Correct Option: , 4

Solution:

(4) Net Power, $P$

$=15 \times 45+15 \times 100+15 \times 10+2 \times 1000$

$=15 \times 155+2000 \mathrm{~W}$

Power, $P=V I$

$\Rightarrow \quad I=\frac{P}{V}$

$\therefore \quad I_{\text {main }}=\frac{15 \times 155+2000}{220}=19.66 A \approx 20 A$