Question

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

• A. $25\, A$
• B. $15\, A$
• C. $10\, A$
• D. $20\, A$

Answer 1

Answer: (D)

$20\, A$

Answer 2

1. Calculate total power:

   • Bulbs (45W): $15 \times 45\, W = 675\, W$
   • Bulbs (100W): $15 \times 100\, W = 1500\, W$
   • Fans (10W): $15 \times 10\, W = 150\, W$
   • Heaters (1kW): $2 \times 1000\, W = 2000\, W$
   • Total Power $P_{total} = 675 + 1500 + 150 + 2000 = 4325\, W$

2. Calculate total current: Using $P = VI$, so $I = P/V$. $I_{total} = \frac{4325\, W}{220\, V} \approx 19.66\, A$

3. Determine minimum fuse capacity: The fuse must be rated higher than the total current. The next standard fuse rating above $19.66\, A$ is $20\, A$.