In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 k W. If the voltage of the
electric main is 220 V, then the minimum fuse capacity (in A ) of the building should be

Options

A. 20 A
B. 15 A
C. `10 A`
D. 25 A

Explanation

To find the minimum fuse capacity in amperes for the marriage hall, we need to calculate the total power consumed by all the electrical appliances and then use the formula for current.

### Step-by-Step Solution:

1. **Calculate the total power consumed by the bulbs:**
- There are 15 bulbs of 45 W each.
- Total power from bulbs (P1) = Number of bulbs × Power per bulb = 15 × 45 W = 675 W.

2. **Calculate the total power consumed by the 100 W bulbs:**
- There are 15 bulbs of 100 W each.
- Total power from 100 W bulbs (P2) = Number of bulbs × Power per bulb = 15 × 100 W = 1500 W.

3. **Calculate the total power consumed by the fans:**
- There are 15 fans of 10 W each.
- Total power from fans (P3) = Number of fans × Power per fan = 15 × 10 W = 150 W.

4. **Calculate the total power consumed by the heaters:**
- There are 2 heaters of 1 kW (1000 W) each.
- Total power from heaters (P4) = Number of heaters × Power per heater = 2 × 1000 W = 2000 W.

5. **Calculate the total power consumed by all appliances:**
- Total power (P_total) = P1 + P2 + P3 + P4
- P_total = 675 W + 1500 W + 150 W + 2000 W = 4325 W.

6. **Convert total power to kilowatts (if needed):**
- P_total = 4325 W = 4.325 kW.

7. **Calculate the current using the formula:**
- The formula for current (I) is given by:
\[ I = \frac{P_{total}}{V} \]
- Where:
- \( P_{total} = 4325 \, \text{W} \)
- \( V = 220 \, \text{V} \)
- Therefore:
\[ I = \frac{4325 \, \text{W}}{220 \, \text{V}} \approx 19.68 \, \text{A} \]

8. **Determine the minimum fuse capacity:**
- Since fuse ratings are typically rounded up, we round 19.68 A to the nearest whole number, which is approximately 20 A.

### Final Answer:
The minimum fuse capacity of the building should be **20 A**.

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