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 :

Current Electricity PYQ — JEE Main 2020 · 7 Jan · Shift 2 · Q50 with Solution

View written solutionFree

Correct answer: B

Find the total power consumed

Given appliances:
- $15$ bulbs of $45\,\text{W}$
- $15$ bulbs of $100\,\text{W}$
- $15$ fans of $10\,\text{W}$
- $2$ heaters of $1\,\text{kW} = 1000\,\text{W}$
Total power: $P = 15(45) + 15(100) + 15(10) + 2(1000)$

Compute each term: $15(45) = 675\,\text{W}$ $15(100) = 1500\,\text{W}$ $15(10) = 150\,\text{W}$ $2(1000) = 2000\,\text{W}$

Therefore, $P_{\text{total}} = 675 + 1500 + 150 + 2000 = 4325\,\text{W}$
Find the total current drawn

Using $P = VI \Rightarrow I = \frac{P}{V}$

Given voltage $V = 220\,\text{V}$ , $I = \frac{4325}{220} \approx 19.66\,\text{A}$
Determine the minimum fuse rating

The fuse rating should be just above the normal current drawn.

Since the current is about $19.66\,\text{A}$ , the nearest higher standard fuse value is: $20\,\text{A}$
Check options
- A: $15\,\text{A}$ — too small
- B: $20\,\text{A}$ — correct
- C: $25\,\text{A}$ — works, but not minimum
- D: $10\,\text{A}$ — too small

Hence, the minimum fuse capacity is $20\,\text{A}$ .