Question

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

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

Answer 1

The correct option is B

$20A$

Step 1: Given

The power rating of first $15bulbs$: $P_1=45W$

The power rating of another $15bulbs$: $P_2=100W$

The power rating of $15fans$: $P_3=10W$

The power rating of first $2heaters$: $P_4=1kW=1000W$

Step 2: Formula Used

$TotalPowerConsumption=Numberofappliances\times Powerratingofeachappliance$

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

Step 3: Calculate the total power consumption

$\begin{array}{rcl}P& =& 15\times 45W+15\times 100W+15\times 10W+2\times 1000W\\ & =& 675W+1500W+150W+2000W\\ & =& 4325W\end{array}$

Step 4: Calculate the minimum current using the formula

$\begin{array}{rcl}I& =& \frac{4325}{220}A\\ & =& 19.66A\\ & \approx & 20A\end{array}$

Hence, the minimum fuse capacity is $20A$.