Question

In a large building , there are $15$ bulbs of $40\text{W}$, $5$ bulbs of $100\text{W}$, $5$ fans of $80\text{W}$ and $1$ heater of $1\text{KW}$. If the voltage of main electric supply is $220\text{V}$ then the minimum current capacity of the main fuse will be:

• A. $8\text{A}$
• B. $10\text{A}$
• C. $12\text{A}$
• D. $14\text{A}$

Answer 1

The correct option is C $12\text{A}$

From the data given in the question,

Total power consumed by the combination of given electrical elements will be,

$P_T=(15\times P_{\text{Bulb}})+(5\times P_{\text{Bulb}})+(5\times P_{\text{Fan}})+(1\times 1000)$

$\Rightarrow P_T=(15\times 40)+(5\times 100)+(5\times 80)+(1\times 1000)$

$\Rightarrow P_T=2500\text{W}$

The fuse should be able to withstand power supplied at given operating voltage $220\text{V}$

Therefore By using, $$P=V\times i$$Or, $P_T=V\times i_{fuse}$

Substituting the data,

$2500=220\times i_{fuse}$

$\therefore i_{fuse}=\frac{2500}{220}=11.3A$

Thus for safe operation of the fuse , its minimum current capacity should be $12\text{A}$.

Hence, option (c) is the correct answer.