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Question: In a large building, there are 15 bulbs of \[40\;{\rm{W}}\], 5 bulbs of \[100\;{\rm{W}}\], 5 fans of...

In a large building, there are 15 bulbs of 40  W40\;{\rm{W}}, 5 bulbs of 100  W100\;{\rm{W}}, 5 fans of 80  W80\;{\rm{W}} and 1 heater of 1  kW1\;{\rm{kW}}. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:
A. 12  A12\;{\rm{A}}
B. 14  A14\;{\rm{A}}
C. 8  A8\;{\rm{A}}
D. 10  A10\;{\rm{A}}

Explanation

Solution

The above problem is resolved by summarizing every value of power rating, as contributed by the appliances in the building electrical board's loading. This power rating value is applied to the mathematical relation for the minimum current that flows through the circuit board by dividing it by the supply voltage throughout the building.

Complete step by step answer:
Given:
The power rating for the 1 bulb is 400 W.
The power rating of 5 other bulbs is, 100 W.
The power rating of the fan is, 80 W.
The power rating of the heater is, 1  kW=1kW×1000  W1  kW=1000  W1\;{\rm{kW}} = 1{\rm{kW}} \times \dfrac{{1000\;{\rm{W}}}}{{1\;{\rm{kW}}}} = 1000\;{\rm{W}}.
The supply voltage is, V=220  VV = 220\;{\rm{V}}.
Now, as the quantity for the power rating of the bulb of 400 W is 15. Then total power rating is,
P1=15×40  W=600  W{P_1} = 15 \times 40\;{\rm{W}} = 600\;{\rm{W}}.
Similarly, the quantity for the power rating of the bulb of 100 W is 5. Then total power rating is,
P2=5×100  W=500  W{P_2} = 5 \times 100\;{\rm{W}} = 500\;{\rm{W}}.
The quantity for the power rating of the fan of 80 W is 5. Then total power rating is,
P3=5×80  W=400  W{P_3} = 5 \times 80\;{\rm{W}} = 400\;{\rm{W}}.
And, the power rating of single heater is, P4=1000  W{P_4} = 1000\;{\rm{W}}.

Then, the net power rating is,

P = {P_1} + {P_2} + {P_3} + {P_4}\\\ P = 600\;{\rm{W}} + 500\;{\rm{W}} + 400\;{\rm{W}} + 1000\;{\rm{W}}\\\ P = 2500\;{\rm{W}} \end{array}$$ The minimum capacity of the main fuse is given as, $$I = \dfrac{P}{V}$$ Solve by substituting the values as, $$\begin{array}{l} I = \dfrac{P}{V}\\\ I = \dfrac{{2500\;{\rm{W}}}}{{220\;{\rm{V}}}}\\\ I \simeq 12\;{\rm{A}} \end{array}$$ Therefore, the minimum capacity of the main fuse of the building will be 12 A **So, the correct answer is “Option A”.** **Note:** The mathematical formula for the current or the power rating in particular circuit analysis is necessary to resolve such conditions. The formula directly provides us the value of the variables like power rating, the current flow, and the voltage. Hence it is important to remember the relation.