Animated Solution for Physics - Current Electricity: In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. Then, the minimum current should be

Calculating Building Load

Imagine you are tasked with designing the electrical panel for a large building. You have a multitude of appliances—bulbs, fans, and heavy-duty heaters—all running simultaneously. How do you determine the correct fuse rating to ensure the building doesn't plunge into darkness the moment everything is turned on? This classic problem is a perfect demonstration of how we apply the fundamental laws of current electricity in real-world scenarios.

In any standard household or building circuit, all electrical appliances are connected in parallel across the main power supply. This parallel configuration ensures that every appliance receives the full mains voltage—in this case, $220\text{ V}$.

Because they are in parallel, the total power drawn from the main source is simply the algebraic sum of the power consumed by each individual appliance.

Let's break down the total load in our building: - 15 bulbs, each consuming $40\text{ W}$ - 5 bulbs, each consuming $100\text{ W}$ - 5 fans, each consuming $80\text{ W}$ - 1 heater, consuming $1\text{ kW}$

Before we add them up, we must ensure all our units are consistent. The heater's power is given in kilowatts ($1\text{ kW}$), which we must convert to watts ($1000\text{ W}$).

Now, we can write the equation for the total power $P_{\text{total}}$:

Let's compute each term:

Adding these together gives us the total power consumption of the building:

We now know that the building draws a total of $2500\text{ W}$ of power from a $220\text{ V}$ supply. To find the total current $I$ flowing through the main line, we use the fundamental power equation:

Rearranging this to solve for current, we get:

Substituting our known values:

When we calculate this fraction, we find:

The total current drawn when all appliances are running is approximately $11.36\text{ A}$. The purpose of a fuse or a main circuit breaker is to protect the wiring from excessive current. However, it must also allow the normal operating current to flow without interrupting the circuit.

If we were to choose a $10\text{ A}$ fuse, it would blow immediately because the operating current ($11.36\text{ A}$) exceeds its capacity. Therefore, the minimum current capacity of the fuse must be an integer value just greater than the maximum drawn current.

Looking at our options, $12\text{ A}$ is the perfect fit. It safely accommodates the $11.36\text{ A}$ load while still providing protection against larger, dangerous surges.