Animated Solution for Physics - Current Electricity: 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

Imagine a building buzzing with activity. We have bulbs lighting up the rooms, fans spinning, and heaters keeping it warm. In any standard household or building, all these electrical appliances are connected in parallel to the main power supply. This parallel configuration ensures that every appliance receives the same full voltage—in this case, $V=220\text{ V}$—and can operate independently of the others.

Our goal is to figure out the minimum capacity of the main fuse required to keep everything running safely. The fuse is the ultimate gatekeeper; it sits in series with the main live wire and protects the entire circuit from drawing dangerously high currents.

To find the required fuse capacity, we first need to know the total current drawn from the mains. And to find the total current, we need the total power.

Because all appliances are in parallel, the total power consumed by the building is simply the algebraic sum of the power consumed by each individual appliance:

Let's break down the power consumption category by category. First, we look at the lighting. We have 15 bulbs rated at $45\text{ W}$ each, and another 15 bulbs rated at $100\text{ W}$ each.

Next, let's account for the fans and heaters. There are 15 small fans, each consuming $10\text{ W}$, and 2 heavy-duty heaters, each rated at $1\text{ kW}$ (which is $1000\text{ W}$).

Now, we bring it all together to find the grand total power consumption for the entire building when everything is turned on simultaneously:

With the total power known, we can find the total current drawn from the $220\text{ V}$ supply. We use the fundamental electrical power relation $P=VI$, which rearranges to:

Substituting our calculated values:

This $19.66\text{ A}$ is the maximum normal operating current that will flow through the main wire.

The fuse is our safety guard. It operates on the principle of Joule heating ($H=I^{2}Rt$). It must allow this $19.66\text{ A}$ to pass continuously without melting its internal wire. Therefore, its rated capacity must be slightly greater than this maximum load value.

Looking at our options ($25\text{ A}$, $10\text{ A}$, $20\text{ A}$, $15\text{ A}$), the $20\text{ A}$ fuse is the perfect fit. It is the minimum safe capacity required.

Why not $25\text{ A}$? A fuse that is rated too high is dangerous. If a minor fault or a slight short circuit causes the current to rise to, say, $22\text{ A}$, a $20\text{ A}$ fuse will quickly blow and protect the wiring from overheating. However, a $25\text{ A}$ fuse would let that dangerous $22\text{ A}$ current pass indefinitely, potentially causing a fire. Always choose the closest standard rating just above your maximum expected load!