Question

A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be used from a household supply. The rating of fuse to be used is:

• A. 2.5 A
• B. 5 A
• C. 7.5 A
• D. 10 A

Hint:

In parallel household circuits, simply add the powers first: \(P_{total} = \sum P_i\).
Then find the total current using \(I = P / V\).
This keeps calculations quick and avoids rounding errors from calculating individual currents.

Answer 1

The Correct Option is D

Step 1: Find the current each appliance draws on its own.
Using \(I = P/V\), the cooler draws \[ I_1 = \frac{1500}{200} = 7.5\text{ A} \] and the fan draws \[ I_2 = \frac{500}{200} = 2.5\text{ A} \]
Step 2: Combine them, since household points are wired in parallel.
Both appliances share the same \(200\) V supply and pull current side by side, so the total current from the supply is \[ I = I_1 + I_2 = 7.5 + 2.5 = 10\text{ A} \]
Step 3: Choose the fuse rating.
A fuse must allow the normal working current to pass through without tripping, so it needs to be rated at least as high as the total load current drawn by both appliances together.
\[ \boxed{10\text{ A}} \]