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: Understanding the Question:
Two electrical appliances (a cooler and a fan) are connected in parallel to a standard household power supply of \(200\text{ V}\). We need to determine the total current drawn and select the appropriate fuse rating.

Step 2: Key Formula or Approach:
Electrical power (\(P\)) is given by:
\[ P = V \cdot I \implies I = \frac{P}{V} \]
Since household appliances are connected in parallel, the total power is the sum of the individual powers, and the total current is the sum of the individual currents.

Step 3: Detailed Explanation:

• Power of the cooler (\(P_1\)): \(1500\text{ W}\)

• Power of the fan (\(P_2\)): \(500\text{ W}\)

• Total power consumed (\(P_{total}\)):
\[ P_{total} = P_1 + P_2 = 1500 + 500 = 2000\text{ W} \]

• Supply Voltage (\(V\)): \(200\text{ V}\)

• Total current (\(I_{total}\)) drawn from the supply:
\[ I_{total} = \frac{P_{total}}{V} = \frac{2000}{200} = 10\text{ A} \]

• A fuse is a safety device designed to break the circuit if the current exceeds a certain safe limit.

• To allow the normal operation of both appliances simultaneously without unnecessarily blowing the fuse, the fuse rating must be at least equal to the total maximum load current, which is \(10\text{ A}\).


Step 4: Final Answer:
The appropriate rating of the fuse to be used is \(10\text{ A}\).