Question

18. An electrician puts a fuse of rating 5A in that part of domestic electrical circuit in which anelectrical
heater of rating 1.5kW, 220V is operating. What is likely to happen in this caseand why? What
change if any needs to be made?.​

Answer 1

Answer:

Given,

Power of electrical circuit = 1.5 kW = 1500 W

Potential Difference of the circuit = 220 V

Rating of the fuse = 5 A

Let the current through the circuit be 'I'.

We know that,

Power = Potential Difference × Current

=> P = V × I

Thus, we can derive,

=> I = P / V

By applying value, we get,

=> I = 1500 / 220 = 6.81 A

So, current through the circuit = 6.81 A

Here,

Current through the circuit > Rating of fuse

=> 6.81 A > 5 A

In this case the fuse wire will melt down and the circuit will break and hence stop working.

We know that , the fuse wires of circuit are rated to operate at a certain limit (current). If that limit (current) exceeds, then the wire melts breaking the circuit. And if the limit (current) is less, then the circuit will not work properly as the components will not get enough current supply.

Here, since the fuse is rated at 5 A but the current through the circuit is 6.81 A, which is more than the rating of fuse. So, the fuse wire will melt breaking the circuit.

The reason because the fuse wire is made up of a certain material which is able to work properly only at a certain range of current. And if the current exceeds, the heat will be produced more and thus the fuse wire will melt down. As, heat is directly proportional to current.

There can be certain changes made. They are :

• A new fuse should be applied in the circuit of 7 A or more rating like 8 A, 10A but not more than 10 A else components will fail in working. As higher rating fuse have have resistance which will oppose the flow of current.

• To decrease your chances of blowing a fuse, compare the wattage and amperage rating of each of these devices with the amount of power your circuits can handle. This information should be easy to find on their packaging, rating plates or tags.

• Also, the electrical appliances along with transformers must be checked that they do not use more current than rating.

'A piece of Supplementary Counsel' :-

• Working of fuses = The fuses work on the principle of the heating effect of the current. Fuse is always connected in series with the electrical circuit. When the excessive current or heat is generated due to heavy current flows in the circuit, the fuse melts down due to the low melting point of the element and it opens the circuit.

• Heating effect of electric current = When current flows through a conductor, heat energy is generated in the conductor. The heating effect of an electric current depends on three factors: The resistance, R of the conductor. A higher resistance produces more heat.

Answer 2

Answer:

Here is your answer:

Explanation:

Given:-

     Power of electric circuit = 1.5 kW

     Potential Difference = 220V

     Rating of fuse = 5A

Let I be the current flowing.

so,

$P=VI$

or,

$I = \frac{P}{V}$

  $= \frac{1500}{220}\\=6.81A$

Since 6.81A is greater than 5A which is the rating of the fuse, the fuse will melt and the circuit breaks

You have to change the fuse to a new fuse of rating bigger than 7A

Hope it helps you and please mark as brainliest