Question

An electrician puts a fuse of rating $5A$, in that part of a domestic electrical circuit in which an electrical heater of rating $1.5kW$, $220V$ is operating. What is likely to happen in this case and why? What change, (if any) should be made?

Answer 1

Hint: In an electrical circuit for safety purposes, a fuse is used in the circuit whose electric current rating is more than the electric current flowing through the circuit. Electric current through a circuit can be calculated from the value of voltage and power consumed using \[P = VI\]where P is power and V is voltage and I is electric current.
Formula used
\[P = VI\](Where P is power and V is voltage and I is electric current)

Complete step by step solution:
We know that in a circuit the fuse will melt and the circuit will break if the electric current passing through the fuse is more than the rating of the fuse.
In the above question, the electric current following in the circuit can be calculated using the formula
 P is power and V is voltage and I is electric current
\[P = VI\]
Which can also be returned as,
 $I = \dfrac{P}{V}$
It’s given in the question that $P = 1.5kW = 1500W$and $V = 220V$ thus,

$I = \dfrac{{1500}}{{220}}A$
$I \simeq 6.82A$
Therefore,
As the current flowing in the circuit is \[6.82A\] which is more than the rating of the fuse which is 5A, therefore, the fuse will melt and the electric heater will not work.
Hence, to operate the given electric heater, a fuse of a rating greater than \[6.82A\] is to be used in the circuit.

Note:
The fuse should not be very minutely greater than \[6.82A\]as there might be fluctuation in the circuit which may lead to rise and fall (oscillation) of the electric current in the circuit and if the rating of fuse is very near to \[6.82A\]then the fuse will melt and the electric heater will not work.