Answer
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Hint:This question is based on the concept of rating of fuse wire. Here rating is done on the basis of current flowing through the fuse wire. If supply current is greater than the rated current of the fuse wire, the fuse will blow off.
Formula used:
Power, ${\text{P = }}\;{{V \times I}}$
Where, ${\text{V = }}$ Voltage
${\text{I = }}$ Current
Complete step by step solution:
It is given that the rating of the fuse wire is ${\text{5A}}$ that means that. If a current of more than ${\text{5A}}$ passes through the fuse wire, the fuse wire will melt & hence the fuse will be blown off. This melting of the fuse wire happens due to the excessive heating caused by the current which flows in excess to the rated current.
We know that the heat produced is directly proportional to the square of current passing through the wire.
In this question the supply current is to be calculated using the power data. Which is given in the question.
After this the supply current value is compared with the value of rated current.
If supply current is more than the rated current, the fuse will blow off keeping all the devices safe which are connected in the circuit.
Power, ${\text{P = }}\;{{V \times I}}$
Where,
${\text{V}}\;{\text{ = }}\;{\text{Voltage}}$
${\text{I}}\;{\text{ = }}\;{\text{Current}}$
$ \Rightarrow {\text{I = }}\dfrac{{\text{P}}}{{\text{V}}}$
Given,
${\text{P = 1}}{\text{.5KW}}\;{\text{ = 1500W}}$
${\text{V = 220V}}$
${\text{I = }}\dfrac{{{\text{1500}}}}{{{\text{220}}}}{\text{ = 6}}{\text{.81A}}$
So supply current is ${\text{6}}{\text{.81A}}$ which is more than the rated current which is ${\text{5A}}$, so the fuse will blow off.
To prevent the fuse from blowing off either of the two things can be done:-
(a) Take a higher rating wire for the fuse probably whose current rating is greater than ${\text{7A}}$.
(b) Don’t use that electrical heater.
Additional information:
It is important to compare the supply current with the rated current and it should be remembered that:- if the supply current is more than the rated current then only the fuse will blow off.
Note:The rated current of the fuse wire is the maximum current the fuse wire can have without melting itself and hence blowing of the fuse.
Formula used:
Power, ${\text{P = }}\;{{V \times I}}$
Where, ${\text{V = }}$ Voltage
${\text{I = }}$ Current
Complete step by step solution:
It is given that the rating of the fuse wire is ${\text{5A}}$ that means that. If a current of more than ${\text{5A}}$ passes through the fuse wire, the fuse wire will melt & hence the fuse will be blown off. This melting of the fuse wire happens due to the excessive heating caused by the current which flows in excess to the rated current.
We know that the heat produced is directly proportional to the square of current passing through the wire.
In this question the supply current is to be calculated using the power data. Which is given in the question.
After this the supply current value is compared with the value of rated current.
If supply current is more than the rated current, the fuse will blow off keeping all the devices safe which are connected in the circuit.
Power, ${\text{P = }}\;{{V \times I}}$
Where,
${\text{V}}\;{\text{ = }}\;{\text{Voltage}}$
${\text{I}}\;{\text{ = }}\;{\text{Current}}$
$ \Rightarrow {\text{I = }}\dfrac{{\text{P}}}{{\text{V}}}$
Given,
${\text{P = 1}}{\text{.5KW}}\;{\text{ = 1500W}}$
${\text{V = 220V}}$
${\text{I = }}\dfrac{{{\text{1500}}}}{{{\text{220}}}}{\text{ = 6}}{\text{.81A}}$
So supply current is ${\text{6}}{\text{.81A}}$ which is more than the rated current which is ${\text{5A}}$, so the fuse will blow off.
To prevent the fuse from blowing off either of the two things can be done:-
(a) Take a higher rating wire for the fuse probably whose current rating is greater than ${\text{7A}}$.
(b) Don’t use that electrical heater.
Additional information:
It is important to compare the supply current with the rated current and it should be remembered that:- if the supply current is more than the rated current then only the fuse will blow off.
Note:The rated current of the fuse wire is the maximum current the fuse wire can have without melting itself and hence blowing of the fuse.
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