
A fuse wire of radius $0.1mm$ melts when a current in $A$ of $10A$ is passed through it. Find the current at which a fuse wire of $0.12mm$ will melt.
Answer
586.8k+ views
Hint: As we all know, the resistance of a wire is given as the product of resistivity of the wire and the length of the wire which is divided by the area of the cross section of wire. First of all find out the resistance of the wire. The heat produced in the wire is the reason for the melting of the wire. Therefore the heat produced in a fuse wire is given by the product of the square of the current passing, resistance of the wire and the area of cross section. From this we will get the relationship between the current flowing and the radius of the wire. Using this find out the current through the fuse wire by comparison.
Complete step by step answer:
The heat produced in a wire is the reason for the melting of the fuse wire. Therefore the heat produced is given as,
$H={{I}^{2}}Rt$
Where $I$ be the current flowing through the wire, $R$ be the resistance of the wire and $t$ be the time taken for the melting.
Also we know that the resistance of a wire is given by the formula,
$R=\dfrac{\rho l}{A}$
Where $\rho $ be the resistivity, $l$ be the length of the wire and $A$ be the area of the cross section.
We can write that,
\[R=\dfrac{\rho l}{\pi {{r}^{2}}}\]
Where \[\pi {{r}^{2}}\] be the area of the wire, in which \[r\] be the radius of the wire.
Substituting this in the equation of heat,
$H={{I}^{2}}\dfrac{\rho l}{\pi {{r}^{2}}}t$
Rearranging the equation will give,
$H\times \pi {{r}^{2}}={{I}^{2}}\rho lt$
From this we can see that,
\[{{I}_{2}}=12A\]
Using this relationship we can write that,
$\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}$
We know that the current through the wire and the radius of the wire in the first condition are
\[\begin{align}
& {{I}_{1}}=10A \\
& {{r}_{1}}=0.1mm \\
\end{align}\]
The radius of wire in the second condition is,
\[{{r}_{2}}=0.12mm\]
And the current through the wire in the second condition is,
\[\dfrac{10}{{{I}_{2}}}=\dfrac{0.1}{0.12}\]
Rearranging the equation will give,
\[{{I}_{2}}=12A\]
Note: Heat produced in a wire carrying current can be found in different ways. One is by taking the product of the voltage, current passing and the time taken for the heating. And another way to find the heat produced is by dividing the product of square of the voltage and time taken by the resistance of the wire.
Complete step by step answer:
The heat produced in a wire is the reason for the melting of the fuse wire. Therefore the heat produced is given as,
$H={{I}^{2}}Rt$
Where $I$ be the current flowing through the wire, $R$ be the resistance of the wire and $t$ be the time taken for the melting.
Also we know that the resistance of a wire is given by the formula,
$R=\dfrac{\rho l}{A}$
Where $\rho $ be the resistivity, $l$ be the length of the wire and $A$ be the area of the cross section.
We can write that,
\[R=\dfrac{\rho l}{\pi {{r}^{2}}}\]
Where \[\pi {{r}^{2}}\] be the area of the wire, in which \[r\] be the radius of the wire.
Substituting this in the equation of heat,
$H={{I}^{2}}\dfrac{\rho l}{\pi {{r}^{2}}}t$
Rearranging the equation will give,
$H\times \pi {{r}^{2}}={{I}^{2}}\rho lt$
From this we can see that,
\[{{I}_{2}}=12A\]
Using this relationship we can write that,
$\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}$
We know that the current through the wire and the radius of the wire in the first condition are
\[\begin{align}
& {{I}_{1}}=10A \\
& {{r}_{1}}=0.1mm \\
\end{align}\]
The radius of wire in the second condition is,
\[{{r}_{2}}=0.12mm\]
And the current through the wire in the second condition is,
\[\dfrac{10}{{{I}_{2}}}=\dfrac{0.1}{0.12}\]
Rearranging the equation will give,
\[{{I}_{2}}=12A\]
Note: Heat produced in a wire carrying current can be found in different ways. One is by taking the product of the voltage, current passing and the time taken for the heating. And another way to find the heat produced is by dividing the product of square of the voltage and time taken by the resistance of the wire.
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