
Two fuse wires of the same length are rated $5A$ and $20A$. Which of the fuse wires are thicker and why?
A) $5A$
B) $20A$
C) Both have the same thickness
D) Cannot be the same
Answer
480.6k+ views
Hint: To solve this question, we must recollect our knowledge about the ohm’s law. Also, the question refers to the dimensions of the wire and the amount of current it can carry. So, we have to find a physical quantity that can relate to the current flowing and the dimensions of the wire.
Formula used:
$R = \rho \dfrac{l}{A}$
Where,
$R$ is the resistance.
$\rho $ is the resistivity.
$l$ is the length of the wire.
$A$ is the area of cross-section wire.
Complete answer:
The resistance is directly proportional to its length, i.e. as the length of the conductor increases, resistance will also increase. Mathematically,
$R \propto l......(1)$
Where,
$R$ is the resistance.
$l$ is the length of the wire.
Also, the resistance is inversely proportional to the area of the cross-section of the conductor. This implies that an increase in the area of the cross-section will decrease the resistance.
$R \propto \dfrac{1}{A}......(2)$
Where,
$R$ is the resistance.
$A$ is the area of cross-section wire.
From equations (1) and (2) we get,
$R \propto \dfrac{l}{A}$
$ \Rightarrow R = \rho \dfrac{l}{A}......(3)$
Where,
$R$ is the resistance.
$\rho $ is the resistivity.
$l$ is the length of the wire.
$A$ is the area of cross-section wire.
Now, we have to analyze ohm’s law which states that the voltage across two points is directly proportional to the current flowing through a conductor between those points.
$V \propto I$
$ \Rightarrow V = IR$
$ \Rightarrow R = \dfrac{V}{I}......(4)$
Where,
$R$ is the resistance.
$V$ is the voltage.
$I$ is the current.
From the question, we can infer that the length of the two wires is the same but the thickness varies. The thickness of the wire is determined by the area of cross-section of the wire. The more the area of cross-section, the thicker the wire will be.
We here assume that all the other quantities except the area of cross-section and current flowing are constant. From equating equations (3) and (4) we get,
$\rho \dfrac{l}{A} = \dfrac{V}{I}$
$ \Rightarrow I \propto A$
Therefore, increasing the thickness of the wire will decrease its resistance and thus an increase in current can be observed.
So, the fuse wire rated $20A$ will be thicker and hence option (B) is correct.
Note:
An electric fuse is a device that is installed in a circuit to stop the flow of excessive current in the circuit. It is based on the principle of heating effect of electric current. A material having high resistivity and a low melting point is ideal to make a fuse.
So, whenever a high current flows through the fuse wire, it melts down due to overheating of wire and thus breaks the circuit. The main function of a fuse is to prevent any damage to the device by restricting excess current flow.
Formula used:
$R = \rho \dfrac{l}{A}$
Where,
$R$ is the resistance.
$\rho $ is the resistivity.
$l$ is the length of the wire.
$A$ is the area of cross-section wire.
Complete answer:
The resistance is directly proportional to its length, i.e. as the length of the conductor increases, resistance will also increase. Mathematically,
$R \propto l......(1)$
Where,
$R$ is the resistance.
$l$ is the length of the wire.
Also, the resistance is inversely proportional to the area of the cross-section of the conductor. This implies that an increase in the area of the cross-section will decrease the resistance.
$R \propto \dfrac{1}{A}......(2)$
Where,
$R$ is the resistance.
$A$ is the area of cross-section wire.
From equations (1) and (2) we get,
$R \propto \dfrac{l}{A}$
$ \Rightarrow R = \rho \dfrac{l}{A}......(3)$
Where,
$R$ is the resistance.
$\rho $ is the resistivity.
$l$ is the length of the wire.
$A$ is the area of cross-section wire.
Now, we have to analyze ohm’s law which states that the voltage across two points is directly proportional to the current flowing through a conductor between those points.
$V \propto I$
$ \Rightarrow V = IR$
$ \Rightarrow R = \dfrac{V}{I}......(4)$
Where,
$R$ is the resistance.
$V$ is the voltage.
$I$ is the current.
From the question, we can infer that the length of the two wires is the same but the thickness varies. The thickness of the wire is determined by the area of cross-section of the wire. The more the area of cross-section, the thicker the wire will be.
We here assume that all the other quantities except the area of cross-section and current flowing are constant. From equating equations (3) and (4) we get,
$\rho \dfrac{l}{A} = \dfrac{V}{I}$
$ \Rightarrow I \propto A$
Therefore, increasing the thickness of the wire will decrease its resistance and thus an increase in current can be observed.
So, the fuse wire rated $20A$ will be thicker and hence option (B) is correct.
Note:
An electric fuse is a device that is installed in a circuit to stop the flow of excessive current in the circuit. It is based on the principle of heating effect of electric current. A material having high resistivity and a low melting point is ideal to make a fuse.
So, whenever a high current flows through the fuse wire, it melts down due to overheating of wire and thus breaks the circuit. The main function of a fuse is to prevent any damage to the device by restricting excess current flow.
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