Question

Which fuse wire is thick: $5A$ or $15A$?

Answer 1

Hint: To solve this question, we must recollect our knowledge about the ohm’s law. Also, it is known to us that resistance is an extrinsic property of the wire i.e. it depends on the wire’s dimensions. So, we have to relate the thickness of the wire and the current using the property of resistance and ohm’s law to reach the answer.

Formula used:
1. $R = \rho \dfrac{l}{A}$
Where,
$R$ is the resistance.
$\rho $ is the resistivity.
$l$ is the length of the wire.
$A$ is the area of cross-section.

2. $V=IR$
Where,
$R$ is the resistance.
$V$ is the voltage.
$I$ is the current.

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.
$R \propto \dfrac{1}{A}......(2)$
From equations (1) and (2) we get,
$R \propto \dfrac{l}{A}$
$ \Rightarrow R = \rho \dfrac{l}{A}......(3)$

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)$
Let the length of the two wires be 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 at $15A$ will be thick.

Note:
An electric fuse is a device that is installed in a circuit to stop the flow of excessive current in the circuit which is based on the heating effect principle. 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 the 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.