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Question: A fuse wire with circular cross section and having a diameter of \(0.4mm\) blows with a current \(I\...

A fuse wire with circular cross section and having a diameter of 0.4mm0.4mm blows with a current II of 3A3A . The value of current for which another fuse wire made of the same material but having circular cross section with diameter of 0.6mm0.6mm will blow is:
A) 3amp3amp
B) 332amp3\sqrt {\dfrac{3}{2}} amp
C) 3(32)3/2amp3{(\dfrac{3}{2})^{3/2}}amp
D) 3(32)amp3(\dfrac{3}{2})amp

Explanation

Solution

A fuse is basically a piece of wire that is used for safety purposes in a circuit. This wire is made up of a material with a high resistance and a very low melting point. It protects the circuit from short circuit or damage.

Complete step by step solution:
Step I:
As it is said that fuse wire has a high resistance, therefore when a large current flows through the wire it heats up and breaks the circuit. This protects the circuit.

Step II:
Given there are two fuse wires used.
Let l1{l_1} be the length and r1{r_1} be the radius of the first fuse wire.
Let l2{l_2} be the length and r2{r_2} be the radius of the second fuse wire.
For a fuse wire, the current ‘I’ flowing varies directly with the radius ‘r’ of the wire.
It can be written as
I(r)3/2I \propto {(r)^{3/2}}

Step III:
Let I1{I_1} be the current flowing in the first wire and I2{I_2} be the current flowing in the second wire.
Given
I1=3A{I_1} = 3A
I2=?{I_2} = ?
d1=0.4mm{d_1} = 0.4mm
r1=0.42=0.2mm{r_1} = \dfrac{{0.4}}{2} = 0.2mm
d2=0.6mm{d_2} = 0.6mm
r2=0.62=0.3mm{r_2} = \dfrac{{0.6}}{2} = 0.3mm

Step IV:
The current through the second wire can be calculated by taking the ratio of both the currents.
I1I2=(r1r2)3/2\dfrac{{{I_1}}}{{{I_2}}} = {(\dfrac{{{r_1}}}{{{r_2}}})^{3/2}}
Substituting the values and solving,
\Rightarrow 3I2=(0.20.3)3/2\dfrac{3}{{{I_2}}} = {(\dfrac{{0.2}}{{0.3}})^{3/2}}
\Rightarrow I2=(32)3/2.3{I_2} = {(\dfrac{3}{2})^{3/2}}.3
\Rightarrow I2=3(32)3/2A{I_2} = 3{(\dfrac{3}{2})^{3/2}}A

Step V:
The current flowing through the wire if diameter 0.6mm0.6mm will be 3.(32)3/23.{(\dfrac{3}{2})^{3/2}}

Option C is the right answer.

Note: It is to be noted that the current flowing through the wire can vary according to length, radius and resistance of the material used in the wire. All since the energy is always conserved, therefore it changes its form from electrical energy to heat energy. When electrons move, some current reflects in the form of heat. It is also a type of energy.