Answer
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Hint: 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 ${l_1}$ be the length and ${r_1}$ be the radius of the first fuse wire.
Let ${l_2}$ be the length and ${r_2}$ 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
$I \propto {(r)^{3/2}}$
Step III:
Let ${I_1}$ be the current flowing in the first wire and ${I_2}$ be the current flowing in the second wire.
Given
${I_1} = 3A$
${I_2} = ?$
${d_1} = 0.4mm$
${r_1} = \dfrac{{0.4}}{2} = 0.2mm$
${d_2} = 0.6mm$
${r_2} = \dfrac{{0.6}}{2} = 0.3mm$
Step IV:
The current through the second wire can be calculated by taking the ratio of both the currents.
\[\dfrac{{{I_1}}}{{{I_2}}} = {(\dfrac{{{r_1}}}{{{r_2}}})^{3/2}}\]
Substituting the values and solving,
$\Rightarrow$ $\dfrac{3}{{{I_2}}} = {(\dfrac{{0.2}}{{0.3}})^{3/2}}$
$\Rightarrow$ ${I_2} = {(\dfrac{3}{2})^{3/2}}.3$
$\Rightarrow$ ${I_2} = 3{(\dfrac{3}{2})^{3/2}}A$
Step V:
The current flowing through the wire if diameter $0.6mm$ will be $3.{(\dfrac{3}{2})^{3/2}}$
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.
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 ${l_1}$ be the length and ${r_1}$ be the radius of the first fuse wire.
Let ${l_2}$ be the length and ${r_2}$ 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
$I \propto {(r)^{3/2}}$
Step III:
Let ${I_1}$ be the current flowing in the first wire and ${I_2}$ be the current flowing in the second wire.
Given
${I_1} = 3A$
${I_2} = ?$
${d_1} = 0.4mm$
${r_1} = \dfrac{{0.4}}{2} = 0.2mm$
${d_2} = 0.6mm$
${r_2} = \dfrac{{0.6}}{2} = 0.3mm$
Step IV:
The current through the second wire can be calculated by taking the ratio of both the currents.
\[\dfrac{{{I_1}}}{{{I_2}}} = {(\dfrac{{{r_1}}}{{{r_2}}})^{3/2}}\]
Substituting the values and solving,
$\Rightarrow$ $\dfrac{3}{{{I_2}}} = {(\dfrac{{0.2}}{{0.3}})^{3/2}}$
$\Rightarrow$ ${I_2} = {(\dfrac{3}{2})^{3/2}}.3$
$\Rightarrow$ ${I_2} = 3{(\dfrac{3}{2})^{3/2}}A$
Step V:
The current flowing through the wire if diameter $0.6mm$ will be $3.{(\dfrac{3}{2})^{3/2}}$
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.
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