A wire is drawn through a die, stretching it to four times its original length. By what factor does its resistance increase?

In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega\({\rm{(\Omega )}}\)is used to represent resistance in ohms.

The electrical resistance of the value\(R\)of a resistive circuit element depends on its geometric structure (its length symbolized as\(L\)and cross-sectional area symbolized as\(A\)) and on the resistivity\(\rho \)of the material of which it is made:

\(R{\rm{ }} = {\rm{ }}\rho \frac{L}{A}\)

• Assuming the original length of the string be\({L_i}\)and the final length of the string be\({L_f}\).
• The final length is four times the original length, so: \({L_f} = 4{L_i}\).

The initial resistance\({R_i}\)of the wire can be calculated from the above equation as:

\({R_i}{\rm{ }} = {\rm{ }}\rho \frac{{{L_i}}}{{{A_i}}}\)……………….(I)

Similarly, the final resistance\({R_f}\)of the wire is:

\({R_f}{\rm{ }} = {\rm{ }}\rho \frac{{{L_f}}}{{{A_f}}}\)

The value of\({L_f} = 4{L_i}\)and\({A_i}\):

\({R_f}{\rm{ }} = {\rm{ }}\rho \frac{{{L_i}}}{{{A_f}}}\)………………(II)

As the volume of the wire remains constant:

\(\begin{aligned}{c}{V_i}{\rm{ }} &= {\rm{ }}{V_f}\\{A_i}{L_i}{\rm{ }} &= {\rm{ }}{A_f}{L_f}\end{aligned}\)

Substituting the value of\({L_f}\)in the above equation, we get:

\(\begin{aligned}{c}{A_i}{L_i}{\rm{ }} &= {\rm{ }}4{A_f}{L_f}\\{A_i}{\rm{ }} &= {\rm{ }}4{A_f}\end{aligned}\)

Substituting the value\({A_i}\)into the first equation as:

\({R_i}{\rm{ }} = {\rm{ }}\rho \frac{{{L_i}}}{{4{A_f}}}\)………………….(III)

Dividing the second equation by the third equation, we obtain:

\(\begin{aligned}{}\frac{{{R_f}}}{{{R_i}}}{\rm{ }} &= {\rm{ }}\frac{{\rho \frac{{4{L_i}}}{{{A_f}}}.}}{{\rho \frac{{{L_i}}}{{4{A_f}}}}}\\ &= {\rm{ }}16\end{aligned}\)

Therefore, the resistance of the wire increases by a factor of \(16\).