Given:
The resistivity of the nichrome is
[tex]\rho\text{ = 1.5}\times\text{ 10}^{-6}\text{ }\Omega\text{ m}[/tex]
The radius of the wire is
[tex]\begin{gathered} r\text{ = 0.642 mm} \\ =6.42\text{ }\times10^{-4}\text{ m} \end{gathered}[/tex]
The potential difference is
[tex]\Delta V\text{ = 12 V}[/tex]
The length of the wire is l = 1 m
Required:
(a) Resistance per unit length
(b) Current in the wire
(c) Find the new resistance in terms of old resistance, if the wire is melted and the wire has 3 times the original length.
Explanation:
(a) Resistance per unit length can be calculated as
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ \frac{R}{l}=\frac{\rho}{A} \\ =\frac{1.5\times10^{-6}}{3.14\times(6.42\times10^{-4})^2} \\ =1.16\text{ }\frac{\Omega}{m} \end{gathered}[/tex]
(b) Current in the wire can be calculated as
[tex]\begin{gathered} I=\frac{\Delta V}{R} \\ =\frac{12}{1.16} \\ =10.35\text{ A} \end{gathered}[/tex]
(c) When the wire is melted and recast into the new wire, its volume will be the same.
[tex]\begin{gathered} V_N=V_o \\ 3l\times A_N=l\times A \\ A_N=\frac{A}{3} \end{gathered}[/tex]
The new resistance will be calculated as
[tex]\begin{gathered} R_N=\rho\frac{3l}{(\frac{A}{3})} \\ =9\times\rho\frac{l}{A} \\ =9R_o \end{gathered}[/tex]
Thus, the new resistance is three times the old resistance.
Final Answer:
(a) Resistance per unit length is 1.16 ohm/meter
(b) Current in the wire is 10.35 A
(c) The new resistance is nine times the old resistance, if the wire is melted and the wire has 3 times the original length.
