Question

Please help me solve The resistivity of nichrome is 1.5 x 10^(-6) ohm meter.

Answer 1

Given:

The resistivity of the nichrome is

`\rho\text{ = 1.5}*\text{ 10}^(-6)\text{ }\Omega\text{ m}`

The radius of the wire is

`\begin{gathered} r\text{ = 0.642 mm} \\ =6.42\text{ }*10^(-4)\text{ m} \end{gathered}`

The potential difference is

`\Delta V\text{ = 12 V}`

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.

Step-by-step explanation:

(a) Resistance per unit length can be calculated as

`\begin{gathered} R=(\rho l)/(A) \\ (R)/(l)=(\rho)/(A) \\ =(1.5*10^(-6))/(3.14*(6.42*10^(-4))^2) \\ =1.16\text{ }(\Omega)/(m) \end{gathered}`

(b) Current in the wire can be calculated as

`\begin{gathered} I=(\Delta V)/(R) \\ =(12)/(1.16) \\ =10.35\text{ A} \end{gathered}`

(c) When the wire is melted and recast into the new wire, its volume will be the same.

`\begin{gathered} V_N=V_o \\ 3l* A_N=l* A \\ A_N=(A)/(3) \end{gathered}`

The new resistance will be calculated as

`\begin{gathered} R_N=\rho(3l)/(((A)/(3))) \\ =9*\rho(l)/(A) \\ =9R_o \end{gathered}`

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.