A wire with a circular cross section has radius 0.16 mm and is lying along the x axis from x=0 to x=0.20 m. The wire is made of an alloy that varies with its length in such a way that the resistivity is given by rho(x)=3x5 Ω⋅m, for x in meters. What is the resistance of this wire?

Respuesta :

Answer:

Resistance = 68.23 Ω

Explanation:

Let's start off by remembering the fact that when resistors are connected in series, their resistances is added up to find the total resistance.

The equation for resistance using resistivity is given below:

Resistance = Resistivity * Length / Area

where Length = x

Resistivity = 3x^5

and Area = [tex]\pi *0.00016^2[/tex] = 8.04 * 10^(-8)   meter squared

Substituting in the value of resistivity, length and area we get:

Resistance = [tex]\frac{ (3x^5) * (x) }{(8.04*10^-^8)}[/tex]

Resistance = [tex]\frac{ (3x^6) }{(8.04*10^-^8)}[/tex]

Since resistance in series is added, we can simply integrate this formula over the length (x = 0 to x = 0.2) to get the total resistance.

Resistance = [tex]\int\ {\frac{ (3x^6) }{(8.04*10^-^8)}} \, dx[/tex]

Resistance = [tex](5.4857*10^-^6) /(8.04*10^-^8)[/tex]

Resistance = 68.23 Ω

Answer: 398 ohm

Explanation: Please see the attachments below

Ver imagen Abdulazeez10
Ver imagen Abdulazeez10