Question

A 2.0-mm-long, 1.0-mmmm-diameter wire has a variable resistivity given by rho(x)=(2.5×10−6)[1+(x1.0m)2]Ωmrho(x)=(2.5×10−6)[1+(x1.0m)2]Ωm where xx is measured from one end of the wire.What is the current if this wire is connected to the terminals of a 17.0V battery?

Answer 1

1.144 A

Step-by-step explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R =
`\rho (x) =(2.5*10^(-6))[1+((x)/(1.0 \ m))^2]`

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as =
`(\rho dx)/(A)`

`dR = ((2.5*10^(-6))[1+((x)/(1.0))^2])/((\pi)/(4)(10^(-3))^2)`

Taking the integral of both sides;we have:

`\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx`

`R = 3.185 [x + \frac {x^3}{3}}]^2__0`

`R = 3.185 [2 + \frac {2^3}{3}}]`

R = 14.863 Ω

Since V = IR

`I = (V)/(R)`

`I = (17)/(14.863)`

I = 1.144 A

∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A