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

Answer:

1.144 A

Explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R = $\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]$

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as = $\frac{\rho dx}{A}$

$dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}$

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

$I = \frac{V}{R}$

$I = \frac{17}{14.863}$

I = 1.144 A

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