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 = [tex]\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2][/tex]
Voltage of the battery = 17.0 v
Now; the resistivity of the variable (dR) can be expressed as = [tex]\frac{\rho dx}{A}[/tex]
[tex]dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}[/tex]
Taking the integral of both sides;we have:
[tex]\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx[/tex]
[tex]R = 3.185 [x + \frac {x^3}{3}}]^2__0[/tex]
[tex]R = 3.185 [2 + \frac {2^3}{3}}][/tex]
R = 14.863 Ω
Since V = IR
[tex]I = \frac{V}{R}[/tex]
[tex]I = \frac{17}{14.863}[/tex]
I = 1.144 A
∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A
