A 2.0-mm-long, 1.0--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?

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 =

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as =

Taking the integral of both sides;we have:

R = 14.863 Ω

Since V = IR

I = 1.144 A

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

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