Question

A wire is 0.79 m long and 0.77 mm² in cross-sectional area. It carries a current of 4.1 A when a 2.0 V potential difference is applied between its ends. Calculate the conductivity ? of the material of which this wire is made.

Answer 1

apply R = rho L/A

and use ohms law V = iR

so R = V/i = 2/4.1 = 0.487 ohms

so rho = RA/L

rho = 0.487 *0.77 *10^-6/0.79

rho = 0.475 *10^-6

conductivity ? of the material is = 1/rho

conductivity ? of the material = 1/0.476 = 2.1*10^6 mhos/m

Answer 2

Resistance is given by ,

R = rho x l / A

R = V/I

2/4.1 = rho x 0.79 /0.77 x 10^-6

=> rho = 4.75 x 10^-7

Inverse of rho is conductivity ,

Conductivity = 2.10 x 10^6 (ANSWER)

Answer 3

resistance= 2/4.4=0.487ohm

conductivity=length/area*Resistance= 0.79/0.77*10^-6*0.487=2.106*10^6simon

Answer 4

Resistance

R=V/I =2/4.1

R=0.4878 ohms

since

R=pL/A

=>p=R*A/L =0.4878*(0.77*10^-6)/0.79

p=4.75*10^-7 ohm/m

conductivity is inverse of resistivity

sigma =1/p =2.1*10^6 m/ohm

Answer 5

Resistance = V / I = 2 / 4.1 ohms

Resistivity =