Question

An experiment is conducted to measure the electrical resistivity of Nichrome in the form of wires with different lengths and cross-sectional areas. For one set of measurements, a student uses 30 -gauge wire, which has a cross-sectional area of 7.30×10⁻⁸ m². The student. measures the potential difference across the wire and the current in the wire with a voltmeter and an ammeter, respectively.(b) What is the average value of the resistivity?

Answer 1

The average value of the resistivity is 4.7 *10^-7 Ωm.

What is resistivity?

Resistivity, can be explained as one that is quantitatively equal to the resistance R of a  a substance which can be a wire,  along with cross-sectional area A,  and division of  length l.

From ohms law,

V=IR

R=V/I

We can get the corresponding values for the resistance as :

$R_{1} =\frac{V_{1} }{I_{1} }$

=5.22/0.540 =2.66OHM

$R_{2} =\frac{V_{2} }{I_{2} }$

5.82/1.28

=5.66

$R_{1} =\frac{V_{1} }{I_{1} }$

5.24/1.543

=3.84Ohm

We can calculate the resistivity as :

$e_{1} =\frac{AR_{1} }{l_{1} }$

$e_{2} =\frac{AR_{2} }{l_{2} }$

$e_{3} =\frac{AR_{3} }{l_{3} }$

Then the equations can be used to calculate the the three resistivity respectively,

as

First resistivity.,

(7.3* 10^-8)*(2.66)/ (0.540)= 7.05*10^-7Ωm

Second resistivity =

(7.3* 10^-8)*(5.66)/ (1.028)= 4*10^-7 Ωm

Third resistivity:

(7.3* 10^-8)*(3.84)/ (1.543)= 1.8 *10^-7 Ωm

Then we can find the average resistivity as :

= $\frac{(7.05*10^-7)+( 4*10^-7)+( 1.8 *10^-7 )}{3}$

=4.7 *10^-7 Ωm

Therefore,  the average value of the resistivity is 4.7 *10^-7 Ωm.

Learn more about Resistance at:

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The table that  complete question is:

L(M) ;    ΔV(V)     I(A)

0.54        5.22      0.72

1.028        5.82       0.414

1.543         5.94       0.281