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A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m length of the wire?
B) What would the potential difference in part A be if the wire were silver instead of copper, but all else was the same?

Respuesta :

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A  where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω

and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω

Finalle we calculate the potential difference (V) for both wires:

Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V

V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

stigawithfun

Answer:

(a) 1.788 x 10⁻⁴V

(b) 1.524 x 10⁻⁴V

Explanation:

The resistance (R) of a wire is related to the resistivlty (ρ) of the wire material, the length (L) of the wire and the cross-sectional area (A) of the wire as follows;

R = ρL / A.          ---------------------(i)

And;

The current (I) flowing through a wire is directly proportional to the potential difference (V) across the wire according to Ohm's law as follows;

V = I x R          ---------------------(ii)

(a) Now, according to the first part of the question, the following are given;

L = 1.80m

diameter d = 1.628mm = 0.001628m

current (I) = 12.0 mA = 12.0 x 10⁻³A

From the diameter, we can get the area (A) of the wire as follows;

A = π d² / 4           [Take π = 3.142 ]

A = 3.142 x 0.001628² / 4

A = 2.08 x 10⁻⁶ m²

Substitute the values of L and A into equation (i) as follows;

R = (ρ x 1.80) / (2.08 x 10⁻⁶)

where ρ = resistivity of copper wire = 1.72 x 10⁻⁸ Ωm

R = 1.72 x 10⁻⁸ x 1.80 / (2.08 x 10⁻⁶)

R = 1.49 x 10⁻² Ω

Substitute the value of R into equation (ii) as follows;

V =  I x R

V = 12.0 x 10⁻³ x 1.49 x 10⁻²

V = 17.88 x 10⁻⁵

V = 1.788 x 10⁻⁴V

Therefore, the potential difference across the length of copper wire is 1.788 x 10⁻⁴V

(b) If the wire were a silver, we will go through the same process as above except that the resistivity (ρ) will be that of silver which is 1.47 x 10⁻⁸ Ωm.

Substitute the values of L and A into equation (i) as follows;

R = (ρ x 1.80) / (2.08 x 10⁻⁶)

where ρ = resistivity of silver wire = 1.47 x 10⁻⁸ Ωm

R = 1.47 x 10⁻⁸ x 1.80 / (2.08 x 10⁻⁶)

R = 1.27 x 10⁻² Ω

Substitute the value of R into equation (ii) as follows;

V =  I x R

V = 12.0 x 10⁻³ x 1.27 x 10⁻²

V = 15.24 x 10⁻⁵

V = 1.524 x 10⁻⁴V

Therefore, the potential difference across the length of silver wire is 1.524 x 10⁻⁴V

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