Can the following equation be used to solve the question?
R = ρ L / A ; find L.. or is there another way?
Question: A rectangular piece of gold is 2.2 cm long and 2.4 cm wide. How deep is the piece when it displays a resistance of R = 1.06 μω for a current that flows along the depth-direction?
That is the way.
Hi, thanks for answering.
Apparently our professor wants us to do this question without using density. Is there another possible way to solve this question that you can direct me to?
ρ is not density in this case. Refer to Table 14.3 on page 455 in the textbook, it states the resistivity value of gold.
sick you knew the textbook we're using
Yes, the equation R = ρL/A can be used to solve the question. In this equation, R represents resistance, ρ represents resistivity, L represents the length, and A represents the cross-sectional area. By rearranging the equation, we can solve for L:
L = R * A / ρ
To apply this equation to the given question, we need to know the resistivity of gold. The resistivity of gold is typically 2.44 x 10^-8 Ωm.
The cross-sectional area (A) of the gold piece can be calculated by multiplying its length (2.2 cm) and width (2.4 cm):
A = length * width
A = 2.2 cm * 2.4 cm
Once we have the resistivity and cross-sectional area, we can substitute these values into the equation to find L:
L = R * A / ρ
L = 1.06 µΩ * (2.2 cm * 2.4 cm) / (2.44 x 10^-8 Ωm)
By performing this calculation, we can determine the depth (L) of the gold piece.