Consider a c-Si solar cell whose fingers have a resistance R=10ohms. Which would be the finger´s resistance (in ohms) if the finger´s width is doubled and the finger´s height is one third of its initial value
double width, height 1/3?
cross sectional area is not 2/3 of the original, so resistance is now 3/2 * 10 ohms or 15 ohms
To determine the finger's resistance in the given scenario, we need to first understand the relationship between the finger's resistance and its dimensions.
In a rectangular-shaped finger, the resistance (R) is given by the resistivity (ρ) multiplied by the length (L) divided by the cross-sectional area (A).
Mathematically, R = ρ * (L / A)
Since we are given that the finger's resistance (R) is 10 ohms, we can rearrange the equation to solve for the resistivity:
ρ = R * (A / L)
Now, let's consider the changes in dimensions. We are told that the finger's width (W) is doubled and its height (H) is one third of its initial value.
Let's assume the initial width as W₀ and the initial height as H₀. Therefore, the new width (W') will be 2 * W₀, and the new height (H') will be (1/3) * H₀.
To calculate the new resistivity (ρ'), we need to find the new cross-sectional area (A') and the new length (L').
The new cross-sectional area (A') would be the product of the new width and new height:
A' = W' * H' = (2 * W₀) * ((1/3) * H₀) = (2/3) * (W₀ * H₀) = (2/3) * A₀
The new length (L') would remain the same as the initial length (L₀) since we are only altering the dimensions of the finger, not its length.
Now, we can substitute the values into ρ = R * (A / L) to find ρ':
ρ' = R * (A' / L') = 10 ohms * [(2/3) * A₀ / L₀]
Next, we substitute the formulas for A₀ and L₀ in terms of W₀ and H₀:
ρ' = 10 ohms * [(2/3) * (W₀ * H₀) / L₀]
Simplifying further:
ρ' = (20/3) * (W₀ * H₀ / L₀)
Therefore, the finger's resistance in the new scenario would be (20/3) * (W₀ * H₀ / L₀) ohms.