Consider a c-Si solar cell whose fingers have a resistance R=0.1Ω. What would be the finger's resistance (in Ω) if the finger's width is doubled and the finger's height is one third of its initial value?
0.15
.15
0.15
To find the new resistance of the finger, we need to consider the changes in dimensions and how they affect the resistance of the finger.
The resistance of a wire is given by the formula: R = ρ * (L / A)
Where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
In this case, the finger's width is doubled and the height is reduced to one third. Let's denote the initial width as W1, initial height as H1, and initial resistance as R1. The new width and height will be W2 and H2 respectively.
When the width is doubled and the height is reduced to one third, we have:
W2 = 2 * W1
H2 = (1/3) * H1
To find the new resistance R2, we need to find the new dimensions (length and cross-sectional area) of the finger.
The length of the finger remains the same, so L2 = L1.
The cross-sectional area of the finger is given by the product of the width and height.
A = W * H
Using the new dimensions (W2 and H2), we can find the new cross-sectional area A2:
A2 = W2 * H2
Substituting the values of W2 and H2:
A2 = (2 * W1) * (1/3) * H1
Simplifying:
A2 = (2/3) * W1 * H1
Now that we have the new dimensions of the finger, we can find the new resistance R2 using:
R2 = ρ * (L2 / A2)
Substituting the values of L2 and A2:
R2 = ρ * (L1 / ((2/3) * W1 * H1))
Simplifying:
R2 = (3/2) * ρ * (L1 / (W1 * H1))
So the new resistance R2 of the finger, when its width is doubled and height is reduced to one third, is given by:
R2 = (3/2) * R1
Given the initial resistance R1 as 0.1Ω:
R2 = (3/2) * 0.1Ω
R2 = 0.15Ω
Therefore, the new resistance of the finger is 0.15Ω.