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?
To find the new resistance of the finger, we need to consider the changes in width and height and how they affect the resistance.
1. Original resistance: R = 0.1 Ω
2. Doubling the width: If the width is doubled, we can consider the new width as 2 times the initial width (W1).
3. Reducing the height: If the height is one third of its initial value, we can consider the new height as one third of the initial height (H1).
4. Resistance is proportional to the resistivity, the length, and inversely proportional to the cross-sectional area. Since the resistivity and length of the finger remain constant, the only change that affects the resistance is the change in the cross-sectional area.
5. The cross-sectional area of the finger is given by A = width × height.
6. Therefore, the new cross-sectional area (A2) would be (2W1) × (H1/3) = 2/3 × (W1 × H1).
7. The new resistance (R2) can be calculated using the formula: R2 = R1 × (A1/A2), where R1 is the original resistance and A1 is the original cross-sectional area.
8. Plugging in the values, we have: R2 = 0.1 Ω × [(W1 × H1)/(2/3 × W1 × H1)].
9. Simplifying, we get: R2 = 0.1 Ω × (3/2) = 0.15 Ω.
Therefore, the finger's resistance would be 0.15 Ω if the width is doubled and the height is one third of its initial value.
To find the finger's resistance after the changes in width and height, we need to consider the relationship between resistance, resistivity, and dimensions.
1. Begin by determining the initial dimensions of the finger. Let's assume the finger has an initial width of W and an initial height of H.
2. The resistance of the finger can be calculated using the formula: R = ρ * (L / A), where R is the resistance, ρ is the resistivity, L is the length of the finger, and A is the cross-sectional area of the finger.
3. For a finger with dimensions W and H, the initial cross-sectional area A can be written as W * H.
4. Given that the initial resistance R is 0.1 Ω, we can rewrite the formula as: 0.1 Ω = ρ * (L / (W * H)).
5. Now, we are asked to calculate the finger's resistance after the width is doubled and the height is reduced to one third of its initial value. We can denote the new width as 2W and the new height as H/3.
6. The new cross-sectional area of the finger (A') can be calculated as (2W) * (H/3) = (2/3) * W * H.
7. We can now rewrite the resistance formula for the new dimensions: R' = ρ * (L / A').
8. We want to find the new resistance R'. Plugging in the values we have, we get: R' = ρ * (L / ((2/3) * W * H)).
9. We can simplify this expression: R' = (3/2) * (ρ * (L / (W * H))).
10. From step 4, we know that the initial resistance R is 0.1 Ω. Substituting this value, we get: R' = (3/2) * 0.1 Ω.
11. Calculating this expression, we find that the finger's resistance after the changes in width and height is 0.15 Ω.
So, the finger's resistance would be 0.15 Ω after doubling the width and reducing the height to one third of its initial value.