Consider a c-Si solar cell whose fingers have a resistance \(R=0.1\Omega\). What would be the finger's resistance (in \(\Omega\)) if the finger's width is doubled and the finger's height is one third of its initial value?
To calculate the new resistance of the finger, we need to consider the changes in its dimensions.
Let's assume the initial width of the finger is \(W\) and the initial height is \(H\). According to the problem, the finger's width is doubled (2W) and the finger's height is one third of its initial value (\(\frac{1}{3}H\)).
The resistance of a conductor can be calculated using the formula:
\[R = \rho \cdot \left(\frac{L}{A}\right)\]
Where:
- \(R\) is the resistance of the conductor.
- \(\rho\) is the resistivity of the material (a constant value for a given material).
- \(L\) is the length of the conductor.
- \(A\) is the cross-sectional area of the conductor.
In this case, the finger's length (\(L\)) and the material's resistivity (\(\rho\)) remain the same.
First, let's find the cross-sectional area of the initial finger:
\[A_{initial} = W \cdot H\]
Next, let's find the cross-sectional area of the new finger:
\[A_{new} = 2W \cdot \frac{1}{3}H\]
Now, we can calculate the new resistance using the updated cross-sectional area:
\[R_{new} = \rho \cdot \left(\frac{L}{A_{new}}\right)\]
Substituting the values:
\[R_{new} = \rho \cdot \left(\frac{L}{2W \cdot \frac{1}{3}H}\right)\]
Simplifying the expression:
\[R_{new} = \rho \cdot \left(\frac{3L}{2WH}\right)\]
Therefore, the finger's resistance with the new dimensions is given by:
\[R_{new} = \frac{3 \rho L}{2WH}\]
Note: Remember that in this explanation, we assumed the resistivity (\(\rho\)) and the length (\(L\)) remained the same.