A wire of length L and cross-sectional area A has resistance R. What will be the resistance R_stretched of the wire if it is stretched to twice its original length? Assume that the density and resistivity of the material do not change when the wire is stretched.
Express your answer in terms of the wire's original resistance R.
The product L*A remains constant after stretching. If L doubles, A decreases by 1/2
Resistance is proportional to L/A.
R2/R1 = L2/L1 *A1/A2 = 4
R1 is the original resistance and R2 is the final resistance.
A)becomes double B)remains same C)becomes half C)becomes square
Remain same
Consider a c-Si solar cell whose fingers have a resistance R=0.1 Ohm. What would be the finger's resistance (in Ohm) if the finger's width is doubled and the finger's height is one third of its initial value?
To find the resistance R_stretched of the wire when it is stretched to twice its original length, we can use the formula for resistance:
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.
Since the resistivity and cross-sectional area of the wire do not change when it is stretched, we only need to focus on the change in length.
Let's denote the original length as L_0 and the stretched length as L_stretched. According to the problem, L_stretched = 2 * L_0.
Now we can find the resistance R_stretched when the wire is stretched:
R_stretched = (ρ * L_stretched) / A
= (ρ * (2 * L_0)) / A
= (2 * ρ * L_0) / A
Therefore, the resistance R_stretched of the wire when it is stretched to twice its original length is equal to 2 * R, where R is the original resistance.
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