The resistance \(R\) of a wire is given by the formula:
\(R = \frac{\rho L}{A}\)
where:
For the first wire, we have \(R_1 = 2 \Omega\), \(L_1 = L\), and \(A_1 = A\).
For the second wire, we have \(L_2 = 2L\) and \(A_2 = A/2\). We want to find \(R_2\).
Using the formula for resistance:
\(R_1 = \frac{\rho L_1}{A_1} = \frac{\rho L}{A} = 2 \Omega\)
\(R_2 = \frac{\rho L_2}{A_2} = \frac{\rho (2L)}{A/2} = \frac{4\rho L}{A}\)
Since \(R_1 = \frac{\rho L}{A} = 2 \Omega\), we can substitute this into the expression for \(R_2\):
\(R_2 = 4 \times \frac{\rho L}{A} = 4 \times R_1 = 4 \times 2 \Omega = 8 \Omega\)
Therefore, the resistance of the second wire is 8 Ω.
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