Three wires A, B, and C of the same material have lengths and area of cross-sections as \( (2l, \frac{A}{2}) \), \( (l, A) \) and \( (\frac{l}{2}, 2A) \), respectively. If the resistances of these wires are \( R_A, R_B, R_C \) respectively, then:
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The resistance of a wire depends directly on its length and inversely on its cross-sectional area. Larger lengths and smaller cross-sections result in higher resistance.
The resistance of a wire is calculated using the formula:\[R = \rho \frac{l}{A}\]where:- \( \rho \) denotes the material's resistivity,- \( l \) represents the wire's length,- \( A \) signifies the wire's cross-sectional area.Applying this formula to three distinct wires yields:For wire A:\[R_A = \rho \frac{2l}{\frac{A}{2}} = \rho \frac{4l}{A}\]For wire B:\[R_B = \rho \frac{l}{A}\]For wire C:\[R_C = \rho \frac{\frac{l}{2}}{2A} = \rho \frac{l}{4A}\]Comparing these resistances:\[R_A = \frac{4 \rho l}{A}, \quad R_B = \frac{\rho l}{A}, \quad R_C = \frac{\rho l}{4A}\]The established order of resistance is \( R_A>R_B>R_C \). Therefore, the correct conclusion is:\[R_A>R_B>R_C\]