91影视

The initial resistance \(R_1\) is 3惟. When the conductor is stretched uniformly until its length is doubled, the cross-sectional area A will be halved. According to the resistance formula, \[R = \rho \frac{l}{A},\] where R is resistance, 蟻 is the resistivity of the conductor, l is the length, and A is the area. Since the length doubles and the area halves, the new resistance becomes, \[R_2 = \rho \frac{2l}{\frac{1}{2}A} = 4 \rho \frac{l}{A}.\] From this equation, we have \(R_2 = 4R_1\), which means \(R_2 = 4 \cdot 3\Omega = 12\Omega\).

To bend the wire in the form of an equilateral triangle, we need to divide the wire into three equal parts. Each part will have a resistance \(R_3 = \frac{1}{3}R_2 = \frac{1}{3}(12\Omega) = 4\Omega\).

As we formed an equilateral triangle with three equal resistances, we can consider two resistors in series and the third one in parallel with that combination. Let's call the resistance between the ends of any side of the triangle \(R_4\). To calculate the effective resistance between any two ends, we have the equivalent resistance in the following configuration: two 4惟 resistors in series (that's 8惟) followed by a 4惟 resistor in parallel. To find the effective resistance, we'll use the formula for parallel resistors: \[\frac{1}{R_4} = \frac{1}{R_{\text{series}}} + \frac{1}{R_{\text{parallel}}} = \frac{1}{8\Omega} + \frac{1}{4\Omega} = \frac{3}{8\Omega}\] Now, to find \(R_4\), we just need to invert the equation: \[R_4 = \frac{8}{3}\Omega.\] The effective resistance between the ends of any side of the triangle is \[\boxed{\frac{8}{3}\Omega}\] which corresponds to option (B).