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(B) Let the resistances of the two wires be $R_1$ and $R_2$.When connected in series: $R_1 + R_2 = 14$  --- $(1)$When connected in parallel: $\frac{R_

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$8$

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(B) Let the resistances of the two wires be $R_1$ and $R_2$.When connected in series: $R_1 + R_2 = 14$  --- $(1)$When connected in parallel: $\frac{R_1 R_2}{R_1 + R_2} = 3.43$  --- $(2)$Substituting $(1)$ into $(2)$: $\frac{R_1 R_2}{14} = 3.43$$R_1 R_2 = 14 	imes 3.43 = 48.02 \approx 48$We have $R_1 + R_2 = 14$ and $R_1 R_2 = 48$. These are roots of the quadratic equation $x^2 - 14x + 48 = 0$.Solving the quadratic equation: $(x - 8)(x - 6) = 0$.Thus,$R_1 = 8\, \Omega$ and $R_2 = 6\, \Omega$.The resistance of the wire with the higher value is $8\, \Omega$.

When two conducting wires are connected in series,the equivalent resistance is $14\, \Omega$,and when they are connected in parallel,the equivalent resistance is $3.43\, \Omega$. The resistance of the wire with the higher value is ................. $\Omega$.

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