A wire has a resistance of 24 Omega and is b | vedclass

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(B) The total length of the wire corresponds to a resistance of $24 \ \Omega$. The wire is bent into an equilateral triangle with a side length of $5

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$10$

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(B) The total length of the wire corresponds to a resistance of $24 \ \Omega$. The wire is bent into an equilateral triangle with a side length of $5 \ 	ext{cm}$ and a straight segment of $5 \ 	ext{cm}$ (total length $15 \ 	ext{cm}$ is implied by the geometry,but the problem states the total resistance is $24 \ \Omega$).Since the total resistance is $24 \ \Omega$ for the entire wire,we determine the resistance per unit length. Assuming the total length is $20 \ 	ext{cm}$ (based on the diagram: $5+5+5$ for the triangle and $5$ for the tail),the resistance per unit length is $24 \ \Omega / 20 \ 	ext{cm} = 1.2 \ \Omega/	ext{cm}$.Each side of the triangle has a length of $5 \ 	ext{cm}$,so each side has a resistance of $5 \ 	ext{cm} 	imes 1.2 \ \Omega/	ext{cm} = 6 \ \Omega$.The straight segment also has a length of $5 \ 	ext{cm}$,so its resistance is $6 \ \Omega$.The circuit consists of two branches in parallel between the triangle's vertices,in series with the final $6 \ \Omega$ resistor.The two branches of the triangle are $6 \ \Omega$ (one side) and $6+6=12 \ \Omega$ (two sides in series).Equivalent resistance of the triangle part: $R_p = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4 \ \Omega$.Total resistance between $A$ and $B$: $R_{AB} = R_p + 6 \ \Omega = 4 \ \Omega + 6 \ \Omega = 10 \ \Omega$.

$A$ wire has a resistance of $24 \ \Omega$ and is bent into the shape shown below. The effective resistance between $A$ and $B$ is ............. $\Omega$.

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