Four conductors of same resistance are connected to form a square. If the resistance between diagonally opposite corners is 8 ohm, the resistance between any two adjacent corners is

32 o h m

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8 o h m

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1 / 6 o h m

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6 o h m

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Solution

The correct option is B $6 o h m$
$R_{e q} = \frac{2 R \times 2 R}{4 R} = 8$
$\Rightarrow R = 8 Ω$
So, $R_{A B} = \frac{R \times 3 R}{4 R}$ $= \frac{3}{4} R$
or, $R_{A B} = 6 Ω$

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Four conductors of same resistance are connected to form a square. If the resistance between diagonally opposite corners is 8 ohm, the resistance between any two adjacent corners is

The correct option is B 6 ohmReq=2R×2R4R=8⇒ R=8ΩSo, RAB=R×3R4R =34Ror, RAB=6Ω

The correct option is B $6 o h m$
$R_{e q} = \frac{2 R \times 2 R}{4 R} = 8$
$\Rightarrow R = 8 Ω$
So, $R_{A B} = \frac{R \times 3 R}{4 R}$ $= \frac{3}{4} R$
or, $R_{A B} = 6 Ω$