Question 154 of 380

A uniform metallic wire has a resistance of $18\Omega$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is:

A$2\Omega$B$12\Omega$C$8\Omega$D$4\Omega$

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Answer: (D) $4\Omega$

Resistance,
$R=\dfrac{\rho l}{A} R\alpha l say{R}_{eff}=l=18 as{l}_{1}={l}_{2}={l}_{3}=l/3 \Rightarrow {R}_{1}={R}_{2}={R}_{3}=R=18/3 =6$
So from diagram
$2R=12andR=6 alsoweknow {R}_{AB}=\dfrac{{R}_{2R}{R}_{R}}{{R}_{2R}+{R}_{R}} {R}_{AB}=\dfrac{6\times 12}{6+12}=4\Omega$

JEE Main Physics practice question on Current Electricity (Combination of Resistances), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.