Five equal resistances are connected in a network as shown in figure. The net resistance between  the points A and B is

Correct Option - 2


Detailed Solution:

Given circuit can be re-drawn and it becomes case of balanced Wheatstone bride

$R_{AB}=\frac{\left(2R\right)\left(2R\right)}{2R+2R}=R$

<div class="photo-widget-full"><div class="figure"><p>Given circuit can be re-drawn and it becomes case of balanced Wheatstone bride</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1755153428phpEYOK0h.jpeg" alt="" width="421" height="86" loading="lazy"></picture></div><div><span contenteditable="false"> <math> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mi>A</mi> <mi>B</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mi>R</mi> </mrow> <mo>)</mo> </mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mi>R</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> <mi>R</mi> <mo>+</mo> <mn>2</mn> <mi>R</mi> </mrow> </mfrac> <mo>=</mo> <mi>R</mi> </mrow> </math> </span></div></div></div></div>

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