Question 112 of 380

The equivalent resistance of the circuit shown below between points $a$ and $b$ is :

A$16\Omega$B$3.2\Omega$C$24\Omega$D$20\Omega$

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Answer: (B) $3.2\Omega$

The resistance in $ab$ is parallel to all other resistances. Since all the resistances other than ${R}_{b}$ is $4\Omega$ the potential is zero$\left({V}_{x}={V}_{y}\right)$ as the upper branch will become a case of balanced wheatstone bridge between {V}_{x}&{V}_{y}. So, the simplified circuit looks like the one below.

Hence, the equivalent resistance is
$\dfrac{1}{{R}_{eq}}=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{16}=\dfrac{5}{16} \Rightarrow {R}_{eq}=\dfrac{16}{5}\Omega =3.2\Omega$

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.