Question 137 of 380

If you are provided a set of resistances, $2\Omega ,4\Omega ,6\Omega$ and $8\Omega .$ Connect these resistances to obtain an equivalent resistance of $\dfrac{46}{3}\Omega$.

A$2\Omega$ and $6\Omega$ are in parallel with $4\Omega$ and $8\Omega$ in seriesB$4\Omega$ and $6\Omega$ are in parallel with $2\Omega$ and $8\Omega$ in seriesC$2\Omega$ and $4\Omega$ are in parallel with $6\Omega$ and $8\Omega$ in seriesD$6\Omega$ and $8\Omega$ are in parallel with $2\Omega$ and $4\Omega$ in series

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Answer: (C) $2\Omega$ and $4\Omega$ are in parallel with $6\Omega$ and $8\Omega$ in series

We can arrange the circuit like above now on solving we get a simplified circuit,

here, both the resistance are connected in series then resultant resistance will be,
${R}_{eq}=\dfrac{4}{3}\Omega +14\Omega =\dfrac{46}{3}\Omega$, in the other case we can not get this result.

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