If you are provided a set of resistances 2 Ω, 4 Ω, 6 Ω and 8 Ω. Connect these resistances so as to obtain an equivalent resistance of 46/3Ω. (a) 4 Ω and 6 Ω are in parallel with 2 Ω and 8 Ω in series (b) 6 Ω and 8 Ω are in parallel with 2 Ω and 4 Ω in series (c) 2 Ω and 6 Ω are in parallel with 4 Ω and 8 Ω in series (d) 2 Ω and 4 Ω are in parallel with 6 Ω and 8 Ω in series

Answer: (d). Try (d): 2 Ω parallel 4 Ω gives (2 × 4)/(6) = 4/3 Ω. Add 6 Ω and 8 Ω in series: 4/3 + 14 = 46/3 Ω So the answer is (d).

If you are provided a set of resistances 2 Ω, 4 Ω, 6 Ω and 8 Ω… | JEE Main 2021 (26 Aug, Shift 2)

Q46JEE Main 2021 · 26 AUG · SHIFT 2Easy

Combination of Resistors

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

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Answer: (d)

Try (d): $2\ \Omega$ parallel $4\ \Omega$ gives $\dfrac{2 \times 4}{6} = \dfrac{4}{3}\ \Omega$ .
Add $6\ \Omega$ and $8\ \Omega$ in series: $\dfrac{4}{3} + 14 = \dfrac{46}{3}\ \Omega$
So the answer is (d).

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Combination of Resistors question from JEE Main 2021 (26 Aug, Shift 2)

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