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 $\frac{46}{3} \Omega$.

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

(A) $4 \Omega$ and $6 \Omega$ are in parallel with $2 \Omega$ and $8 \Omega$ in series.

(B) $6 \Omega$ and $8 \Omega$ are in parallel with $2 \Omega$ and $4 \Omega$ in series.

(D) $2 \Omega$ and $6 \Omega$ are in parallel with $4 \Omega$ and $8 \Omega$ in series.

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Current Electricity

If you are provided a set of resistances

2Ω, 4Ω, 6Ω

and

8Ω

. Connect these resistances so as to obtain an equivalent resistance of

\frac{46}{3} Ω

Answer: (C)

Split the target value:

\frac{46}{3} = \frac{42 + 4}{3} = 14 + \frac{4}{3}

.
A series pair gives

6 + 8 = 14 Ω

, and a parallel pair gives

\frac{2 \times 4}{2 + 4} = \frac{4}{3} Ω

. Placing these two combinations one after the other,

R = 14 + \frac{4}{3} = \frac{46}{3} Ω

so the

2 Ω

and

4 Ω

resistors are joined in parallel, and that pair is put in series with the

6 Ω

and

8 Ω

resistors joined in series.

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option