If you are provided a set of resistances 2 ohm, 4 ohm, 6 ohm and 8 ohm. Connect these resistances, s

Q: Why did we split $\frac{46}{3}$ into $14 + \frac{4}{3}$ instead of something else?

In circuit problems involving a mix of series and parallel connections, the total resistance is often the sum of a series part (which gives an integer) and a parallel part (which gives a fraction). Splitting it this way helps us isolate which resistors are in series and which are in parallel.

Q: How do we know the $2\Omega$ and $4\Omega$ resistors must be in parallel?

The required resistance for that section is $\frac{4}{3} \Omega$, which is approximately $1.33 \Omega$. A key rule of parallel circuits is that the equivalent resistance is always less than the smallest individual resistor. Since $1.33 < 2$, they must be in parallel.

Q: Could we have solved this by just checking all the options?

Absolutely! Checking the options one by one is a valid and often fast strategy for MCQs. However, learning to deduce the circuit by splitting the fraction builds stronger intuition for more complex subjective problems.

Q: What if the target resistance was greater than the sum of all resistors ($20 \Omega$)?

If the target resistance was greater than $20 \Omega$, it would be physically impossible to achieve with this set of resistors, because the maximum possible resistance is obtained when all of them are in series ($2+4+6+8 = 20 \Omega$).

Q: Is there a general formula to find the combination for any target resistance?

There isn't a single plug-and-chug formula. It requires logical deduction, testing combinations, and understanding that series connections increase resistance while parallel connections decrease it.

Question

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

Options:

(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.

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

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

Sigma Percentile · Accepted Answer

### Decoding the Circuit Puzzle

Imagine you are an electrical engineer handed a box with exactly four resistors: $2\Omega$, $4\Omega$, $6\Omega$, and $8\Omega$. Your mission? To wire them together to create a highly specific equivalent resistance of $\frac{46}{3} \Omega$. At first glance, this fraction might look intimidating, but let's break it down using a bit of mathematical intuition.

### The Power of Splitting Fractions

When faced with an awkward fractional resistance like $\frac{46}{3} \Omega$, the best strategy is to convert it into a mixed fraction. Let's do the math:

$$ \frac{46}{3} = 15.33 \Omega $$

This can be perfectly split into an integer and a smaller proper fraction:

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

Why is this split so powerful? Because in circuit design, adding resistances directly implies a **series combination**. This tells us that our final circuit will likely consist of a block with $14 \Omega$ resistance connected in series with another block having $\frac{4}{3} \Omega$ resistance.

### Building the Blocks

Now, let's look at our available resistors: $2\Omega$, $4\Omega$, $6\Omega$, and $8\Omega$.

**Block 1: The $14 \Omega$ Series Section**
How can we get exactly $14 \Omega$? If we take the $6\Omega$ and $8\Omega$ resistors and connect them end-to-end (in series), their resistances simply add up:

$$ R_s = 6\Omega + 8\Omega = 14\Omega $$

**Block 2: The $\frac{4}{3} \Omega$ Parallel Section**
We are now left with the $2\Omega$ and $4\Omega$ resistors. We need them to produce $\frac{4}{3} \Omega$. Since $\frac{4}{3}$ is smaller than both $2$ and $4$, they must be connected in parallel. Let's verify this using the parallel resistance formula:

$$ R_p = \frac{R_1 	imes R_2}{R_1 + R_2} $$

Substituting our values:

$$ R_p = \frac{2 	imes 4}{2 + 4} = \frac{8}{6} = \frac{4}{3} \Omega $$

It's a perfect match!

### The Final Assembly

To achieve our grand total of $\frac{46}{3} \Omega$, we simply connect our two blocks in series.

We place the $2\Omega$ and $4\Omega$ resistors in parallel, and then connect that entire parallel group in series with the $6\Omega$ and $8\Omega$ resistors.

$$ R_{eq} = R_p + R_s = \frac{4}{3} + 14 = \frac{46}{3} \Omega $$

Looking at our options, this corresponds exactly to option (d): **$2\Omega$ and $4\Omega$ are in parallel with $6\Omega$ and $8\Omega$ in series.**

Animated Solution for Physics - 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 $\frac{46}{3} Ω$ .

Select Answer:

Visualized Solution

$R_{e q} = \frac{46}{3} Ω$

\text{Splitting the Target}

\text{Forming } 14 \Omega

\text{Forming } \frac{4}{3} \Omega

\text{Final Combination}

\text{The Way Forward}

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Decoding the Circuit Puzzle

The Power of Splitting Fractions

Building the Blocks

The Final Assembly

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