Find the equivalent resistance between two ends of the following circuit: % Circuit Description The circuit consists of

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Find the equivalent resistance between two ends of the following circuit: % Circuit Description The circuit consists of three resistors, two of r3 in series connected in parallel with another…

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Given: The circuit consists of three resistors. Two resistors of value $\frac{r}{3}$ are in series, and this series combination is connected in parallel with another resistor of $r$ .

Find: The equivalent resistance between the two ends.

From the description, the two series resistors give

R_s = \frac{r}{3} + \frac{r}{3} = \frac{2r}{3}

This is in parallel with the resistor $r$ . Therefore,

\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{\frac{2r}{3}} = \frac{1}{r} + \frac{3}{2r} = \frac{5}{2r}

So,

R_{eq} = \frac{2r}{5}

However, the solution explicitly concludes that all three resistors of value $\frac{r}{3}$ are effectively in parallel, giving

\frac{1}{R_{eq}} = \frac{3}{r} + \frac{3}{r} + \frac{3}{r} = \frac{9}{r}

Hence,

R_{eq} = \frac{r}{9}

The solution states the final answer is $\frac{r}{9}$ . This matches option C. There is a discrepancy between the question text description and the solution-page conclusion, and the answer is taken from the solution as required.

Given: the solution labels the correct option as D, but both solution approaches conclude the final equivalent resistance is $\frac{r}{9}$ .

Find: The option corresponding to the concluded resistance.

From the first approach,

\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

with

R_1 = R_2 = R_3 = \frac{r}{3}

so

\frac{1}{R_p} = \frac{3}{r} + \frac{3}{r} + \frac{3}{r} = \frac{9}{r}

and therefore

R_p = \frac{r}{9}

Now compare with the options:

A: $\frac{r}{6}$
B: $r$
C: $\frac{r}{9}$
D: $\frac{r}{3}$

The concluded value $\frac{r}{9}$ corresponds to option C. Therefore, the correct option is C.

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