Study the following circuit. Values of r₁, r₂ and r₃ are 1 ohm, 2 ohm and 3 ohm respectively. A resistor R is connected across the points C and D. What should be the value of R for which the resistance of the network across AB is R?

Solution

Let us consider the extreme right square of the loop.
Resistance across EF = (r₁+R+r₂) and r₃ in parallel

$= \frac{r_{3} (r_{1} + r_{2} + R)}{(r_{1} + r_{2} + r_{3} + R)}$

This value should be equal to R, so that by the repeated operation of this type, we will be left with only one square which will be the left extreme one and it will have a value R

i.e., $\frac{r_{3} (r_{1} + r_{2} + R)}{(r_{1} + r_{2} + r_{3} + R)} = R$

Now, Substituting the numerical values

$\frac{3 (1 + 2 + R)}{(1 + 2 + 3 + R)} = R$

$\Rightarrow$ $\frac{3 (3 + R)}{(6 + R)} = R$

$\Rightarrow$ $9 + 3 R = 6 R + R^{2}$

$\Rightarrow R^{2} + 3 R - 9 = 0$

i.e., $R = \frac{- 3 \pm \sqrt{9 + 36}}{2} = \frac{- 3 \pm 3 \sqrt{5}}{2}$

$∴$ $R = \frac{3 (\sqrt{5} - 1)}{2} Ω$ is the required value of R.

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Study the following circuit. Values of r₁, r₂ and r₃ are 1 ohm, 2 ohm and 3 ohm respectively. A resistor R is connected across the points C and D. What should be the value of R for which the resistance of the network across AB is R?

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Study the following circuit. Values of r₁, r₂ and r₃ are 1 ohm, 2 ohm and 3 ohm respectively. A resistor R is connected across the points C and D. What should be the value of R for which the resistance of the network across AB is R?