Consider an infinite ladder network as shown in the figure. A voltage V is applied between the points A and B. This applied value of voltage is halved after each section. Find the value of $\frac{R_{1}}{R_{2}}$ :

1

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\frac{1}{2}

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2

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3

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Solution

The correct option is A $\frac{1}{2}$
Here, consider the part of circuit in middle where $R_{e q}$ and $R_{2}$ are parallel in series with $R_{1}$ .
Since the circuit is same to original circuit, the equivalent resistance is again $R_{e q}$ .
$R_{e q} = R_{1} + (\frac{1}{R_{2}} + \frac{1}{R_{e q}})^{- 1}$
Therefore, $(R_{e q})^{2} - R_{1} R_{e q} - R_{1} R_{2} = 0$
Taking positive solution we get,
$R_{e q} = \frac{(R_{1}) \pm \sqrt{(R_{1})^{2} + 4 R_{1} R_{2}}}{2}$ (1)
Now, consider a link of voltage difference $V_{i}$ between top and bottom.
The current through this equivalent circuit will also be the current through $R_{1}$ and is $\frac{V_{i}}{R_{e q}}$ .
Hence, $V_{i + 1} = V_{i} - R \frac{V_{i}}{R_{e q}}$
Now, the applied value of voltage is halved after each section hence, $\frac{R_{1}}{R_{e q}} = \frac{1}{2}$
Using this in equation (1), we get:
$4 R_{1} = R_{1} + \sqrt{(R_{1})^{2} + 4 R_{1} R_{2}}$
$R_{1} = 2 R_{2}$
$\frac{R_{1}}{R_{2}} = \frac{1}{2}$

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Consider an infinite ladder network as shown in the figure. A voltage V is applied between the points A and B. This applied value of voltage is halved after each section. Find the value of R1R2 :

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