Consider an infinite ladder network shown in | vedclass

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(B) Let the voltage at the input be $V$. According to the problem, the voltage across the shunt resistor $R_2$ in the first section is $V/2$. Applying

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$R_1/R_2 = 1/2$

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(B) Let the voltage at the input be $V$. According to the problem, the voltage across the shunt resistor $R_2$ in the first section is $V/2$. Applying Kirchhoff's Current Law at the first node: The current entering the first section is $I = (V - V/2) / R_1 = V / (2R_1)$. The current flowing through the first shunt resistor $R_2$ is $I_2 = (V/2) / R_2$. The current flowing into the rest of the infinite ladder is $I' = (V/2 - V/4) / R_1 = V / (4R_1)$. Since $I = I_2 + I'$, we have: $V / (2R_1) = V / (2R_2) + V / (4R_1)$. Subtracting $V / (4R_1)$ from both sides: $V / (4R_1) = V / (2R_2)$. Dividing by $V$ and rearranging: $1 / (4R_1) = 1 / (2R_2) \implies 2R_2 = 4R_1 \implies R_1 / R_2 = 2/4 = 1/2$.

Consider an infinite ladder network shown in the figure. $V$ voltage $V$ is applied between the points $A$ and $B$. This applied voltage is halved after each section.

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