What is the current I shown in the given circ | vedclass

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(C) In the given ladder network,let the current flowing through the first series resistor $R$ be $I_1$. At each node,the current splits between the sh

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$\frac{V}{16 R}$

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(C) In the given ladder network,let the current flowing through the first series resistor $R$ be $I_1$. At each node,the current splits between the shunt resistor $2R$ and the next series resistor $R$. By analyzing the circuit from right to left,we observe that the current halves at each stage due to the symmetry of the ladder network. Specifically,if $I_1$ is the current entering the first series resistor,the current flowing through the subsequent shunt resistors will be $I_1/2, I_1/4, I_1/8$,and so on. The current $I$ in the final branch is given by $I = I_1/8$. Since the voltage across the first shunt resistor $2R$ is $V$,the current $I_1$ flowing through the first series resistor is $I_1 = V / (2R)$. Substituting this value into the expression for $I$,we get: $I = \frac{1}{8} 	imes \frac{V}{2R} = \frac{V}{16R}$.

What is the current $I$ shown in the given circuit?

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