Consider the circuit shown in the figure wher | vedclass

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(A) Let the resistance of each resistor be $R = 1 k\Omega = 1000 \Omega$. The circuit is a ladder network. Let the current in the rightmost vertical r

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$34$

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(A) Let the resistance of each resistor be $R = 1 k\Omega = 1000 \Omega$. The circuit is a ladder network. Let the current in the rightmost vertical resistor $X$ be $I_4 = 1 mA$. The current in the horizontal resistor connected to it is also $1 mA$. Applying Kirchhoff's Current Law $(KCL)$ at the node above $X$,the current in the vertical resistor to the left of $X$ is $I_v = 1 mA + 1 mA = 2 mA$. Moving left,the current in the next horizontal resistor is $I_h = 1 mA + 2 mA = 3 mA$. The current in the next vertical resistor is $I_v = 3 mA + 2 mA = 5 mA$. Continuing this process: - Current in the next horizontal resistor: $I_h = 3 mA + 5 mA = 8 mA$. - Current in the next vertical resistor: $I_v = 8 mA + 5 mA = 13 mA$. - Current in the first horizontal resistor: $I_h = 8 mA + 13 mA = 21 mA$. - Current in the first vertical resistor: $I_v = 21 mA + 13 mA = 34 mA$. The potential difference between $A$ and $B$ is the voltage across the first vertical resistor: $V_{AB} = I_{v1} 	imes R = 34 mA 	imes 1 k\Omega = 34 V$.

Consider the circuit shown in the figure where all the resistances are of magnitude $1 k \Omega$. If the current in the extreme right resistance $X$ is $1 mA$,the potential difference between $A$ and $B$ is: (in $V$)

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