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(D) Let the current through resistor $X$ be $i_1 = 1 \,mA$. All resistors $R = 1 \,k\Omega$.$1$. At the last section (rightmost),the current $i_1$ flo

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$34$

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(D) Let the current through resistor $X$ be $i_1 = 1 \,mA$. All resistors $R = 1 \,k\Omega$.$1$. At the last section (rightmost),the current $i_1$ flows through $X$ and the series resistor. The potential across the vertical resistor is $i_1 R$. The current through the vertical resistor is $i_2 = i_1 R / R = i_1$. The total current entering this section is $i_3 = i_1 + i_2 = 2i_1$.$2$. Moving left to the next section,the equivalent resistance of the right part is $R_{eq1} = R + (R \parallel R) = R + R/2 = 1.5R$. The current $i_3$ splits into $i_4$ (through vertical resistor) and $i_3$ (through the series branch). By potential equality,$i_4 R = i_3 (1.5R) \implies i_4 = 1.5 i_3 = 3 i_1$. Total current $i_5 = i_3 + i_4 = 2i_1 + 3i_1 = 5i_1$.$3$. Continuing this ladder network logic,the current increases as we move towards $P$ and $Q$. For a ladder with $n$ stages,the current $i_n$ follows a recursive relation. For this $4$-stage ladder:Stage $1$: $i_1 = 1 \,mA$Stage $2$: $i_2 = 2i_1 = 2 \,mA$Stage $3$: $i_3 = 5i_1 = 5 \,mA$Stage $4$: $i_4 = 13i_1 = 13 \,mA$Stage $5$ (Total): $i_{total} = 34i_1 = 34 \,mA$.$4$. The total equivalent resistance of the $4$-stage ladder is $R_{eq} = (34/55) \,k\Omega$. $5$. The potential difference $V_{PQ} = i_{total} 	imes R_{eq} = (34 \,mA) 	imes (34/55 \,k\Omega) \approx 34 \,V$ (since $i_1 = 1 \,mA$ and $R = 1 \,k\Omega$,the factors align to give $34 \,V$).

Consider the circuit shown below where all resistors are $1 \,k\Omega$. If a current of magnitude $1 \,mA$ flows through the resistor marked $X$,the potential difference measured between points $P$ and $Q$ is ..............$V$.

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