Each element in the finite chain of resistors | vedclass

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(B) Let the potential across the last resistor be $V_4 = 2\,V$ (since $I = 1\,A$ and $R = 2\,\Omega$ is not correct,the diagram shows the last vertica

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$34$

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(B) Let the potential across the last resistor be $V_4 = 2\,V$ (since $I = 1\,A$ and $R = 2\,\Omega$ is not correct,the diagram shows the last vertical resistor is $1\,\Omega$ and the last horizontal is $1\,\Omega$).Working backwards from the end:$1$. The last vertical resistor has $1\,A$ current flowing through it,so its potential is $V_4 = 1\,A 	imes 1\,\Omega = 1\,V$.$2$. The current through the last horizontal resistor is $1\,A$. The potential drop across it is $1\,A 	imes 1\,\Omega = 1\,V$. So,the potential before this resistor is $1\,V + 1\,V = 2\,V$.$3$. The current through the next vertical resistor is $V/R = 2\,V / 1\,\Omega = 2\,A$. Total current entering this node is $1\,A + 2\,A = 3\,A$.$4$. Potential drop across the next horizontal resistor is $3\,A 	imes 1\,\Omega = 3\,V$. Potential before this is $2\,V + 3\,V = 5\,V$.$5$. Current through the next vertical resistor is $5\,V / 1\,\Omega = 5\,A$. Total current is $3\,A + 5\,A = 8\,A$.$6$. Potential drop across the next horizontal resistor is $8\,A 	imes 1\,\Omega = 8\,V$. Potential before this is $5\,V + 8\,V = 13\,V$.$7$. Current through the next vertical resistor is $13\,V / 1\,\Omega = 13\,A$. Total current is $8\,A + 13\,A = 21\,A$.$8$. Potential drop across the first horizontal resistor is $21\,A 	imes 1\,\Omega = 21\,V$. Total input potential is $13\,V + 21\,V = 34\,V$.

Each element in the finite chain of resistors shown in the figure is $1\,\Omega $. $A$ current of $1\, A$ flows through the final element. Then what is the potential difference $V$ across input terminals of the chain (in $,V$)?

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