In the ladder network shown, the current thro | vedclass

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(B) Starting from the rightmost loop, the current through the $1\, \Omega$, $3\, \Omega$, and $2\, \Omega$ resistors is $0.25\, A$. The voltage across

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$20$

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(B) Starting from the rightmost loop, the current through the $1\, \Omega$, $3\, \Omega$, and $2\, \Omega$ resistors is $0.25\, A$. The voltage across this branch is $V_1 = 0.25 	imes (1 + 3 + 2) = 0.25 	imes 6 = 1.5\, V$.This voltage $V_1$ appears across the $6\, \Omega$ resistor. Thus, the current through the $6\, \Omega$ resistor is $I_2 = 1.5 / 6 = 0.25\, A$.The total current entering the middle node is $I_{total} = 0.25\, A + 0.25\, A = 0.5\, A$.This current $0.5\, A$ flows through the $5\, \Omega$ resistor. The voltage drop across the $5\, \Omega$ resistor is $V_2 = 0.5 	imes 5 = 2.5\, V$.The total voltage across the $6\, \Omega$ resistor and the $5\, \Omega$ resistor is $V_{mid} = 1.5 + 2.5 = 4.0\, V$.This voltage $V_{mid}$ appears across the $8\, \Omega$ resistor. Thus, the current through the $8\, \Omega$ resistor is $I_3 = 4.0 / 8 = 0.5\, A$.The total current entering the first node is $I_{in} = 0.5\, A + 0.5\, A = 1.0\, A$.Finally, the input voltage $V$ is the sum of the voltage drops across the $7\, \Omega$ resistor, the $8\, \Omega$ branch, and the $9\, \Omega$ resistor:$V = I_{in} 	imes 7 + V_{mid} + I_{in} 	imes 9$$V = (1.0 	imes 7) + 4.0 + (1.0 	imes 9) = 7 + 4 + 9 = 20\, V$.

In the ladder network shown, the current through the resistor $3\, \Omega$ is $0.25\, A$. The input voltage $V$ is equal to: (in $V$)

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