Discussion
In the given network, \(Z_{11}\) is equal to
Answer: B
The input impedance with the output open (\(I_2 = 0\)) is \(Z_{11} = E_1/I_1\), which is \(\frac{6}{11}\) ohm (option B). With the output open, the loop holds the 2 ohm and 4 ohm resistors and the dependent source, whose plus terminal is at the bottom.
\[ E_1 - 2I_1 - 4I_1 + 10E_1 = 0 \]
\[ 11E_1 = 6I_1 \]
\[ Z_{11} = \frac{6}{11}\ \Omega \]
The value \(-\frac{6}{11}\) (A) comes from taking the source with the opposite polarity, which would give \(1 - 10 = -9\) in the denominator, not 11. The other two values do not follow from the loop.
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