This is a 2nd part of this thread.

The circuit is the same

but this time I tried to determine

\(h_{21} = \left . \frac{I_2}{I_1} \right |_{V_2 = 0} = \frac{g_m - sC_2}{G_1 + s(C_1 + C_2)}\)

and my doubt is related with the VCCS in parallel with a short-circuit.

It seems it's turned off in calculating \(h_{11}\).

At first glance, I would say that a VCCS in parallel with a short-circuit has no effect on the rest of the circuit (same as a resistor in parallel with short-circuit) so the following circuit

disagrees with the book's solution, so while working on this circuit:

applying the current divider the result agrees with the following reasoning:

\(

\begin{align}

I_2 = I_1\left(\frac{\frac{1}{sC_1} \parallel R_1 }{\left(\frac{1}{sC_1} \parallel R_1 \right ) + \frac{1}{sC_2}} \right ) - g_mV_1 = I_1\left(\frac{\frac{1}{sC_1} \parallel R_1 }{\left(\frac{1}{sC_1} \parallel R_1 \right ) + \frac{1}{sC_2}} \right ) - g_m I_1\left( R_1 \parallel \frac{1}{sC_1} \parallel \frac{1}{sC_2} \right )

\end{align}

\)

So the question is: if the VCCS has been turned of off calculating \(h_{11}\), why shouldn't be turned off calculating \(h_{21}\) ? This is a non-sense because these are two different behaviours for the same situation (VCCS in parallel with a short-circuit) , and of course I am wrong and my reasoning is not correct.

Could you help me to understand ?

Thank you in advance,

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