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Given the below diagram.

enter image description here

The 2 identities are given.

$$ V_{1}=AV_{2}+BI_{2} $$

$$ I_{1}=CV_{2}+DI_{2} $$

$$ A,B,C,D~\text{are constant coefficients.} $$

As \$V_{2}=0~\$

\$V_{1}=BI_{2}\$

\$I_{1}=DI_{2}\$

the above circuit is congruent with the below diagram.

enter image description here

Hence,

\$V_1=r_3I_2\$

\$V_1=r_1(I_1-I_2)\$

These statements are given from the textbook.

The problem for me is that why the pre position of \$r_2\$ was shortened in the second diagram.

I think that as \$V_2=0\$ then any current doesn't flow between the endpoints of \$r_2\$ so this part must be opened,not be shortened.

Is the second diagram incorrect?

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The second diagram is correct, because imposing \$V_2=0\$ means that no voltage drops across that portion of the circuit and such a condition is obtained indeed by short-circuiting \$r_2\$. This condition does not imply no current flow, and in fact \$I_2\$ flows through \$r_3\$, down into the shorted branch (by-passing \$r_2\$ and thus avoiding any voltage drop across it) and back towards the negative "side" of \$V_1\$, where it is combined with \$I_1-I_2\$ correctly yielding \$I_1\$.

Put it another way, if you actually opened your rightmost branch, then no current would flow through neither \$r_2\$ or \$r_3\$, but \$V_2=V_1\neq0\$ would hold and contradict your starting \$V_2=0\$ assumption.

Summarizing:

  • \$I=0\$ --> open circuit
  • \$V=0\$ --> short circuit
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Is the second diagram incorrect?

The 2nd diagram has the \$V_2\$ port shorted out hence \$V_2 = 0\$. There is no other way to make \$V_2 = 0\$ other than by shorting it out (or making \$r_3\$ infinity) hence, the 2nd diagram is correct.

If you made \$r_3\$ infinity then it doesn't help you progress the problem because \$I_2\$ becomes zero.

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