# Is this diagram wrong as the congruent circuit?

Given the below diagram.

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.

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?

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

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.