Electronic – Mechanical- Electrical circuit equivalence

As always, it's helpful to first draw the DC and AC circuits.

DC circuit:

^{simulate this circuit – Schematic created using CircuitLab}

The operating point is evident by inspection:

$$I_C = \frac{\beta}{1 + \beta}I_2 = \alpha I_2 $$

$$V_C = I_C(\frac{75\Omega}{\alpha} + \frac{100k\Omega}{\beta}) + V_{BE} $$

Update to address comment:

I can't perfectly grasp your equation for Vcc.I think understand you divide resistance with beta and alpha to make them equivalent resistance looking through C.

Assuming you meant \$V_C\$ rather than \$V_{CC}\$, by KVL we have

$$V_C = V_E + V_{BE} + V_{R1}$$

We have

$$V_E = I_E R_S = \frac{I_C}{\alpha}R_S $$

and

$$V_{R1} = I_B R_1 = \frac{I_C}{\beta}R_1$$

Thus

$$V_C = I_C(\frac{R_S}{\alpha} + \frac{R_1}{\beta}) + V_{BE} $$

AC circuit:

^{simulate this circuit}

The small-signal circuit is thus

^{simulate this circuit}

This is a straightforward circuit to solve. What have you tried so far?

In your two circuits, the dependent voltage source branch is transformed to a dependent current source paralleled with a resistor. It's "source transformation" and is an application of "Norton's Theorem". Source transformation also applies to dependent sources. So, the two circuits are equivalent.