Cascaded Amplifier Circuit Analysis Problem
In order to use KVL equations, you should be wary of current sources. While it is still possible to include them in the equations, they tend to complicate things. There is one in your circuit, part of the EFGH two-port.



It is possible to convert this to a Thévenin equivalent. For this you determine the open-circuit voltage, and the closed loop current to calculate the equivalent resistance:

$$V_{thev} = V_{oc} = G\cdot v_3 \cdot R_4$$

$$R_{thev} = \frac{V_{oc}}{I_{cc}} = \frac{G\cdot v_3 \cdot R_4}{G\cdot v_3} = R_4$$

The Thévenin equivalent then becomes







simulate this circuit &ndash; Schematic created using CircuitLab

Leading up to the following circuit, where you can apply KVL laws more easily:







simulate this circuit

This schematic nicely defines 3 distinct loops (I chose from left to right \$\Gamma_1\$, \$\Gamma_2\$ and \$\Gamma_3\$, all clockwise) for which we can write the KVL equations:

$$\begin{align}
-V_{IN} + R_5\cdot \Gamma_1 &amp;= 0 \\
&amp;\Downarrow \\
\Gamma_1 &amp;= \frac{V_{IN}}{R_5}
\end{align}$$

(\$i_s = \Gamma_1\$)

$$\begin{align}
Z\cdot \Gamma_1 + R_1\cdot \Gamma_2 + R_7\cdot \Gamma_2 + R_3\cdot \Gamma_2 + R_6\cdot \Gamma_2 &amp;= 0\\
&amp;\Downarrow \\
Z\cdot \Gamma_1 + (R_1 + R_7 + R_3 + R_6)\cdot \Gamma_2 &amp;= 0
\end{align}$$

(\$v_1 = R_1\cdot \Gamma_2\$, \$v_3 = -R_3\cdot \Gamma_2\$)

$$\begin{align}
-kR_1\cdot \Gamma_2 + R_2\cdot \Gamma_3 + R_8\cdot \Gamma_3 - GR_4(-R_3)\cdot \Gamma_2 + R_7\cdot \Gamma_3 &amp;= 0 \\
&amp;\Downarrow \\
(-kR_1 + GR_4R_3)\cdot \Gamma_2 + (R_2 + R_8 + R_7)\cdot \Gamma_3 &amp;= 0
\end{align}$$

These equations can be solved relatively easily. Finally, \$V_{OUT} = R_8\cdot \Gamma_3\$.

Similar to current sources for KVL, voltage sources turn out to be a bit tricky for KCL laws too. KVL is preferred here because:


There are more voltage sources (3) than current sources (1), so we can save work
The only current source is shunted with a resistor, making conversion to the Thévenin equivalent easy
KVL laws do not need a ground node to be solved, KCL would require you choose grounds in the three loops