Problem-Solving Strategy for applying the Thevenin's Theorem
STEP 1. Remove the load and find the voltage across the open-circuit terminals, $V_{oc}$. All the circuit analysis techniques presented here can be used to compute this voltage.
STEP 2. Determine the Thévenin equivalent resistance of the network at the open terminals with the load removed. Three different types of circuits may be encountered in determining the resistance, $R_{Th}$.
(a) If the circuit contains only independent sources, they are made zero by replacing the voltage sources with short circuits and the current sources with open circuits. $R_{Th}$ is then found by computing the resistance of the purely resistive network at the open terminals.
(b) If the circuit contains only dependent sources, an independent voltage or current source is applied at the open terminals and the corresponding current or voltage at these terminals is measured. The voltage/current ratio at the terminals is the Thévenin equivalent resistance. Since there is no energy source, the open-circuit voltage is zero in this case.
(c) If the circuit contains both independent and dependent sources, the open-circuit
terminals are shorted and the short-circuit current between these terminals is determined. The ratio of the open-circuit voltage to the short-circuit current is the resistance $R_{Th}$.
STEP 3. If the load is now connected to the Thévenin equivalent circuit, consisting of $V_{oc}$ in series with $R_{Th}$, the desired solution can be obtained.
PROBLEM:
Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed.
If there are any errors in solution, please let me know in the comments
SOLUTION:
Since $V_3=2V$, $V_2=2-2V_x$, $V_1=2-V_x=6V$
$V_x=-4V$
Applying KCL, we get
$\frac{V_{oc}-V_1}{1k}+\frac{V_{oc}-V_3}{1k}=0$
$V_{oc}=4V$
Since $V_6=2V$, $V_5=2-2V_x$, $V_4=2-V_x=6V$
$V_x=-4V$
Applying KCL, we get
$I_{sc}=\frac{V_4}{1k}+\frac{V_6}{1k}=8mA$
Because of the presence of the dependent source, $R_{Th}$ must be determined from the equation
$R_{Th}=\frac{V_{oc}}{I_{sc}}=0.5k\Omega$
Therefore, $I_o=\frac{4}{0.5k+2k}=1.6mA$
Related Posts: Norton's Theorem
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