Finding the Thevenin Equivalent Circuit for a Network with a Load Resistor

[icesalmon]

Homework Statement

I just want to check to see if my understanding in anything here has gaps.

Q: Find the Thevenin equivalent circuit for the network external to he load resistor $R_L$:

Homework Equations

Ohm's Law: $V = IR$
Kirchoff's Voltage Law: Σ$V_{drops}$ + Σ$V_{gains}$ = 0
Kirchoff's Current Law: Σ$I_{closed loop}$ = 0

The Attempt at a Solution

I started by redrawing the circuit as best as I could

For the thevenin resistance I removed $R_L$ with an open circuit and then short circuit the 12V source and place an ohm-meter at the open circuit where $R_L$ was removed:

Calculating the thevenin resistance: I get $R_{th} = ( R_1 // R_2 ) + R_3 = 4.04kΩ$

Placing the $12V$ source back in and calculating the thevenin voltage $V_{th}$ would be the voltage drop across the resistor $R_2$.

Using superposition I can determine the effects each voltage source has on $R_2$
starting by shorting out the $4V$ source I get the network on the left

Since there is an open circuit where $R_L$ was removed current will not flow through that branch and I can calculate $I'$ by using KVL:
Σ$V_{drops}$ = Σ$V_{gains}$: $-12V + I_1R1 + I_1R2 = 0$ $I_1$ => $I_1'R_2$ => $V_{th}' = 8V$

Now removing the 12V Source $E_1$ and seeing that $E_2$ is in parallel with $R_2$ I can automatically determine $V_th$'' to be $E_2 = 4V$

From superposition: $V_{th} = V_{th}' + V_{th}''$ $V_th = 8V + 4V = 12V$

 

[NascentOxygen]

Correct.