Thévenin equivalent of circuit with two voltage sources

\$R_1\$ and \$R_2\$ form a simple resistor potential divider, which is trivial to describe algebraically, but we need to be really careful with \$V_2\$, because even though \$V_2=+5V\$, the absolute potential ("absolute" meaning "with respect to ground") at the bottom of \$R_2\$ is actually \$-V_2=-5V\$. It can be a bit confusing when you have the positive side of a positive voltage source connected to ground. Just keep that in mind.

By contrast, the potential at the top of \$R_1\$ is unambiguous; it's clearly +5V relative to ground, or \$+V_1\$.

It's probably better if I redraw the situation with node names, and start using absolute potentials (with respect to ground):

^{simulate this circuit – Schematic created using CircuitLab}

We are interested in the potential \$V_X\$ at node X, the junction of the pair of resistors. That will be the Thevenin voltage, since we leave the output open, disregarding load \$R_L\$. Consequently there's no current through \$R_3\$, and the open circuit output voltage \$V_O\$ will be the same as \$V_X\$

We already have

$$ V_A = 0V + V_1 = +5V $$ $$ V_B = 0V - V_2 = -5V $$

If we treat this using Ohm's law, to find the current \$I\$ through \$R_1\$ and \$R_2\$, it would go as follows. Find the potential difference across the pair, and divide by the total resistance of the pair:

$$ I = \frac{V_A - V_B}{R_1 + R_2} $$

Then you apply Ohm's law again to work out the voltages \$V_{R1}\$ and \$V_{R2}\$ across each resistor:

$$ \begin{aligned} V_{R1} &= I R_1 \\ \\ &= R_1 \frac{V_A - V_B}{R_1+R_2} \end{aligned} $$

$$ \begin{aligned} V_{R2} &= I R_2 \\ \\ &= R_2 \frac{V_A - V_B}{R_1+R_2} \end{aligned} $$

Now you must find the potential at the junction of the resistors, node X. This is an application of KVL. Start at some point where you know the potential (either at the top of the chain or the bottom), and jump over elements along the path towards the node of interest, summing the changes in potential across each element as you go.

If you started at the top, node A, you would be subtracting voltages across the resistors as you go, because according to my annotated direction of \$I\$, the lower end of all the resistors in the chain must be lower in potential than the upper end. Therefore each time we jump downwards across a resistor, this must incur a fall in potential.

I want to do this starting at B, though, going upwards. We want to know the potential \$V_X\$ at node X, so starting at node B, where absolute potential is \$V_B\$, work your way up the chain adding each potential change as you go. Here there's only one jump, across \$R_2\$:

$$ \begin{aligned} V_X &= V_B + V_{R_2} \\ \\ &= V_B + R_2 \frac{V_A - V_B}{R_1+R_2} \\ \\ &= V_B + (V_A - V_B) \frac{R_2}{R_1+R_2} \end{aligned} $$

You could apply this procedure for any number of resistors in the chain. KVL will always work out.

_{Note: If you continued the journey to node A, your "accumulated" potential must necessarily equal \$V_A\$. The sum of voltages across all resistors in the chain must equal the potential difference between the two ends. That's just another way of stating KVL.}

That's all. There's nothing left to do except to plug in the numbers:

$$ \begin{aligned} V_X &= (-5V) + \left[(+5V) - (-5V)\right] \frac{10k\Omega}{10k\Omega+10k\Omega} \\ \\ &= (-5V) + (+10V)\frac{1}{2} \\ \\ &= 0V \end{aligned} $$

This was an obvious result from the start, since the two resistors are equal, meaning that their junction would have a potential exactly half-way between the two ends at −5V and +5V.

\$V_X = 0V\$ is the Thevenin voltage, so the Thevenin equivalent is this:

^{simulate this circuit}