# Thevenin Voltage BJT two DC sources

I'm trying to find the thevenin voltage at the base of Q1, but I can't seem to properly work out the value. It's just something in the basics I have a misunderstanding of:

That's the circuit, and I've tried to draw the circuit to find the Thevenin voltage but I don't understand what the total voltage is to find the voltage across R2. V should equal V = 10V, but I calculate it as 0V. I just don't understand why it's 10V?

If I use V=10V then I calculate that the Thevenin voltage is -1.67V.

A simple voltage divider using two resistors could just use the following formula for the midpoint voltage:

$$V_{_\text{TH}}=\frac{V_2\cdot R_1+V_1\cdot R_2}{R_1+R_2}$$

That's where $$\V_2\$$ is the voltage tied to $$\R_2\$$ and $$\V_1\$$ is the voltage tied to $$\R_1\$$.

It may (among other ways) follow this approach:

\require{cancel}\begin{align*} I_{_\text{TOT}} &=\frac{V_1-V_2}{R_1+R_2} \\\\ V_{_\text{TH}} &= V_2+I_{_\text{TOT}}\cdot R_2 \\\\ &=V_2+\frac{V_1-V_2}{R_1+R_2}\cdot R_2 \\\\ &=V_2\cdot \frac{R_1+R_2}{R_1+R_2}+\frac{V_1-V_2}{R_1+R_2}\cdot R_2 \\\\ &= \frac{V_2\cdot\left(R_1+R_2\right)}{R_1+R_2}+\frac{\left(V_1-V_2\right)\cdot R_2}{R_1+R_2} \\\\ &= \frac{V_2\cdot\left(R_1+R_2\right)+\left(V_1-V_2\right)\cdot R_2}{R_1+R_2} \\\\ &= \frac{V_2\cdot R_1\cancel{+V_2\cdot R_2}+V_1\cdot R_2\cancel{-V_2\cdot R_2}}{R_1+R_2} \\\\ &=\frac{V_2\cdot R_1+V_1\cdot R_2}{R_1+R_2} \end{align*}

Note that the voltage difference is used to work out the current through the two resistors, in series. Not the voltage sum.

In the final equation the voltages are summed, but not before first multiplying them by the resistance opposite to them.