# Could a voltage divider with more than two resistors be used?

As a step in an exercise, I'm trying to find voltages $$\V_1\$$ and $$\V_2\$$ below

Could a voltage divider (with more than two resistors) be used here? So that, for example:

$$V_2 = \frac{2\ \mathrm{k\Omega}}{2\ \mathrm{k\Omega}+1\ \mathrm{k\Omega}+2\ \mathrm{k\Omega}}\cdot 4\ \mathrm{V} = 1.6\ \mathrm{V}$$

• Well, the current in the three resistors would be $I=\frac{4\:\text{V}}{2\:\text{k}\Omega\,+\,1\:\text{k}\Omega\,+\,2\:\text{k}\Omega}$, assuming that the opamps don't draw anything away from the series branch. So wouldn't $V_1$ be what you say $V_2$ is? Commented Jan 22 at 20:11

You can work this backwards to design a divider, say you have a 12 V source and you want 5 V, so you think 'I want 5 parts of 12', then you can just make the total resistance some multiple of 12, say 12$$\\times\$$100 for 1200$$\\Omega\$$, and you know the bottom resistor must be 5 parts of that so 5$$\\times\$$100 = 500$$\\Omega\$$, then just subtract that from the total to get 700$$\\Omega\$$ for the top resistor.