# How can I calculate what voltage I have in the middle of 2 series resistors in an electrical circuit?

For example: VCC 5 V - resistor 1.5 kΩ - resistor 860 Ω - GND

• What current flows through the resistors <--- use ohms law. Then use ohm's law to calculate the voltage across the 860 ohm. Commented Aug 18, 2022 at 20:49
• This is basic circuit theory. A few moments googling would provide you with many versions of the answer. Commented Aug 18, 2022 at 21:11
• Downvoted because this is a trivial question with no attempt to show a simple circuit, and the OP is not a new contributor (which might excuse such behavior). Commented Aug 19, 2022 at 2:30

simulate this circuit – Schematic created using CircuitLab

Think of the resistor chain as being like a ladder of height R1 + R2. Now what fraction of the height is the R1 - R2 junction? Answer: $$\frac {R_2} {R_1 + R_2} = \frac {860} { 1500 + 860} = 0.364$$

Since you know your VCC you can now calculate the junction voltage.

Current through both resistors: I = V / R = 5 V / (1.5 kOhm + 860 Ohm) = 2.1 mA

Voltage across 860 Ohm: V = R * I = 860 Ohm * 2.1 mA = 1.82 V

Another common way: https://en.wikipedia.org/wiki/Voltage_divider#Resistive_divider

This requires an understanding that the current flowing through all the resistors in a series connected chain of them is the same in every one. This simplifies things a lot, since when you apply Ohm's law $$\V=I \times R\$$ to each resistor, it's the same $$\I\$$ in every case, and a lot cancels out.

The consequence is that the voltage across each resistor is proportional to the resistance. Below are a couple of illustrations of that. Notice the left one is a chain connected to 0V (ground) at the bottom, and the right one has a non-zero voltage at its base, but in both cases I want you to see how the the total voltage across all the resistors in the chain is divided amongst the resistors in proportion to their individual resistances:

simulate this circuit – Schematic created using CircuitLab

On the left, then, the total voltage across the entire chain is $$\10V - 0V = 10V\$$, and you have to share a part of those 10V across each resistor in the chain. Check for yourself that those voltmeters are reading values that agree with this.

In the right circuit, I've complicated things by having a non-zero voltage at the bottom of the chain, -12V. Now the voltage across the whole chain is the difference between the top and bottom voltages $$\+12V - (-12V) = 24V\$$. The same rule applies though, you have to divide up the 24V total into chunks in proportion to the resistances. Check that those voltmeters are indeed showing values that agree with this idea, and that they all add up to 24V.

All this works even if the voltages at the top and bottom are reversed.

The last part of the puzzle is how to work out the voltage at any particular node. We'll start out at the bottom of the chain on the left, and apply a variant of Kirchhoff's voltage law. At the bottom, the voltage is obviously 0V. When I move upwards to node A at the top of R1, I encountered a rise in potential by the amount shown on VM1, +7V. Therefore, at node A, the voltage is:

$$V_A = 0V + 7V = +7V$$

Simple. Jump upwards across R2 now. This incurs another additional change in potential, a rise of 2V. Now we are at node B, and I can say with confidence that the voltage at that node must be:

$$V_B = 0V + 7V + 2V = +9V$$

On the right, starting at the bottom, at −12V, we cross over R4, which has 12V across it, to node C. Therefore:

$$V_C = (-12V) + (+12V) = 0V$$

Moving upwards to node D we encounter a rise in potential of +4V, so:

$$V_D = V_C + (+4V) = (-12V) + (+12V) + (+4V) = +4V$$

In general, a formula for working out the voltage across any single resistance $$\R_X\$$ of a group of N resistors connected in series, with a total voltage difference from end to end of $$\V_{TOTAL}\$$ is:

$$V_{RX} = V_{TOTAL}\times \frac{R_X}{R_1 + R_2 + R_3 +\dots+ R_N }$$

To work the absolute potential at the top of $$\R_X\$$, you sum the voltages across all the resistors up to and including $$\R_X\$$, and add the bottom potential. That operation simplifies to:

$$V_X = V_{BOTTOM} + V_{TOTAL} \times \frac{R_1 + R_2 + \dots\ + R_X}{R_1 + R_2 + R_3 +\dots+ R_N }$$

For your simple 2-resistor divider, this means (please note that your resistors are numbered top-to-bottom, so keep that in mind):

\begin{aligned} V_{CENTER} &= V_{BOTTOM} + \left(V_{TOP} - V_{BOTTOM}\right) \times \frac{R2}{R2 + R1} \\ \\ &= 0 + \left(V_+ - 0\right) \times \frac{860}{860 + 1500} \\ \\ &= V_+ \times 0.364 \\ \\ \end{aligned}