# Voltage divider explanation

I am self studying linear electrical circuits and I have a problem using the voltage divider. To my understanding if I have voltage source and two or more resistors connected in series I can use the voltage divider to calculate the voltage drop across the resistor that is in my interest. So for example I have the below circuit:

the voltage drop across the resistor R1, using the voltage divider is: $$v_{1} = \frac{R_1}{R_1+R_2} \times V$$

and similarly the voltage drop across the resistor R2, using the voltage divider is: $$v_{2} = \frac{R_2}{R_1+R_2} \times V$$ Makes sense to me.

Now I came across one problem regarding the Thevenin equivalent in the following circuit with the use of superposition theorem:

For the R_th 5||20 is in series with 3||6.

Rth = 6 Ω. Done that.

For the V_th now I keep each voltage source when the other are short circuits and at the end I add up all the voltages according to superposition theorem.

I start with the 50 V source to act alone in the circuit:

The green region does not play any role here because voltage does not flow that way.So my interest is on A with the positive polarity.

And here is where i want an explanation for the voltage divider. It takes $$V_{50} = \frac{20}{20+5}\times 50=40V$$. Wait a second.

R1 and R2 are in series. Therefore I can apply the voltage divider. Now I want to calculate what is the drop across R1=5 Ω so I have to take $$V_{50} = \frac{5}{20+5}\times 50$$. But instead the solution says that you have to take $$V_{50} = \frac{20}{20+5}\times 50$$. Why? Why It takes the opposite resistor drop? What I am missing here?

And continues with the 100 V source to act alone and all the other voltage sources as short circuits with:

And again it takes the opposite : $$V_{100}=\frac{5}{20+5} \times 100 = 20V$$

Why?

I don't care about the Thevenin equivalent here. Just the voltage divider issue. I think I am missing something basic here. Can anyone please explain to me?

To calculate with superposition you want the voltage at point A relative to point B.

In the part of the calculation in question, the sources are shorted on the right so the potential at B is the same as the potential at the (-) terminal of the 50V and 100V supplies.

You want the voltage at point A relative to that point.. so you have this:

simulate this circuit – Schematic created using CircuitLab

Which is the same as this:

simulate this circuit

Of course you could also calculate the voltage across the 5Ω resistor and subtract it from 50V.

• So it is because of the junction that is between the resistors? In my very first picture It calculates the voltage divider v1 on the junction A?That is why it take the first R_1 divided by the sum? Commented May 18, 2023 at 17:26
• Remember that voltage is always relative to some point. Your calculation is correct, it's just referred to the wrong point in this case (and the sign is reversed). Commented May 18, 2023 at 17:39
• In which picture and what sign are you referring to?Thank you for your clear answer I appreciate it a lot Commented May 18, 2023 at 17:42
• In your third image there is a -50V source applied to a voltage divider (picking the + terminal of the 50V supply as the reference), so there's -10V at A relative to the (+) terminal of the 50V source. That means 50V - 10V = +40V relative to the (-) terminal (which is at the same potential as B. Commented May 18, 2023 at 17:54
• Clear.Thank you very much for your effort and your answer. Commented May 18, 2023 at 17:58

First, I'd like to welcome you to the site and thank you for putting effort into writing a detailed question.

Only voltage differences are physically meaningful, so whenever you're talking about voltage in a circuit, you have to pick a node to be zero volts ("ground"). You might think that B is the obvious choice, but in this case, a better choice is the bottom node that connects the four voltage supplies:

simulate this circuit – Schematic created using CircuitLab

Your Thevenin voltage is then $$\V_A - V_B\$$. No current flows in the right half of the circuit (voltage doesn't "flow"), so there's no voltage across either resistor. That means that $$\V_B = 0 \mathrm V\$$. $$\V_A\$$ is equal to the voltage across the 20-ohm resistor, which you get with the voltage divider equation you mentioned:

$$V_A = 50\mathrm V \frac {20\Omega}{5\Omega + 20\Omega} = 40 \mathrm V$$

which means that the 50V supply's contribution to the Thevenin voltage is:

$$V_A - V_B = 40\mathrm V - 0 \mathrm V = 40 \mathrm V$$

It works the same way for the other three supplies. Note that if choose the bottom node as ground the supplies on the right side will make a negative contribution to the Thevenin voltage.

EDIT: You asked what would be different if you chose the node between the 20-ohm resistor and the 100V source as ground. For the first part (with only the 50V source active) nothing would change -- deactivating the 100V source shorts the bottom node to ground, giving you the same voltage divider as above.

For the second part (with only the 100V source active) you have a very different-looking situation:

simulate this circuit

There's a 100V source between the bottom node and ground, so the bottom node is now at -100V. There's still no current through the right half of the circuit, so $$\V_B\$$ is also -100V. On the left side, because you moved the ground $$\V_A\$$ is now equal to the voltage across the 20-ohm resistor, which (starting from ground) is -80V. But despite all the negative numbers, the Thevenin voltage is positive:

$$V_A - V_B = -80\mathrm V - -100\mathrm V = 20 \mathrm V$$

which it should be given the way the 100V source is pointing.

Likewise, if you choose node A as ground, you will find that $$\V_B\$$ becomes -20V, giving the same result. If you choose the bottom node as ground, $$\V_A\$$ becomes 20V and $$\V_B\$$ becomes 0V. So the answer is always the same. Which node you choose for ground is arbitrary, so choose the one that makes the problem easiest to solve!

(You could even pick a different ground node for each source in the superposition, although I wouldn't recommend that.)

• Thank you for your kind welcome.I appreciate it.Question:If I had take as a ground the node- junction below the 20 ohm resistor what would have changed ? Commented May 18, 2023 at 17:31
• @HomerJaySimpson I updated my answer. Commented May 18, 2023 at 17:52
• Clear.Thank you very much for your effort and your answer. Commented May 18, 2023 at 17:59

To clarify, lower resistance will have less voltage over it, and higher resistance has more voltage over it.

The formula is derived by combining two formulas together.

That is why you calculate it as having some total voltage over sum of both resistors, and from there you get the current which is equal for both resistors.

And then you use that current to calculate the voltage over the resistance you want.

By combining those, you don't have the current in the formula any more.

So with the 100V example, you want the output voltage referenced to ground, so the voltage over the 5 ohm that is on ground side. Not the 20 ohm one. And due to what I said earlier, there is 5/25 of the voltage over 5 ohms and 20/25 over 20 ohms.

Well you can't change and calculate a voltage divider like that.

So you change the circuit with two supplies by removing a supply, you get a new resuly for the new circuit with only one supply, not the original with two supplies.

If one end of the divider is at 50V and the other end of the divider is at 100V, you will know that for all possible resistances, there must be a voltage that's between 50V and 100V at the divider junction.

If you decide to remove a supply (e.g. the 100V supply) it means replacing it with 0V supply, so obviouslu the current through the resistors is different and maybe even direction of current, and then the junction will have different current and different direction of current, and you know the voltage must be between 0V and 50V as the 100V was removed.

What you can do is think in relative voltages. There's simply 50 volts over the resistors, as one is referenced by 50V to ground.

• He's using superposition. It's perfectly legitimate to zero out the 100V supply as part of that process. Commented May 18, 2023 at 17:10
• So it is because the junction is between the two resistors?And you take the last one in series connection?For example when I have only the 100v voltage source active and the junction was (imagine that) between the voltage source and the resistor of 20 ohms how the voltage divider would be calculated ? Commented May 18, 2023 at 17:22
• Clear.Thank you very much for your effort and your answer. Commented May 18, 2023 at 17:59