Voltage drop across a single resistor and across two resistors

I have been having difficulty understanding voltage drops across resistors. Now, I know the theory and how to apply Ohm's law. The question is why does the voltage drop across resistors of the same resistance vary from the first circuit to the second circuit? Does it have anything to do with current? Why does it happen? I am trying to find an intuitive explanation as to why it happens.

Thanks!

• Are you familiar with Kirchoff's laws? Jul 20 '20 at 13:12
• What causes the volt drop? Do you understand ohm's law? Jul 20 '20 at 13:14
• From Ohm's Law, which you understand, compute the current in the first circuit. Compute the current in the second circuit. Now, from Ohm's Law, which you understand, given the current in each circuit, compute the voltage across each resistor. Add the results to your question. Now explain which bit you don't understand. Jul 20 '20 at 13:21
• Note that when you use the CircuitLab button on the editor toolbar and "Save and Insert" on the editor an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. Jul 20 '20 at 13:38
• First of all, it is obvious that the voltage drop must be 16 V in both cases because the voltage source provides 16 V. However, it is indeed not obvious that the voltage drop over a single part (e.g. a resistor) is not constant! Think about a Zener diode: Such parts have a (more or less) constant voltage drop. If you connect two of them in series, the voltage drop will be twice as high (as long as current is flowing). Jul 21 '20 at 12:36

Your 2 × 100 Ω resistors are in series so your total circuit resistance is 200 Ω and this will restrict the current to half of the value obtained in the single resistor circuit. simulate this circuit – Schematic created using CircuitLab

Figure 1. An equivalent circuit using a potentiometer.

Here we've replaced the 2 × 100 Ω resistors with a 200 Ω potentiometer with its wiper in mid position. It should be clear that:

• When the wiper is at the bottom of the resistance track the output will be 0 V.
• when the wiper is at the top of the track the output will be 16 V.
• When the wiper is anywhere in between the output voltage will be proportional to the fractional distance from the bottom to the top.

In your example you have equal resistances so the voltage will be 8 V.

• Thanks for the answer, though I am looking for something slightly different. Not sure if it was clear in the question, but, I want to know why the voltage drop in the second circuit across each resistor is only half of the voltage drop across the resistor in the first circuit. And I dont want the answer in regards to Ohm's law or the fact that the voltage drops need to add to 16V. I already am aware of that. I am looking for more of an intuitive explanation than anything. Jul 20 '20 at 13:46
• I guessed that so I didn't mention Mr. Ohm other than in the resistance units. Read my answer again. I think that if you grasp the operation of the potentiometer that clarity will come. Jul 20 '20 at 13:50
• @CauanKazama, Well, the intuitive way to look at it is, if the voltage drop across the one resistor is 16v, and somehow it stayed 16v across each of the two resistors, the total voltage drop would add to 32v, but your supply voltage is 16v. So if you are only supplying 16v, where the heck could 32v come from? Jul 20 '20 at 23:11
• I appreciate that tiny smudge on the image. Made me try and clean my screen. Jul 21 '20 at 10:35
• @orithena: If you're referring to the $\color {green}{t }$ it's a trick I use to force the imgur engine to scale the schematic to a reasonable size. It also, oddly enough, helps me identify my own schematics years later! Jul 21 '20 at 10:44

In the first circuit, you have one (single) voltage source and one (single) resistor. This one (single) resistor is connected directly across the voltage source terminals ( terminals $$\A\$$ and $$\B\$$). Thus, from point $$\B\$$ to $$\\$$A the voltage is equal to the battery terminal voltage $$\V_B\$$ and because our single resistor is also connected directly between these two-point (B and A), The resistor must "see" the same voltage across his terminals as is "given" by the battery. And this is why $$\V_B = V_1\$$. The voltage across the battery equal the voltage across the resistor.

But for the second case, we have a different situation. Again we have one (single) voltage source but this time we have two resistors connected in series. And again the voltage across terminals $$\A\$$ and $$\B\$$ is equal to the battery voltage. But now neither of resistors is connected directly across the battery terminal voltage. So the voltage drop across the resistors will split because our two resistors are connected in series thus in a series circuit, the current that flows through each of the components is the same (only one path for current to flow).

$$\V_B = V_1 + V_2 = IR_1 + IR_2\$$

How can i calculate Vs in this circuit knowing Vo=2?

And some water analogy example of a series circuit. And some water analogy for parallel connection. Notice that this time all resistor will see the same voltage (VB) but the current will split betwen resistors. • If they're not your drawings then you need to credit the author. (This is site policy.) Jul 20 '20 at 13:54
• The first two are mine. But I do not know the Author of a "water analogy" drawings. I found on the web, they probably come from a Polish book used in primary school.
– G36
Jul 20 '20 at 14:16
• @G36 well where did you find them on the web? Jul 21 '20 at 12:51
• @user253751 I found it here elektroda.pl
– G36
Jul 21 '20 at 15:22

here you have to apply voltage divider rule to understand voltage drop distribution. here is some reference link: - https://www.electricalclassroom.com/voltage-division-rule-potential-divider-circuit/

In ur 1st case when a load is only 100ohm, the voltage drop across the Resistor is 16V. but in 2nd case when you have two resistors in series, so total Resistance is R = 200ohm.

Remember one thing that, the current is always constant in a series circuit, and voltage is constant in case of a parallel circuit.

As this is our series circuit current is constant in this case.

so the voltage drop for each resistor is different in that case, according to V = IR, V = 16V and Total R = 200ohm, so I = V/R , I = 0.08A.

so, the voltage across the 100ohm Resistor is, V = IR , I = 0.08A and R = 100ohm V = 8V. so the voltage across the 100ohm resistor is 8V.

• Thanks for the answer! Although not exactly what I was looking for. I have a good understanding of Ohm's law and can calculate the voltage and the current flowing. What I really want is an answer as to why the voltage drop in the second circuit across each resistor is half, even though they have the same resistance as the one in the first circuit. Jul 20 '20 at 13:42
• @Cauan Kazama you've got answers from the ones of the most expert people here and you seem not to get the answer you want..at this point you should consider having the wrong question in your head.. shouldn't you ? Jul 20 '20 at 20:07

It is because there is half the current.

The amount of voltage dropped by a resistance is directly related to how much current is flowing across it. It is a 1 to 1 relationship.

• "It is a 1 to 1 relationship." No, it's an R:1 relationship (but I know you know that). Jul 20 '20 at 14:48
• @Transistor heheh good point! I was trying to avoid anything that sounded like Ohm's law, to appease the needs of the OP. Jul 20 '20 at 14:50

Being sarcastic is not my habit so, even if very good answers have already been posted, I'll try it too.

You seem confused by the fact that in both cases the resistors are the same but not the voltage across them. Mhh..without saying anything about what you do not want to hear (ohm..my god I said it !) R3 is not alone: R4 has its influence. So you cannot think of it like you do and compare it to the circuit were the resistor is alone.

To answer your question precisely: yes it has something to do with current. R4 participates with R3 to lower the current (higher total resistance). R3 (or R4) sees less current and smaller current gives smaller voltage across same resistance (sorry Ohm's law has been invoked here).

I am sure one answer here will bring light to you :)

• An interesting way of presenting the answer ... which does nоt make me yawn out of boredom... even though it is already midnight here:) Jul 20 '20 at 20:56
• I did my best though. Are you sure you are not testing creativity of people to get the more fantastic answer ? May be to find a wonderful way to explain Ohm's law to children ?^^ I start to doubt .. Jul 20 '20 at 21:03

It is simple algebra V=IR or R=V/I or I=V/R.

On the left, current is I=V/R=16/10 = 1.6 amps so V=IR=1.6*10=16 volts (drop)

For both resistor on the right, current (I)=V/R=16/20=.8 For EACH resistor on the right, voltage drop=IR=10*.8=8 volts.

• A beautiful little story about the ubiquitous resistors... But since we are more technicians than mathematicians, let's convert the "algebra" to "physics":) On the left, R acts as a 'voltage-to-current converter'. Both resistors on the right first act as a composed 'voltage-to-current converter'; then each of them acts as a "current-to-voltage converter'. Thus, as a whole, they act as a "voltage-to-voltage converter' (aka 'voltage divider') with two possible outputs. One of them is floating and the other grounded. Usually, we use the latter as an output but in some cases we can use even both. Jul 21 '20 at 6:18

An intuitive way to look at is that all the voltage is dropped across two resistors, and since the resistors are the same, the voltage drop across each will be the same, each taking half. This is called “symmetry”.

I just stumbled across this on a suggested reading list, and read because it seemed odd on my list.

Teaching IT I've kind of developed a feel for when students aren't sure how to ask the question they really want to know. You mentioned "intuition" so I think you're looking for analogies to your own actions.

Rather than an Ohm's Law question, maybe you have a Drift Velocity question, how fast the electrons actually move.

One way to put this is current arises from change in charge quantity per unit time (I = dQ/dt), a bunch of algebra later we can get to number of electrons passing by at drift Velocity (distance = Velocity*time), research "Drift Velocity" for more details.

I'm on a mobile device which affects my ability to type all the math clearly, sorry.

In short with movement of electrons producing current the difference between wire and resistor gives rise to a current and there's twice as much of that difference in your second circuit, then that current value goes into Ohm's Law to give us a voltage drop for each resistor, instead of the conventional voltage drop goes in to give us current.

The voltage drop across a resistor in a circuit is determined by the current flowing through it (product of resistance and current).

The current through the resistor in the first circuit is double that in the second. It's likewise with the voltage drops.

First I will say that the OP's question and all the answers here (including the latest one from a minute ago) are great and I rate them with +1:) I will only supplement them with a few more extravagant but "thought-provoking" considerations...

"The question is why does the voltage drop across resistors of the same resistance vary from the first circuit to the second circuit? Does it have anything to do with current? Why does it happen? I am trying to find an intuitive explanation as to why it happens."

"What I really want is an answer as to why the voltage drop in the second circuit across each resistor is half, even though they have the same resistance as the one in the first circuit."

If you really want the voltage drops across resistors with the same resistance to be the same, I can offer you a solution - just replace the voltage sources with current sources. This is not just a joke but a very real circuit configuration that we can observe in some well-known electronic circuits (e.g., in the so-called "common-emitter stage with emitter degeneration" or "phase splitter").

But let's go back to the OP 1- and 2-resistor circuits powered by voltage sources and draw some interesting conclusions.

The first is that we may not be interested in the current flowing through the resistors and their resistance. In both circuits the voltage does not depend on either the current or the resistance. In the second circuit, the voltage drop across a resistor depends only on the ratio of its resistance to the total resistance.

A second interesting conclusion we can draw with respect to the Transistor's potentiometer. Although this is a variable resistor, when we rotate its wiper, we do not actually change anything - neither the resistance ... nor the current ... nor the voltage. We simply measure (choose) the voltage at one point on its internal resistive layer... but all other points have linearly decreasing voltages.

Of course, we can imagine that when rotating the wiper, the one partial resistance increases when the other decreases so their sum stays constant... and, as a result, the current is constant as well. We can see such "electronic potentiometers" in CMOS stages, current-feedback amplifers (CFA), etc.