# What does voltage drop mean in a circuit? [duplicate]

Let's calculate the voltage drop per resistor:

total R = 1000 + 680 = 1680 Ω.

circuit I = 5/1680 = 2.97 mA.

v1 = 0.00297 × 1000 = 2.97 V = voltage drop across R1

v2 = 0.00297 × 680 = 2.01 V = voltage drop across R2

Here are my questions:

1- Across R1, is voltage dropped BY 2.97 V or TO 2.97 V (I'm guessing it's the former)?

2- If I connect a multi-meter in parallel to the poles of R1 I get the same reading, so the multi-meter is also showing the voltage difference (drop) between the two poles which is essentially the drop from the original source.

3- I'm trying to drop the voltage from 5 V to 3 V to be used by a 3 V device. The device specifies it needs 150 mA to work. How can I use a single resistor to drop the voltage to 3 V?

I think I can replace R2 in my diagram with my device being R2= V/I = 3/0.15 = 20 Ω, correct? R1 then is V1(drop)/0.15 = 2/0.15 = 13.3 Ω. So If R1 becomes a 13.3 Ω resistor my device then will get 3 V and 150 mA?

• your calculation is incorrect .... 5/1680 = 0.00298 Commented Nov 21, 2022 at 0:53

The voltage across R1 is indeed 2.97V, as you have correctly calculated. What this means is that if we declare the voltage at one end of the resistor to be $$\V_X\$$, then the voltage at the other end must be 2.97V higher or lower, either $$\V_X+2.97\$$ or $$\V_X-2.97\$$ depending on the direction of current through R1. In other words the voltage across the resistor, otherwise called the "potential difference", is 2.97V.
Remember that, for a resistor (not all components have this property), the node of a two-terminal element where current enters the element (left side of R1 in this case) always has the higher voltage.

Note: I prefer the term "potential" over "voltage" when referring to the voltage at some point, and generally I use the term voltage in the context of potential difference between two points in a circuit. This is purely a matter of personal preference, and the terms "potential", "voltage" and "E.M.F." are interchangeable.

The term "drop" is synonymous with "potential difference". This means the voltage "dropped" by R1 is also 2.97V. The term "drop" implies a reduction, and you would really only use it in the context where the presence of the element contributes to a reduction, or "loss" of potential, as is the case here, where you go from +5V potential on the left side of R1 down to $$\+5V + (-2.97V) = +2.03V\$$ on the right. However, when describing a journey anti-clockwise around the loop, going from the right side of R1 to the left, there's nothing technically wrong with saying something like "R1 drops 2.97V, so the left side of R1 is 2.97V higher in potential than its right side". In any case, the "drop" always refers to the voltage across an element, and only tells you the difference between absolute potentials at either side.

The term "absolute potential" is often stated simply as the "potential at/of" some point, and is always quoted relative to some mutually-agreed-upon "ground" or "zero-potential node". There's no such zero potential in real life, but to simplify the algebra and more easily understand a circuit's operation, engineers always tell us which node is to be considered to have this theoretical "zero volts potential" using a ground symbol, shown at bottom left here:

simulate this circuit – Schematic created using CircuitLab

There are four voltmeters in that circuit that all show a potential relative to this designated "ground" node. They are VMA, VMB, VMC and VM1. All those meters have one side "grounded", which means they each show the absolute potential at some point, the potential relative to our arbitrarily chosen zero volt node.

VMA is telling you that the potential at node A is +12V, or $$\V_A=+12V\$$. VMB is telling you that the potential at node B is +10V, $$\V_B=+10V\$$, and so on.

There are two voltmeters, VM2 and VM3 which are not grounded on either side. These are connected across resistors R2 and R3, and therefore they are telling you the voltage drop, or voltage across those individual resistors, and they do not show any absolute potential.

Every voltmeter in this circuit is displaying a potential difference, the difference in potential between the two nodes they are connected to. VM1, for example, is measuring the potential difference between nodes G and C. VMB is showing the potential difference between nodes G and B.

You will notice that VMB displays 10V, which is the sum of voltages across R1 and R2. Similarly, VMA is showing +12V, which is the sum of all the potential differences across all the resistors, and which is obviously true, since we have explicitly imposed that potential difference using a 12V source.

In the context of voltage "drops", it should now be clear that the 2V "drop" across R3 is causing potential to fall from +12V (at node A) to +10V (at node B). Another term commonly used is "develop", as in "R2 develops 4V, so that the potential at C is 4V lower than the potential at B".

It's important to note that these concepts are at the heart of Kirchhoff's Voltage Law (KVL), which encapsulates this idea of voltages "stacking up" like this. KVL applied to this circuit would say "the sum of voltage drops around this loop is zero". In other words, if you start at any node and add up the changes in potential (some increases which are positive, some decreases which are negative) as you go from node to node around the loop, when you get back to where you started, you must return to the same potential you had at the start.

Another KEY understanding to be gleaned from my example circuit is that each resistor drops a share of the 12V source in proportion to the resistances in the chain. This is represented algebraically as follows:

$$V_{AB} : V_{BC} : V_{CG} = R_3 : R_2 : R_1$$

In other words, if you doubled all the resistances, the absolute potentials and potential differences would not change. This implies that this circuit has no dependence on current. As long as the ratio of resistances stays the same, the potentials all stay the same.

In relation to your third question, the main phrase in that last paragraph is "as long as the resistance ratios remain the same", which paraphrased says that if you can guarantee that the effective resistance of any device (a "load") in that chain remains the same, then you can be sure that the voltage across that load will also stay the same.

I very much doubt you can make that guarantee about any load you intend to insert in place of R2 in your own circuit:

simulate this circuit

While it's true that you could use a single resistor to provide a "voltage drop" from +5V down to +2V, the voltage across the load (R2) will only be 2V as long as any variation in the load's effective resistance is matched by a proportional change in the resistance of R1, which obviously doesn't happen; R1 is fixed at 1kΩ.

Due to R1 being fixed, any change in the effective resistance of load R2 must necessarily cause a change in the voltage across that load. In practice, any arbitrary load in place of R2 will likely not have a guaranteed, fixed resistance, and in general, a single resistance R1 will not provide a fixed voltage drop from 5V to $$\V_F\$$.

There are components to do what you require, even when the load's effective resistance is changing all the time. They are called "linear voltage regulators", and their purpose is to replace R2 in your circuit. They function by constantly changing their equivalent resistance to maintain a fixed, known potential at their output, the voltage across load R1. The regulator is U1 in these next circuits:

simulate this circuit

U1 in each case is responsible for varying the effective resistance between its "IN" and "OUT" terminals, to maintain "OUT" at the required potential, shown on VM1. In the second circuit (top right), you see that even though I changed R1, $$\V_{OUT}\$$ did not change from 2.03V. Also, in the third circuit (bottom), output $$\V_{OUT}=2.03V\$$ in spite of having doubled battery voltage, hence the name "voltage regulator".

The third terminal "GND" is necessary for the regulator to "know" what 0V is, so that it can regulate its output potential to be exactly 2.03V above this zero-reference at all times.

Other than saying there exist different regulators for different output potentials, and even adjustable ones to produce any arbitrary potential you desire, I won't describe them any further. That's for your own study, but I do hope this clears things up for you.

• Nice story... I have read it with great pleasure... There is only one problem with the last part - the OP does not know what is in the 3-tetminal "box" U1 to imagine how it does it... Commented Nov 22, 2022 at 14:13

All voltage measurements are relative. Saying: "There is 5V here" only makes sense in relation to a 0V reference. I added a few labels to your picture.

If we set C to be the common ground (0V per convention), then there is +5V at A and +2.02V at B.

If B is selected to be 0V, then there is +2.98V at A and -2.02V at C.

And if we select A to be 0V, then there is -2.98V at B and -5V at C.

But if you look closely, then no matter what convention we use, the voltage difference (the voltage drop) across the resistor R1 is 2.98V.

• With regard to "every voltage is relative", I like to view the concept of "voltage" somewhat like "altitude relative to the Earth's center of mass". While it might be possible to measure the stature of a person by measuring the altitude at the top of the head, and the altitude at the bottom of the feet, and subtracting, other means of measuring stature are simultaneously easier and more accurate. Absolute voltage at a point can be defined as the change of energy state produced by moving an electron infinitely far away, but it's usually easier to measure relative voltages. Commented Nov 21, 2022 at 20:48

Your Ohm's law calculations seem to be right, but dropping from 5 to 3 V using a resistor is asking for trouble. The problem is that the load is unlikely to draw a constant 150 mA. So you could well kill the load device with overvoltage.

The best solution is to use a 5 to 3 V buck converter. These are cheap from electronics supply houses.

• They might have one that's set to 3V, but you can also look for adjustable ones. There will most likely be a small screw that changes the output voltage. Put your multimeter on the output and turn the screw until it's the right voltage. Commented Nov 21, 2022 at 13:08
1. Ohm's law says there is 2.97V over the resistor. You calculated it yourself.

2. Connecting multimeter over R1 gives you the voltage over R1.

3. If it is a constant load of 3V 150mA then you can. Otherwise if load current is not constant then you can't and it is a wrong idea.

• But you didn't answer the question, 2.97V over the resistor, what does this mean? voltage is dropping by this amount or it is becoming this amount.
– Dan
Commented Nov 21, 2022 at 0:49
• @Dan If it becomes that amount, you have calculated that a same node has 2.97V and 2.01V simultaneously, which is a bit paradoxal. Of course 2.01V and 2.97V has to add up to 5V, considering the sum does not precisely add up due to limited precision in the calculations. Commented Nov 21, 2022 at 0:56
• Quote Dan: "voltage is dropping by this amount or it is becoming this amount ?". For answering this question, one must ask before: WHICH VOLTAGE is "dropping"? Two answers are possible. (1) As seen from the consumer (R2) it is the DC battery voltage which droppes BY 2.97V. (2) As another formulation, we can say that there is a voltage drop across R2 with an amount of 2.97 volts. I think, historically, the first interpretation is the correct one. However, over time, the second form has also become accepted.
– LvW
Commented Nov 21, 2022 at 17:23
• @Dan, also remember that voltage is always a differential, so if you say the voltage "becomes" this amount, the question is "when compared to which other point". Now, in a circuit like yours, one would usually draw a ground symbol on lower node, and call it the zero reference point, but your diagram doesn't have it, and in principle the choice is arbitrary. Not that choosing any of the nodes there as a zero would give a voltage of 2.97 V for the node between the resistors, but one of 2.01 V, 0 V or -2.97 V. Commented Nov 21, 2022 at 19:52
• @Dan "Voltage across" or "over" a component means "the difference between the voltages at either end". Commented Nov 22, 2022 at 11:03

You can get a good intuitive idea of ​​what a "voltage drop" is from Ohm's famous experiment in 1826 where he studied the voltage distribution along a resistive wire. Also, fluid and other analogies that allow us to transfer our ideas from life to abstract electrical phenomena, are very useful. See, for example, the Wikibooks story about the voltage drop phenomenon that my students and I created in 2008 where we investigated voltage drops across various resistive materials.

In 2013, I dedicated a special question in ResearchGate to "voltage drop" compared with "voltage"; I hope it will be interesting for you.

How can I use a single resistor to drop the voltage to 3 V?

You can... but a more sophisticated trick is to use a "non-linear voltage-stabilizing resistor", e.g. a diode with 2 V forward voltage. And since there is no such diode, you can make one by connecting three 0.7 V silicon diodes in series. I hope you understand the benefits of this solution.

But it also has a drawback - the diode carries the changes of the input voltage if it varies. That is why usually the load is supplied through a resistor and a Zener diode is connected in parallel.

You have to distinguish between theoretical considerations about "resistors" and practical implementations.

Ohm's law is mainly theoretical and talks about "resistors" which have a linear relation between voltage and current.

In practice, a resistor is an electronic device which more or less obeys this relation, but which also can change its temperature and thus can change its resistance, i. e. the relation between current and voltage.

Most device don't have a constant resistance - some draw the same current irrespective which voltage you apply (in a certain range), others can switch parts of it on and off, changing the current they draw. These aren't suitable for your simple circuit, a buck converter is what you need there.