# Why does voltage drop across resistance contradict Ohm's law?

Very basic question here.

I have been learning a ton about Ohm's law, and I don’t seem to ever be able to completely understand it.

One of which I am getting particularly confused about is voltage drop and Ohm's law.

According to what I know about Ohm's law, if you increase the resistance, the current decreases, but the voltage stays the same. See below figures:

From: learn.sparkfun.com

Example:

Here is a graph of the voltages of my 5V battery with resistors in the circuit:

simulate this circuit – Schematic created using CircuitLab

The questions I have are:

• What are the physics behind this?
• Is there another law that covers this?
• Why does this contradict Ohm's law, or is there something that makes this inline with Ohm's law?
• "However, In reality, I know there would be a voltage decrease if I “increased” the Resistance" No, this is wrong. What you "know in reality" isn't reality and that is the problem, not Ohm's law. Unless you explain why you think this, we cannot pick it apart. "if you increase the resistance, the current decreases, but the voltage stays the same" It depends whether you are holding voltage constant, or current constant as you change resistance. If voltage is held constant, increasing resistance decreases current. If current is held constant, increasing resistance increases voltage. – DKNguyen Aug 10 at 15:23
• in your water diagram you have to raise the water level on the right side if you want the same flow out of the two pipes ... electricity is the same way – jsotola Aug 10 at 15:40
• This is exactly why I hate the water analogy. Anyhow, as mentioned in an answer below, the internal impedance of the battery is why you are getting the odd readings you're getting. Your model is too simple. – Kyle B Aug 10 at 18:55
• He is not measuring the voltage in parallel but building a voltage divider with his multimeter isn't he? – Lars Hankeln Aug 10 at 20:37
• Connect the meter across the resistor and measure the voltage. Voltmeters go in parallel, otherwise they become a part of the circuit and you get the voltage potential across the high impedance of the meter. You are measuring a series circuit formed from resistor AND meter. It is an illusion. – StainlessSteelRat Aug 10 at 21:33

Firstly, Ohm's law only states that current through a metallic conductor is directly, proportional to the potential difference across it. There are several cases like semiconductors, electrolyte solutions, gas mediums where ohm's law does not apply.

According to what I know about ohms law, if you increase the resistance, the current decreases, but the voltage stays the same

Yes, if the driving source is an ideal voltage source, the voltage across the resistance will stay the same no matter the magnitude of the resistance. But, if it is an ideal current source,the the voltage WILL change according to the resistance connected across its terminals, but the current will stay as a constant. Both the scenarios satisfy Ohm's law.

However, In reality, I know there would be a voltage decrease if I “increased” the Resistance.

This case, I assume you are talking about a real voltage source, for e.g, a dry cell. And by increasing the "resistance", I can only assume you are talking about increasing the load since in real life increasing the resistance will not decrease the voltage output of a real life voltage source.
Please not that any and all voltage sources in the real world has some internal resistance. See figure below,

Here when current flows in this circuit, due to ohm's some voltage has to be dropped across the internal resistance r, causing the output voltage, i.e, the voltage available at cell terminals across resistance R, to drop or increase as R is decreased or increased.

I hope this clarifies your doubt.

EDIT: Please note that the circuit you provided is an improper method for measuring voltage across an element. Here you are not measuring the voltage across the resistance but you are measuring the voltage a across the cell terminals with the series resistance appearing as the internal resistance of the cell. So, it is wrong to apply Ohm's law the way you stated in this scenario. Remember voltmeter is connected in parallel while ammeter in series. Here in this scenario the internal resistance of the cell increases and it act like a dead cell, with diminishing voltage. The reason for this is that every analog voltmeter has an internal series resistor which can hamper the reading if the internal impedance of the source too high. An analog voltmeter needs a minimum current through it for the pointer to move as it employs electromagnetic effects. If the internal impedance of the source is too high this minimum current will not flow and the meter will show less than it should. For example a voltmeter might need 100microamps for full deflection, if the internal resistance of source limits this current to 95 microamps from the same source, the meter will show a less small value.

In the case of digital voltmeters, there too exists a potential divider network plus the input impedance of the active device(s), which will also give low value readings it the driving source have too high impedance.

• Given the new Information in the question it should be clear, that the internal resistance of the battery is not the cause of the observed behavior. – Lars Hankeln Aug 10 at 20:49
• Ohm's Law can sometimes even fail for metallic conductors, i.e., tungsten filaments in incandescent bulbs where the resistance increases due to heating. – Michael Seifert Aug 11 at 14:28
• Yes Ohm's law fails in case of metallic conductors for eg, superconductivity. But in case of tungsten, its not a failure per se. Ohm's law still holds, only the resistivity of the material change due to self heating. – ASWIN VENU Aug 11 at 14:39
• "Firstly, Ohm's law only states that current through a metallic conductor is inversely proportional to the potential difference across it." By "inversely" did you mean "directly"? Also, I feel like "metallic conductor" only makes it more confusing? Is a resistor a conductor or an insulator? – user541686 Aug 12 at 9:49
• @ASWINVENU - Hi, As required by this site rule, when you include something in an answer (e.g. photo, image or text) which isn't your own original work, you need to properly reference (cite) it. Those images seem to have come from other places, so in order to comply with that rule, can you edit your answer and add a link back to the original source web page for each image, please? Thanks :-) – SamGibson Aug 12 at 9:55

The problem with your test setup is that you are not measuring the voltage across the resistor you are swapping out. You are measuring the drop across the multimeter's internal resistance. You have a finite multimeter resistance. I'm guessing its about 1MΩ. If you compute for the resistive divider of 100kΩ and 1MΩ, you get 4.54V, roughly what you measured.

If you want to measure current, you need to either use an ammeter or a small resistor's voltage drop.

• This is the first and main issue. It's nothing much to do with internal resistance (for once). To measure the voltage drop across the resistor (which is what OP's water analogue calls for) you would need to put the voltmeter in parallel with the resistor (if you like measuring the "water pressure" difference between the top of the thin tube and the bottom). What's that going to tell you? That the voltage is always ~5V. Why doesn't the voltage go up with R goes up? That would be because the current is reducing (not the voltage increasing) with the higher resistance. – abligh Aug 11 at 3:31
• This should be the top answer. The experimental setup is all wrong and I don't understand why none of the above answers have pointed this out... – jeffpkamp Aug 11 at 3:41
• @jeffpkamp most of the other answers were posted at a time when the asker was refusing to answer questions about exactly what experiment they were conducting, hence the mis-design of the experiment was not evident. – Chris Stratton Aug 11 at 3:57
• @jeffpkamp, the test setup was later added by op. – ASWIN VENU Aug 11 at 5:29

However, In reality, I know there would be a voltage decrease if I “increased” the Resistance.

Nope.

In reality, there's a specific voltage to current relation on a device that obeys Ohm's law: $$\E = I \cdot R\$$. $$\E\$$ in this case means voltage; we keep the old notation (E for Electromotive force) because it makes sense in a screwy and subtle way. But that's it -- Ohm's law only states the relation between voltage, current, and resistance, and only in a device that obeys Ohms law (i.e., a resistor's behavior is a very close match to Ohm's law; a diode's behavior isn't, nor is a spark gap's or a neon lamp's).

So if you hold the voltage constant the current is determined by the voltage and the resistance. If you hold the current constant the voltage is determined by the current and the resistance.

If neither the voltage nor the current is constant, then both are determined by the actions of the resistance and of whatever the driving device is (i.e., if you made a constant-power source, where you set some power $$\P\$$ and -- within limits -- $$\i \cdot v = P\$$, then you could combine this with $$\v = i \cdot R\$$. Then for any $$\P\$$ and $$\R\$$ you could solve that system of two equations to find $$\i\$$ and $$\v\$$.

In the case of your measurement, you are assuming that your multimeter (or oscilloscope?) has an infinite input impedance. Real measuring devices have real input impedances, and in general the higher the impedance (for otherwise equivalent performance) the more the instrument costs.

In your case, you are seeing a 0.45V drop across a 100k$$\\Omega\$$ resistor, which indicates that your meter has an input impedance of 1M$$\\Omega\$$ or so.

• So if there really is no voltage drop, is there an explanation for why it appears to be so? @TimWescott – X Builder Aug 10 at 15:54
• @XBuilder, where are you observing this voltage drop? – jsotola Aug 10 at 17:17
• I wrote it in the above question. But, reading these answers explains that it is indeed the inertial resistance of my battery that is the cause. – X Builder Aug 10 at 18:43

According to what I know about ohms law, if you increase the resistance, the current decreases, but the voltage stays the same.

The voltage doesn't stay the same because of Ohm's law.

It stays the same because you connected a constant voltage source across the resistor, and the rule for a constant voltage source is "the voltage is always the same".

In your picture, the "voltage" stays the same because you kept the height of the water column the same. Again, this is not related to Ohm's law.

According to my observations, I see a voltage decrease if the Resistance is “increased”

This is an unexpected result. With most real world voltage sources, you should have seen the voltage across a resistor increase when you changed the resistor's value. This is because real voltage sources are not ideal voltage sources, as I explained in my answer to your previous question.

What are the physics behind this? Is there another law that covers this? Why does this contradict ohms law, or is there something that makes this inline with ohms law?

You are probably not measuring what you think you are. There are no perfect voltmeters; all have some resistance. It might be 1MΩ, 10MΩ, or even something that varies with the voltage range selected.

simulate this circuit – Schematic created using CircuitLab

So you're actually connecting two resistors in series across your battery, call them R1 and R2, and measuring the voltage across one of them - in this case R2.

Look up "voltage divider" for what is happening.

From the voltage drop when you make R1 100K, your meter is approximately 1MΩ.

According to what I know about ohms law, if you increase the resistance, the current decreases, but the voltage stays the same.

That isn't what Ohms law says--Ohm's law is a relationship between variables for current, voltage and resistance.

1. If you increase the resistance and keep the voltage the same you will see a decrease in current.

2. If you increase the resistance and keep the current the same you will see an increase in voltage.

Otherwise, if you increase the resistance and neither current or voltage are kept constant you will simply see the V=IR relationship maintained through whatever is naturally varying within your setup. Any variances you see between the theoretical Voltage, Current and Resistance values and your measured values are explained by real internal resistances, power supply limitations, etc.

The higher the resistance, the more charge flow= current is restricted.

simulate this circuit – Schematic created using CircuitLab

A logical schematic for any language.

Simple answer... Set your meter to measure current, not voltage. Then try the same test again. You will see that higher value resistors give less current flow. That's Ohm's law!

Actually getting the difference between voltage and current can be difficult at first if you're self-taught, it took a while to click in my head. But once it does, Ohm's law seems like it's obvious!

Also, without intending to be patronising, stay the hell away from mains power! Even experienced electronics nerds have managed to literally explode their multimeter. The guy was trying to measure how much current a socket could supply, so he put his meter on "current" and connected it straight across the mains!

The "current" setting on a multimeter essentially measures across a short-circuit, a simple metal bar built into the meter. Connecting a metal bar, with very low resistance, across hundreds of volts, meant enough current to explode the vulnerable parts inside the meter.

Anyway, he lived to tell the tale on a forum online. Once he'd reset the circuit breaker and restored power to the house.

Your next experiment might be putting two resistors in series, and measuring the voltage across each of them. Connect two resistors in series, and their other ends to the terminals of a battery. This time, set meter to voltage, and put the probes on either side of a single resistor. Note the voltage across each one. Then do the same with different values of resistor.