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I am taking my first steps into learning about electronics as a hobbyist and I am struggling to understand something about resistance. I have read about Ohm's law and it seems fairly straightforward so I set up a simple circuit on http://123d.circuits.io.

This circuit consists of a 9V battery and a single resistor (10 Ohms) with its leads connect directly to the 9V battery.


simulate this circuit – Schematic created using CircuitLab

When I place the probes of their simulated voltmeter across the leads of the resistor, it reads 7.83V. I can't make any permutation of Ohm's law produce that result. What would be the expected voltage across the resistor? Based on other posts on this site, I would expect the voltage to be 9V. I don't know how that simulator (123D) calculates its values or how it simulates the circuit.

I am sure that I am missing something fundamental but so far I haven't figured out what.

Is the 123D circuit simulator right? Should there be only 7.83V across that resistor?

I set up a similar circuit (using CircuitLab) and it seemed to show that the voltage would be 9V. So I am confused. Is this just an inaccuracy of the 123D's simulator?

Can someone help me to clear the fog?

share|improve this question
A real 9V battery has some internal resistance and may not be exactly 9V when open. Maybe the model is more like a 9V battery than an ideal 9V source. What happens to the voltage when you make the resistor 500K? – Spehro Pefhany Jul 24 '14 at 19:53
If I change the resistor in the simulator to 500K, their simulated meter then reads 9V. – Chris Dunaway Jul 24 '14 at 19:55
It's really 8.999 volts, rounded up to 9V. The internal resistance is still there, but it becomes "negligible" with such a high load resistor. – gbarry Jul 24 '14 at 21:37
That was really a great question, because you find out about ohm's law, ideal versus real models, battery internal resistance, a way to measure it, circuit analysis, and how to calculate it all. – gbarry Jul 24 '14 at 21:43
A battery's resistance is due to its materials, which are partially liquid or gel. The resistance not only is going to change with temperature, but also with the amount of reactants in the battery vs. end products. In other words, as the battery is used its resistance changes. The resistance of your body is another good example: you are not a simple resistor. Your skin has resistance but after burning through, your internal organs have lower resistance. youtube.com/watch?v=8xONZcBJh5A – Dov Jul 24 '14 at 23:19
up vote 3 down vote accepted

They are simulating a (more or less) real 9V battery. They've modeled the battery as an ideal voltage source of 9V with a series resistance of ~1.5 ohms.


simulate this circuit – Schematic created using CircuitLab

If you work this out, the current is 9V/(11.5 ohms) = 0.783A, so the voltage across the 10 ohm resistor must be 7.83V.

share|improve this answer
Do all (real) 9V batteries have the same internal resistance? Is there a real resistor inside the battery or does it just depend on how the battery was made? – Chris Dunaway Jul 24 '14 at 20:08
There is not a real resistor in the battery, it is just used to model the inherent less-than-ideal properties an actual battery has. An ideal voltage source can easily source a hundred gazillion amps without a sweat, a 9v can't. – whatsisname Jul 24 '14 at 20:27
Chris- it's not a real resistor, just a side effect of the way the battery works. The real internal reactance varies with the type of battery, the temperature, current and especially the state of charge - as the battery discharges, the internal resistance rises. – Spehro Pefhany Jul 24 '14 at 20:31

The 7.83 volts tells you precisely what the internal series resistance of the battery is. Open circuit it is 9 volts but under load it drops to 7.83 volts - the current thru the 10 ohm is clearly 783 mA. This current also flows thru the internal resistance of the battery to lose 9 minus 7.83 volts (1.18 volts).

1.18 volts lost at 783 mA means the internal resistance is 1.507 ohms.

All the above is about ohms law and applying it.

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Chris, and welcome to the world of real electronics. What you are running up against is called output impedance. All circuits (except superconductors) have it. In the case of a battery, you should not look at it as purely a source of 9 volts. Instead, it is a 9-volt source (that varies, among other things, with temperature) which has a resistor in series with its output. In the case of your circuit, you can think of the resistor as having a value of 1.49 ohms. Try it in your simulation and see what you get.

With that established, think about what would happen if it weren't true. You could, for instance, weld metal with a 9 volt battery - not for long, but you could definitely draw a bead. Does this seem reasonable? It's also true that some batteries, like car batteries, have very low output impedances, and you CAN weld stuff with them. It's bad for them and shortens their life, but you can do it. Also, don't get too cocky about knowing the value of the output impedance in other applications. For batteries in particular, the effective output impedance varies with current.

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Thanks for the comment. Is there a way to measure the internal resistance of any battery? I'm guessing if I connected my meter across the terminals of the battery in resistance measuring mode that it would probably damage the meter? – Chris Dunaway Jul 24 '14 at 20:17
Superconductors do have impedance. What they don't have is resistance. Unless you can build a superconducting circuit of zero size, there will be inductance, a kind of impedance. – Phil Frost Jul 24 '14 at 20:53
@ChrisDunaway - Your fears are exactly right. Do not attempt to just hook up a multimeter on "Ohms". You probably won't damage it (these things are pretty well idiot-proofed), but you won't get useful results. You determine the output resistance in two steps. 1 - you read the open-circuit voltage. 2 - you put a known resistance as a load and measure the output voltage. Knowing the load resistance and voltage you calculate current. Knowing the difference between the open-circuit and loaded voltage, and the current, you calculate R = E / I, and Bob's your uncle. – WhatRoughBeast Jul 24 '14 at 20:58

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