# [Beginner]Which direction electricity follows in a breadboard?

Discussion in 'General Electronics Chat' started by John Fraskos, Mar 3, 2015.

Mar 3, 2015
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Hi all

I'm a complete beginner who is waiting his first arduino kit to arrive via mail. Meanwhile, I have started learning and experimenting with how circuits work. I 've no prior experience in electronics whatsoever, so please excuse my probably stupid question and potentially wrong terminology.

So, here is what I am struggling to understand:

I am a bit confused on which direction the electricity flows in a circuit which is powered by a 9V battery. I have created a simulated circuit on a breadboard, to light up a led. You can find my breadboard here:http://123d.circuits.io/circuits/622863-the-unnamed-circuit and run a simulation.

Here is a screenshot as well:

So, from what I understand, the electricity flows from the Battery's positive terminal to the negative. This means that at the top breadboard, the electricity passes from the resistor at first, then from the led, and then goes to the ground. So far so good.

My question arises at the second breadboard, at the bottom. Here the electricity passes first from the led, and then from the resistor, and finally to the ground. So, why does the led lights up normally and does not explode, since the resistor is placed AFTER it and not BEFORE it?

I am missing something quite fundamental here, and so far, the online reading I've done does not seem to help. Any help would be quite appreciated!

Cheers!
John

2. ### tshuck Well-Known Member

Oct 18, 2012
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In the physical world, electrons (which is what makes electricity) move from the negative terminal to the positive. Analyzing circuits with this approach is called 'electron current'.

Due to a wrong assumption during the discovery of electricity, most people view electricity as being comprised of positive charges - using this method to analyze circuits is called 'conventional current'.

Your question, however, seems to be why the current in the circuit is the same. This is because the resistor resists the movement of the electrons and limits the amount of current (the rate at which the electrons move) through the LED, regardless of whether it comes before, or after the LED.

All of the electrons entering the LED go through the resistor as well.

Were the resistor to be omitted, you would quickly have a broken LED.

peppercash likes this.

Mar 3, 2015
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Thanks for this! Really helps man.
I'll do some further reading on this, sounds fascinating.

Cheers!

4. ### MikeML AAC Fanatic!

Oct 2, 2009
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A better way to say it:

the current flows from the Battery's positive terminal to the negative terminal. This means that at the top breadboard, the current passes through the resistor, then the current flows into the anode pin of the led, flows through the LED, out the Cathode pin, and then returns to the Batteries negative terminal (there is no ground in this circuit)

In the lower breadboard circuit, your question is answered by understanding series circuits. There is only one current. It is the same everywhere in the circuit: Out the positive terminal, through the LED, through the resistor, and back to the negative terminal. The order of which comes first in a series circuit doesn't matter; what matters is that the voltage drop across the LED + the voltage drop across the resistor = the battery voltage.

Last edited: Mar 3, 2015
5. ### ErnieM AAC Fanatic!

Apr 24, 2011
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A fair question. Here is a fair answer: the order of the elements of a series circuit does not matter one bit.

A series circuit is where each element (two or more) has two terminals, such as resistors, batteries, LEDs and lots more.

Here with three elements in series there are six different ways to connect them together that all work exactly the same.

Apr 5, 2008
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Hello,

Have a look at the following pdf for the inner connections of the breadboard:

Bertus

ErnieM likes this.
7. ### MikeML AAC Fanatic!

Oct 2, 2009
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Except that current will not flow from Cathode to Anode in an LED, so there only four

8. ### ErnieM AAC Fanatic!

Apr 24, 2011
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1. Battery - LED - Resistor
2. Battery - Resistor - LED
3. LED - Battery - Resistor
4. LED - Resistor - Battery
5. Resistor - Battery - LED
6. Resistor - LED - Battery

No exceptions!

Note the polarity of the LED and Battery must remain the same, as it would any element that has a polarity feature.

Last edited: Mar 3, 2015
9. ### Roderick Young Member

Feb 22, 2015
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Think of the electricity flow as water in a pipe. I can drive a turbine (the LED) from the water flow. But if I put a flow restrictor (that is, a plate with only a tiny hole in it to let water through), the flow of water will be slowed. It does not matter if the flow restrictor is upstream or downstream of the turbine if it's a sealed pipe - the water will still be slowed. That flow restrictor is analogous to your resistor.

Mar 3, 2015
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Ohkay, probably another misunderstanding by my side... I thought that the negative vertical rail at the side of the breadboard is considered the ground... Seems not.
So, what would the ground be (if it existed)?

Thanks!

Mar 3, 2015
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Thanks for the replies guys!
Quite excited, I really appreciate the input from all

12. ### tshuck Well-Known Member

Oct 18, 2012
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Ground is a tricky concept as it means different things. Strictly speaking, it is a connection to the ground/earth, but most often, people mean to refer to it as 'common'.

This is the point that all voltages around the circuit are measured against (since voltage requires two points). The common can be any point, though it is typically the lowest voltage, or in a battery-powered circuit, the negative terminal.

Mar 3, 2015
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Roger that sir.
Thanx

14. ### bwilliams60 Active Member

Nov 18, 2012
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Think of voltage as potential difference between two points. If a water pipe carried flowing water and you measured two points, both measuring 50 PSI, the difference would be zero psi (same as voltage). When you look at a battery, you can think of the negative side of the battery as the most negative point in the circuit (0 volts) and the positive terminal as the most positive side of the circuit (9.0 volts) relatively speaking. Each component in a series circuit will "use" part of that voltage(pressure) as the electrons pass through the circuit because there is only one path to flow. So in your example, the LED may use 3.0 volts and the resistor will use 6.0 volts. Both convert electrical energy into another form (heat and light). If you add the components "voltage drop" together, you should end up with source voltage (9.0 volts). You can find each of these by placing your DVOM (DCV setting) across each of the components with the circuit working.

15. ### WBahn Moderator

Mar 31, 2012
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The problem you are having is a common one. You are thinking that because the LED is connected to 9V that it is seeing a voltage of 9V. This isn't the case. The voltage that the LED is "seeing" is the voltage difference between its positive terminal and its negative terminal. Since those are the two points that it is connected to, it is the voltage at those two points that is "sees". So if your LED had a forward voltage of 2V and you connected it between two voltage supplies one of which was at 1000V and the other of which was at 998V, the LED would be happy (don't do this because a diode is very sensitive to very slight changed in voltage across them such that if it was 1000V and 997V you would destroy the device, but it would be the same if it was simply connected directly across a 3V supply.

As for "ground", that is merely a point in the circuit, more properly called "common" in most cases, the we agree to use as a reference when talking about voltages at different points in the circuit. Voltage, by definition, involves a difference in potential energy between to points in a circuit, just as "height" involves a difference in elevation between two points. So we can talk about the "height" of a hill and either use as a reference the ground around the mountain or we could use the height above mean sea level (or any other point). If we string a cable (or a water pipe or whatever) between the tops of the two hills, then what we care about is the difference in elevation of the tops of the two hills. If we use a specific point on the ground around the hills as our reference then Hill A might be 400 ft tall and Hill B might be 300 ft tall and we conclude that our cable will go downward 100 ft in going from Hill A to Hill B. But if we used mean seal level as our reference then Hill A might be 8700 ft tall and Hill B might be 8600 ft tall. We would still conclude that our cable will go downward 100 ft in going from Hill A to Hill B. We could even use as a reference the top of the highest mountain in the state, in which case Hill A might have a height of -5600 ft and Hill B might be -5700 ft tall. We still draw the same conclusion about our cable. As long as all of our measurements are made relative to the same reference point, it doesn't matter where that reference point happens to be.

So in our circuit, we get to arbitrarily pick ANY point in the circuit and simply assign it a voltage and then use that assignment in describing the voltage at other points. By convention, in battery powered circuits we generally choose to assign the negative terminal of the battery a voltage of 0V. But we could just as correctly assign the positive terminal a voltage of 0V, making the negative terminal -9V. Or we could assign the positive terminal of the battery a voltage of -100V, making the negative terminal -109V. When we calculate the currents through devices or the voltages across devices, we will get exactly the same answers regardless of which choice we make.

Mar 3, 2015
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Thanks for the detailed input Wbahn!

Quite useful, thanks. Is there any voltage drop from the jumper wires? Or the drop comes only from the other components?

Thanks!

17. ### tshuck Well-Known Member

Oct 18, 2012
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Only a superconductor has no voltage across it while current flows, however, we usually neglect the voltage on conductors as it is too small to affect measurements for most intents and purposes.

Last edited: Mar 3, 2015
18. ### bwilliams60 Active Member

Nov 18, 2012
722
88
Everything in the circuit has some amount of voltage drop, including the battery itself. Think of voltage drop as a dynamic(moving) way of measuring resistance in a circuit. Each component no matter what it is made of, will have some form of resistance. A wire will have very little resistance and therefore, a very small voltage drop. So in your example, it may be like 0.003 VDC across the wire. Not enough to cause a problem. But id you had a bad connection due to corrosion, bad solder joint etc, you would have excessive voltage drop and that would be a problem. The idea of a circuit is to take one energy and turn it into another energy. We use electricity to convert to light, heat, mechanical motion etc. We can control or manipulate the voltage and current by using control devices. The end result (by design) is that we make an object do something by controlling the amount of voltage and current we put to it (power).

Mar 3, 2015
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Got it now, thanks

20. ### nsaspook AAC Fanatic!

Aug 27, 2009
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Not directed at anyone but to 'water' analogies in general.

Please don't use water analogies for beginners to describe electricity IMHO. I know it's tempting to do as people assume others understand hydraulics when they don't have a basic understanding of electricity. An analogy is good when people actually have a deep understanding of both subjects and use the analogy to explain a process in one as analogous to the process in another.

Explain electrical circuits in electrical terms.

Last edited: Mar 3, 2015