I have been doing this for years, but I'm afraid my understanding of some theories may have been slewed slightly and wondering if someone could clear up some of it.

For my first post, the most basic of all questions.

I have always thought of EMF or voltage as the pressure behind the electrons. The pressure that moves electrons which in turn is current. Correct?

So, my question is about "voltage drops". If I have resistor in a complex circuit, it has a voltage drop which means that this unit "consumes" a bit of the pressure behind it, and what leaves the resistor has less pressure than before?

Is my analogy correct? And if this is the case, at the end of the circuit where the electrons move to the positive terminal or ground, and the circuit has consumed the voltage in voltage drops, then is there no pressure left before it enters the battery or ground?

Is the explanation clear or should I try differently? I hope somebody can help my mind untwist.

Thanks.
Doc

 

Forget about pressure and electrons.
Simply use Ohm's Law to calculate the current in a resistance. Then use simple math to calculate the current and power.

 

Oh yes, well, I know the calculations, myriads of them actually. I'm trying to understand the physics of what voltage drops are and what happens to the movements of them, on a molecular level.

Thanks for the input.

 

Start with thinking about how a water faucet regulates the amount of water that can flow though it.

 

Heat has a big part to play in resistors voltage drop..or where the pressure goes.

 

Pressure is more similar to voltage (or EMF). Current is very similar to volume (gallons per minute).

Volt = http://en.wikipedia.org/wiki/Volts

...Water flow analogy

Amp = http://en.wikipedia.org/wiki/Ampere

In practical terms, the ampere is a measure of the amount of electric charge passing a point per unit time. Around 6.241 × 1018 electrons, or one coulomb, passing a given point each second constitutes one ampere.

 

Exactly as I thought.

So, lets say we have a simple circuit with a source battery of 3 volts. In the circuit we have two resistors that each have a voltage drop of 1.5 volts, which according to Kirchoff's law would make sense.

So if we follow the electrons when they leave the negative terminal, they have 3 volts of pressure, after the first resistor, they would have 1.5 volts or pressure because of the first resistor voltage drop.

Now if the next resistor drops 1.5 volts also, then wouldn't that mean that as the electrons reach the positive terminal of the battery, there is no pressure left?

How would electrons flow if this is the case? This is what makes no sense to me.

 

Exactly as I thought.

So, lets say we have a simple circuit with a source battery of 3 volts. In the circuit we have two resistors that each have a voltage drop of 1.5 volts, which according to Kirchoff's law would make sense.

So if we follow the electrons when they leave the negative terminal, they have 3 volts of pressure, after the first resistor, they would have 1.5 volts or pressure because of the first resistor voltage drop.

Now if the next resistor drops 1.5 volts also, then wouldn't that mean that as the electrons reach the positive terminal of the battery, there is no pressure left?

How would electrons flow if this is the case? This is what makes no sense to me.

No then each resistor would drop 1 volt each. If all resistors are in series and are equal in value then the voltage drop across each will be the total voltage divided by the number of resistors.

 

There would be a point where you added so much resistance that there is no current flow. That is called an insulator.

 

Even insulators follow Ohm's law. The current is extremely low. There are test equipment designed to verify that insulators are intact and doing their job, but I'm having a senior moment and can't remember their name.

Ohm's law will tell you how much current is flowing through a resistor with set voltage and resistance. It is the main starting point for all else in electronics. Some devices, such as diodes and LEDs, have quantum effects that set the voltage they drop, and are partly exempt from Ohm's Law, but only partly.

 

Even insulators follow Ohm's law. The current is extremely low. There are test equipment designed to verify that insulators are intact and doing their job, but I'm having a senior moment and can't remember their name.

Ohm's law will tell you how much current is flowing through a resistor with set voltage and resistance. It is the main starting point for all else in electronics. Some devices, such as diodes and LEDs, have quantum effects that set the voltage they drop, and are partly exempt from Ohm's Law, but only partly.

Yes I forgot to add that no current flow, assumes a perfect insulator.

 

No then each resistor would drop 1 volt each. If all resistors are in series and are equal in value then the voltage drop across each will be the total voltage divided by the number of resistors.

But, isn't the total sum of voltage drops equal to the source, according to kirchhoff?

Or does the remaining wire or whatever connection drop the extra volt?


Thanks for the replies everyone, I'm almost out of the cloud.

 

What extra volt? You have a supply of 3 volts. You have 3 resistors of equal value. Each resistor will drop 1 volt for a total of 3 volts.

Yes the wire will drop some voltage in the real world but so small you will probably not be able to measure it, assuming your wires are of some reasonable length. But even with the resistance of the wire the total dropped across the whole circuit can't be an more or less than 3V. So if you really wanted to calculate for the resistance of the wire then the voltage drop across each of those three resistors is going to be something less than 3v. Once again so small you are not going to be able to measure it with ordinary instruments.

 

So, lets say we have a simple circuit with a source battery of 3 volts. In the circuit we have two resistors that each have a voltage drop of 1.5 volts, which according to Kirchoff's law would make sense.

In my example I used 2 resistors at 1.5v for total of 3v. But even in your case, if the 3 resistor consumes all 3 volts of pressure, then what is moving the rest of the electrons to the positive terminal?

 

This is actually where the pressure analogy breaks down a bit. There is 3 volts potential between the two battery terminals. Like magnetic attraction there is an attraction of 3V from the negative terminal to the positive terminal, both are attracted to each other via an electric field.

The resistors "resist" the flow of current (by converting it to heat), and this resistance follows Ohm's Law formula. They electrons are not "used" up, they are constrained by the resistance, like water flowing through a choak valve. If the power source is deep enough and the resistance is low enough some serious power can be used. An arc welder only uses a couple of volts, for example, and some very high currents.

This brings us to the second most important formula in electronics, power (in watts).

Power = Volts X Amps

This relates very closely to Ohm's Law, and will tell you how hot a resistor will get for a specific resistance and voltage.

 

Hey doctorvox,

Think of the voltage drops in this manner. In a series circuit, there is only one path for the electrons to flow.

If you were to run a garden hose with relatively high pressure (voltage), then if you could measure the amount of water molecules passing running through the hose at any one point of the hose (current), you are going to read a relatively high amount of water molecules.

Now if you were to crimp the hose at any point (resistance) and measure the amount of water molecules passing through the hose at the points behind and in front of the crimped area, you are going to read the same exact figure. That is, there are the same number of water molecules passing through the hose at any point because there are no alternative paths for the water to go to. Similarly if you measure the force of the water behind and in front of the crimp in the hose, it will be the same also.

Now some changes happen if your were to crimp the hose at another point. Remember though, the water molecule count at a certain point remains the same no matter how much resistance you put on the hose because it is a straight shot path from the supply to the drain.

Now let's step away from the water analogy and go to circuits. Since we have seen that the current must be the same in a series circuit, then in order for the circuit to follow Kirchhoff's laws, something must be dropped. If the voltage did not drop across two resistors, then there would be more electrical potential after the resistors than before them which is quite impossible.

Resistors reduce the amount of current that is drawn by dropping voltage which in turn is released as heat. LEDs act as resistors in that they too dissipate voltage but in the form of light.

If you pass too much current through a 1/4 watt resistor, it wont be able to tolerate the heat dissipated, it glows red and burns out. Much in the same way as if you connect a single LED to a 5V power supply with no resistance at all, the LED cannot tolerate the force being dissipated, the LED blows. Different wattages can tolerate different currents.

If you connect many LEDs in series with a relatively low voltage, you wont need resistors because the voltage is divided among them. The more LEDs you connect in series the smaller the voltage that each tolerates and thus dimmer each is.

 

have been doing this for years, but I'm afraid my understanding of some theories may have been slewed slightly and wondering if someone could clear up some of it.

For my first post, the most basic of all questions.

I have always thought of EMF or voltage as the pressure behind the electrons. The pressure that moves electrons which in turn is current. Correct?

So, my question is about "voltage drops". If I have resistor in a complex circuit, it has a voltage drop which means that this unit "consumes" a bit of the pressure behind it, and what leaves the resistor has less pressure than before?

Is my analogy correct? And if this is the case, at the end of the circuit where the electrons move to the positive terminal or ground, and the circuit has consumed the voltage in voltage drops, then is there no pressure left before it enters the battery or ground?

Is the explanation clear or should I try differently? I hope somebody can help my mind untwist.

Stick with this and you'll make good headway.

 

Voltage indicates a separation of electric charge. Such a separation can come about in many ways: rubbing cats on a balloon (I have a hilarious story about that that I'll tell one of these days), chemical techniques, changing magnetic flux, moving raindrops in a thunderstorm (which we really don't understand yet), ion beams, etc. The existence of the voltage (or potential as the physicists like to call it) means that charges experience a force, familiar to all students of electrostatics, where this force is in the direction that indicates the biggest change per unit distance in the potential. It is entirely analogous to a bowling ball on a mountain side -- the ball will roll in the direction of the steepest descent. This force, if the charges are free to move, tends to reduce the potential because it tries (and often succeeds) to move the separated charges back together again. Since charge movement in conductors doesn't occur without collisions with atoms in most situations, the work done on these charges transfers energy to the atoms in the conductor, leading to increased temperatures on a macroscopic scale. We alternately bless this phenomenon, like I do at the moment while my little electric heater keeps me warm, or curse it, when it robs our designed circuit of useful energy that we would like it to do some useful work somewhere else.

Now then, consider a proud EE papa who has designed his low power baby to work at a paltry 1 μA. This is a tiny millionth of a coulomb per second. But in terms of the teeny electron, this is more than a million million electrons per second. And that is the same number of electrons passing a given point as the number of people on Earth, but at a thousand times per second. That hopefully gives you an idea of the number of electrons involved. We see the macroscopic ensemble averages.

And, at the end of the electron's journey, it gets back together with an opposite charge, leading to no more charge separation and, thus, no more voltage. I think that's the way to think about voltage because that's essentially what's going on physically.