voltage dividers

Thread Starter

pilotnmech

Joined Feb 26, 2005
27
I am having a problem understanding voltage divider problems in my book. I understand, but don't know why voltage (pressure) is dropped when electrons go through a resistance. It seems that if it's harder for electrons to move through a resistance they would get backed up and the pressure behind would rise since there is a lot of voltage (push) behind, but the volume would fall (water analogy). Anyway, it's easy to figure just the voltage drops. What my problem is is when you tap off of the resistors. My book is showing voltage drop in one direction, but current flow in the opposite, which goes against all of the other problems in the book so far. It's a DC circuit. My other problem is that when you tap off of the resistors, as in my first figure it seems that you have now turned the circuit into a series-parallel setup which means if you start at the bottom with 300v some of the current goes through R1, and some goes through the circuit A at 300v. That would mean that there is a voltage drop through R1 and a voltage drop through the circuit A, and whatever is left joins at point B and then goes through R2 (90v dropped in R1 and what's left of the voltage in circuit A), then there is a voltage drop in R2 and what's left of the 300v in circuit B, etc. If you start at the top current would be going in the opposite direction from the figure, and the process makes more sense. Is the current direction in the book wrong? All of the other problems keep on showing current opposite of voltage too. I asked an engineer about the problem and he couldn't give me any good answer except that he thought the current direction is wrong. I also asked other people and they said that voltage does go in the opposite direction of current, which makes no sense to me since voltage is what causes current (electrons) to move. Also, how can you get a negative voltage? It seems if you've used up the voltage there can be none left.
 

n9xv

Joined Jan 18, 2005
329
I am having a problem understanding voltage divider problems in my book. I understand, but don't know why voltage (pressure) is dropped when electrons go through a resistance. It seems that if it's harder for electrons to move through a resistance they would get backed up and the pressure behind would rise since there is a lot of voltage (push) behind, but the volume would fall (water analogy).
Voltage is dropped through the resistance because fewer electrons are exiting the resistor than origionally entered it. Resistors resist the flow of current. A givin level of current flows into the resistor but a lesser amount of current flows out the other end due to the resistance. This difference of quantity of electrons between the entering end and the exiting end can be measured as an actual voltage. Imagine a wire several miles long with a current flowing through it. if you measure with a voltmeter a point a few inches apart you would not see a voltage because the resistance is insignificant at that point. Now, move the positive test lead several thousand feet down the wire --- with the ground lead at the starting point of the wire --- and you will see a voltage rise in proportion to the distance down the wire. (for the sake of discussion, neglect the resistance of the test leads).



What my problem is is when you tap off of the resistors. My book is showing voltage drop in one direction, but current flow in the opposite, which goes against all of the other problems in the book so far. It's a DC circuit. My other problem is that when you tap off of the resistors, as in my first figure it seems that you have now turned the circuit into a series-parallel setup which means if you start at the bottom with 300v some of the current goes through R1, and some goes through the circuit A at 300v. That would mean that there is a voltage drop through R1 and a voltage drop through the circuit A, and whatever is left joins at point B and then goes through R2 (90v dropped in R1 and what's left of the voltage in circuit A), then there is a voltage drop in R2 and what's left of the 300v in circuit B, etc. If you start at the top current would be going in the opposite direction from the figure, and the process makes more sense. Is the current direction in the book wrong? All of the other problems keep on showing current opposite of voltage too. I asked an engineer about the problem and he couldn't give me any good answer except that he thought the current direction is wrong. I also asked other people and they said that voltage does go in the opposite direction of current, which makes no sense to me since voltage is what causes current (electrons) to move. Also, how can you get a negative voltage? It seems if you've used up the voltage there can be none left.
Some text books will indicate electron current flow (from negative to positive). Other text books will indicate conventional current flow (from positive to negative - sometimes refered to as "hole" flow as in the space in an atom where an electron used to be). Unless otherwise stated, any schematic is understood to indicate conventional current flow. The arrows on diodes, transistors etc. etc - are indicating conventional current flow. The current still splits and recombines in the same manner, just considering it from a different direction. Some books do a poor job of explaining which method their using and will even switch back and forth without making it clear to the reader. I suspect you are the victim of such a book!


I also asked other people and they said that voltage does go in the opposite direction of current, which makes no sense to me since voltage is what causes current (electrons) to move.
To the "other people" - - - Voltage DOES NOT move. Only the current is moving in any circuit. Voltage is a difference of potential that is necessary to set up the condition for current to flow.

Mathematically (and in reallity too) the sum total of all voltage drops will equal the supply voltage or, the supply voltage minus all voltage drops will equal 0 - always.
 

David Bridgen

Joined Feb 10, 2005
278
pilotnmech,

The diagram shows electron flow, not conventional current flow, but that shouldn't be a problem, all the rules (Ohm's law, Kirchoff's theorem and the superposition theorem) still apply. I recommend that you study these until you are comfortable with them.

Your engineer who "couldn't give me any good answer except that he thought the current direction is wrong" is a very strange "engineer" indeed.
And the other people who "said that voltage does go in the opposite direction of current" have a rather strange idea of voltage.

Voltage doesn't "go" anywhere, it is applied across a system or component, and it is the resultant current which goes somewhere. A current through a component in a system will cause a voltage to be develped, or dropped, across that component, not in it.
 

Thread Starter

pilotnmech

Joined Feb 26, 2005
27
I appreciate the quick responses from the knowledgeable people. Here is my analogy of how voltage is produced, please tell me if I'm correct. There are many electrons in one end of the battery (the negative side), or power source, and the electrons hate each other and want to go to the other side (positive) which is lacking electrons. This is what causes the pressure, or voltage to build. The electrons won't be happy until they are neutralized with an even number of protons. When a path is provided for the electrons to move one bumps and repels the next guy and so on but they can never rest because the power source is still providing the imbalance so as long as the path exists the cycle never ends. Getting back to the problem, the book always showed current and voltage drops starting at the same place, it threw me when they showed current coming in the bottom, and the voltage drops starting at the top. I have already done MANY problems using the theorems, but just because you can figure something out because you're taught how doesn't mean you completely understand it. I want to be highly knowledgeable. Thanks.
 

n9xv

Joined Jan 18, 2005
329
I think you have a good hardy understanding of it so far. Its the domino effect of electrons repelling each other from atom to atom etc.


but just because you can figure something out because you're taught how doesn't mean you completely understand it. I want to be highly knowledgeable.

I admire/respect that attitude. Many people in technical feilds can calculate things but dont really understand them, at least not enough to explain them.
 

Thread Starter

pilotnmech

Joined Feb 26, 2005
27
I have to understand things well enough that I can teach them. I find it hard to believe that someone can go through all of the calculus, etc. in engineering classes, but still not know the basics well enough to teach it. I appreciate all of the help.
 

Tekker

Joined Apr 22, 2005
33
Originally posted by pilotnmech@Apr 20 2005, 09:38 PM
Also, how can you get a negative voltage? It seems if you've used up the voltage there can be none left.
[post=7121]Quoted post[/post]​
It looks like this was the only part of your post that hasn't been covered yet....

Negative voltages are created by a reference point. As you probably already know, ground is a common point in the circuit that is often used as a reference point for making measurements. There can be voltages that are "above" ground (positive voltages) and voltages that are below than ground (negative voltages).

For instance, say you have two resistors connected to a battery and you make ground be between the two resistors. If you connect the black lead of your meter to ground and used the red lead to measure across the first resistor you would see a positive voltage displayed. Then if you were to move the red lead to the bottom of the second resistor (leaving black lead at ground) you would see a negative voltage displayed. But if you connected the red lead to the top of the first resistor and the black lead at the bottom of the second, you would still see a positive voltage (because you're not referenced to ground). So it's basically all relative. :D

Hope that helps. B)

-tkr
 

erice1984

Joined Jun 9, 2007
16
Electrons = energy
Heat = energy

Electrons meet Resistance, and pass through, others leave in the form of Heat.

Just remember energy can not be created nor destroyed, only transfered
 

mOOse

Joined Aug 22, 2007
20
Voltage is dropped through the resistance because fewer electrons are exiting the resistor than origionally entered it. Resistors resist the flow of current. A givin level of current flows into the resistor but a lesser amount of current flows out the other end due to the resistance. This difference of quantity of electrons between the entering end and the exiting end can be measured as an actual voltage.
How can fewer electrons exit the resistor than entered it?
Do you mean the current leaving is less than the current entering?
That seems incorrect to me.
 

beenthere

Joined Apr 20, 2004
15,819
Same here - the current flow consistos of as many electrons anywhere in a series circuit. The voltage drop in a resistor means that the electrons passing through lost some energy in doing so - showing up as heat.
 

chuckey

Joined Jun 4, 2007
75
Think about the water analogy, your house hold tap is supplied from some reservoir. This reservoir is some feet above your tap, this is the water pressure, like voltage. It exists whether or not you draw any water. Static electricity is just like that, loads of volts NO current. As you turn on the tap some water flows (like current). So what stops your tap delivering one million gallons of water an hour?, its the resistance of the the complete circuit. ( I= V/R). Put a hose on the tap and take it outside and point it upwards, the jets heights will change as you turn the tap. This is because as maximum water flow (low resistance) the pressure drop in the pipework uses up all the pressure from the reservoir, at lower flows, the pressure drop in the pipe goes down so there is more energy left in the water to push the jet higher. So the pressure drops around the circuit always add up to that available (height of reservoir)
Historically it was thought that the flow of electricity was by positive particles, so batteries were labelled + cos they thought that the particles were coming out of that terminal. They were wrong!!!!
Frank
 

Dave

Joined Nov 17, 2003
6,970
Historically it was thought that the flow of electricity was by positive particles, so batteries were labelled + cos they thought that the particles were coming out of that terminal. They were wrong!!!!
Frank
Many texts still discuss current flow in the conventional sense, i.e. positive to negative. For the basics of circuit analysis it is often best to look at in this way. Whereas we know that in reality current flow is due to migration of negative (electron) charge and hence is negative to positive. From an analytical point of view, as long as you are consistent both are equally valuable.

Dave
 

FredM

Joined Dec 27, 2005
124
It is sad that we are 'stuck' with 'incorrect' direction of current flow due to the initial error - but for all practical purposes it makes no difference.. Conventional (incorrect) current flow has the big advantage that component symbols (diodes, BJTs, SCR's etc) have their arrows pointing to correspond to current flow from +ve to -Ve.. This makes circuit analysis a lot less confusing..

Trying to 'think correctly' about current flow in a circuit can be a real brain bender, I suppose if one ONLY ever thought about it in the 'correct' sense, the problem would reduce - but using (or switching between) both systems will confuse one and lead to errors.. My feeling is that its best to do analysis using conventional current flow, and to use text books which adopt this approach.. One can always, once the circuit is understood, go back and 'think correctly' about it.. But the only time I have ever needed to do this was with some physics equipment (ionisation chambers and ion wind generators) where the actual direction of electron travel was important.
 
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