I'm trying to understand the concept of voltage and I've reached section six of chapter one (http://www.allaboutcircuits.com/vol_1/chpt_1/6.html) of the book on this website.

The author explains that voltage is basically "potential" and can be thought of as the same concept of dropping a rock from a certain height.

I think I understand what he is saying, but I'm confused when the author says this:

In a normal lamp circuit, the resistance of a lamp will be much greater than the resistance of the connecting wires, so we should expect to see a substantial amount of voltage between points 2 and 3, with very little between points 1 and 2, or between 3 and 4.

(Please search the page in the link above to find the context for this quote).

If my source voltage is 9v, I don't understand what the author means by saying "we should expect to see a substanial amount of voltage between points 2 and 3".

Isn't the voltage 9v but then DROPS between points 2 and 3? (eg. the "potential" energy [voltage] starts at 9v, but encounters resistance [a light-bulb] so it now is only 7v after the battery?) I don't understand how there is "substanial voltage" between 2 and 3?!

Please help!

Thanks!

 

The sum of all your voltage drops (in a circuit) cannot exceed the total voltage in the circuit.

Resistance in a circuit causes the voltage to drop (increase) just like a kink in a water hose.

The kink in a water hose increases the pressure around the kink, just like resistance of the light bulb in a circuit.

The resistance of the light bulb causes the voltage drop across the light. There is very little resistance across the wires so little voltage drops across the wires in the circuit.

You might be confusing the term "Drop" The voltage drop indicates a voltage as a result of the resistance across the component in the circuit. No resistance in the circuit, no voltage drop. If you measure the terminals across a 9V battery you will get 9V. No voltage drop.

Thanks

Joe

 

The battery or other voltage source is like a pump, it pumps electrons and gives them "pressure" (voltage). In the complete circuit the pressure generated by the voltage source has to be *exactly* equal to the voltage drop arou8nd the circuit. Or, to put it another way, the total voltage around the complete circuit is zero. If the battery gives 9 V then the sum of the voltage drops has to be 9V.

Most of the voltage drop occurs in the lightbulb because it has higher resistance but a bit of voltage drop also happens in the wires as their resistance is not zero.

Voltage drop equals intensity times resistance (ohm's law).

 

I'm trying to understand the concept of voltage and I've reached section six of chapter one (http://www.allaboutcircuits.com/vol_1/chpt_1/6.html) of the book on this website.

The author explains that voltage is basically "potential" and can be thought of as the same concept of dropping a rock from a certain height.

I think I understand what he is saying, but I'm confused when the author says this:


(Please search the page in the link above to find the context for this quote).

If my source voltage is 9v, I don't understand what the author means by saying "we should expect to see a substanial amount of voltage between points 2 and 3".

Isn't the voltage 9v but then DROPS between points 2 and 3? (eg. the "potential" energy [voltage] starts at 9v, but encounters resistance [a light-bulb] so it now is only 7v after the battery?) I don't understand how there is "substanial voltage" between 2 and 3?!

Please help!

Thanks!

You should familiarise yourself with Kirchoff's Voltage Law: http://www.allaboutcircuits.com/vol_1/chpt_6/2.html

Essentially, the voltage of the source is equal to the sum of the voltage drop across all series elements in the circuit. In the example you have referenced these series elements can be broken down into the points 1-2, 2-3, 3-4. For each of these the voltage drop across each element is derived by Ohm's Law: V = IR. A point to note for series circuits is that the current through each element is the same. if you are unsure of this, post back and we can give you some reasoning behind this.

Therefore, we can see that since element 1-2 {R(1-2)} and 3-4 {R(3-4)} are just wires the resistance will be practically zero (it is important to note for this explanation that they are not actually zero). Furthermore the resistance of the bulb {R(2-3)} is substantially greater than that of the wires.

Therefore the voltage drops are given by Kirchoff's Voltage Law:

V(source) = IR(1-2) + IR(2-3) + I(3-4)

Without putting the numbers into the equation we can see that if R(2-3) is large compared to R(1-2) and R(3-4) then V(source) ~ IR(2-3) - i.e. a substantial amount of the source voltage is dropped between points 2 and 3 - its is not exactly equal because as we said before the resistance of the wires is practically zero and not actually zero.

Dave

 

You might be confusing the term "Drop" The voltage drop indicates a voltage as a result of the resistance across the component in the circuit. No resistance in the circuit, no voltage drop. If you measure the terminals across a 9V battery you will get 9V. No voltage drop.

I think the term "voltage drop" is exactly what I'm confusing If I have a 9V source, then is my "potential voltage" 9V, but the actual voltage is 0V if nothing is connected?

I'm actually thinking about voltage as a starting quantity. For example, if I start with a 9V source, the battery will "use up" part of the voltage and it will "drop" by 3V (for example) because of the battery requiring voltage to push the electrons through it.

From what you are saying, it sounds like voltage is created (eg. caused by) the resistance across the battery (or whatever). It sounds like you are saying that voltage starts off at 0V...

Does voltage start at 0V or 9V? (or neither?)

Or, to put it another way, the total voltage around the complete circuit is zero. If the battery gives 9 V then the sum of the voltage drops has to be 9V.

Most of the voltage drop occurs in the lightbulb because it has higher resistance but a bit of voltage drop also happens in the wires as their resistance is not zero.

It sounds like you're saying that voltage starts off at 0V but has the potential for 9V? (see my above question/comment)

Essentially, the voltage of the source is equal to the sum of the voltage drop across all series elements in the circuit.

Can you define "voltage drop"? From the book it seems to be defined as "the voltage between two points", but can that be defined as "the voltage "used up" between two points"?

I think I may be incorrect in thinking that voltage starts a 9V but then in reduced to zero...

Thanks for all of the help everybody! I really, really appriciate it!

 

Voltage drop is the observed change in voltage across a resistance. It's easier to see with two or more resistances in series. Use two 100 ohm resistors as the resistors, and a 12 volt batttery putting voltage across them.

Measuring from the - terminal, we see +12 VDC at the + terminal and at the end of the 100 ohm resistor connected to it. The voltage at the point that the resistors join together is 6 volts - we see a 6 volt drop across the 100 ohm resistor. Same for the other one - it has 6 volts on one side, and zero on the other - another 6 volt drop.

 

Voltage is a measurement of a physical phenomenon. It is not a tangible object - hence it cannot be "used up" in that sense. Current, however, is a tangible, physical object - namely electrons flowing through a circuit. The flow of current through a resistance can be quantified by measuring the voltage the current flow creates across the resistance.

So, voltages are created by current flowing through a resistance.

A voltage source, like a battery, is unique in that it creates energy (voltage.) Almost every other electronic component USES UP energy. So, a battery ADDS voltage to a circuit, whilst most components SUBTRACT voltage from the circuit. Hence GS3's comment about the total circuit voltages adding up to 0.

Looking at that battery/lamp circuit. Point 1 is 0V. The battery provides the energy to put point 4 at 9V. Point 3, for most intents and purposes, is exactly the same as point 4. Likewise, point 2 is the same as point 1. The ENTIRE 9V is now across the lamp - if you like, there's a 9V voltage "drop" across the lamp.

So, the battery provides 9V, the lamp has 9V across it. Remember, batteries create energy, the lamp takes it away, so 9V (battery) + -9 (lamp) = 0.

The term voltage drop is redundant - ignore the word "drop" and it still makes sense. A voltage is always measured between two points, so yes, your definition of voltage drop is correct.

 

Let's say we've got 1.2 inches of AWG20 wire between point 1 and point 2. Resistance will be about one micro-Ohm. Say we have 2.4 inches of AWG20 wire between points 3 and 4. Resistance will be about two micro-Ohm. Just for fun, let's put the resistance of the bulb at 27 Ohms. Total circuit resistance is 27.003 Ohms. (Give or take...)

We'll make the voltage across the battery 9Vdc, between points 1 and 4. If we measure between points 1 and 3, we've "dropped" about 667 microVolts from the 9 volts, leaving 8.999333 Volts between 1 and 3. (Our meter is not that accurate, though. We see "0" and "9.") Measuring between 1 and 2 we "drop" an additional 8.999000 volts (we see it as "9") leaving only 333 microvolts. We drop the last three microvolts measuring between point 1 and itself.

So, voltages are created by current flowing through a resistance.

Chicken before the egg? Maybe. I learned "egg before chicken." We can't have any current without voltage (unless we count those anomalous superconductor thingies.) We can indeed have voltage without current.

 

The voltage across the lamp is equal to the product of the current passing through the lamp and resistance of the lamp. This voltage is called a voltage drop.

 

I really appriciate all of the help... I don't understand why I just can't grasp this concept! (I can graduate from college with a four year degree, but have trouble understanding the simple [?] concept of voltage?!)

Would I be correct in saying that a battery generates 9 volts and each component with resistance uses up (eg. see niftydogs comment) part of that voltage?

It makes sense to me that if we start with 9v and 0.1v is required to push electrons through a resistance (eg. a battery). It also makes sense to me that to push electrons through any resistance it will require more and more voltage. (eg. a larger resistor would require more voltage to achieve the same rate of current?)

I guess I'm just not comprehending the whole "drop" terminology

I'll have to re-read the beginning of the book on this site (and the posts from this thread) and try really hard to picture what is going on...

EDIT: I also think that using the term "drop" was a bad idea

 

Would I be correct in saying that a battery generates 9 volts and each component with resistance uses up (eg. see niftydogs comment) part of that voltage?

Yes. "Drops" means "uses up."

 

When a voltage is applied across a resistance current is lost. From Ohms law (V=I*R) we see that when current is reduced, voltage drops.

SO.....

so a little math should clear up all your confusion.

Example using Ohms Law: V = I * R

10V = I * 200ohm
I = .05Amps

The voltage drop across one 200ohm resistor is 10V, 100% of the voltage is applied to 1 resistor and we have .05Amps of current

Let's say instead your 200ohm was made up of a 50ohm resistor and 150ohm resistor in series. Since the sum of 50ohm and 150 ohm is 200 and we know the total voltage across is 10V, we can find the voltage drop across each resistor by Ohms law again.

With a 150ohm resistor, how many volts would you need to have the same .05Amp current?

V1 = .05Amps * 150ohm
V1 = 7.5V

With a 50ohm resistor, how many volts would you need to have the same .05Amp current?

V2 = .05Amps * 50ohm
V2 = 2.5V

V1 + V2 = 10V, the same as if there was one 200ohm resistor, the voltage was just spread out over different resistors each with it's own voltage drop.

An analogy would be if you tried to push two boxes at the same time across the floor. if one box was heavier than the other than obviously the heavier one would have the majority of the resistance to you pushing it.

 

You also have to keep in mind there is no "left over" voltage in a circuit, 100% of it is applied, regardless if the energy is dissipated in heat as in a resistor, or light like a blub or mechanical energy like a motor or any combination thereof.

 

about your question whether to start from 0 or 9.
the voltage is a potential which is always measured with some ref like 9 m is height from ground (or from Michael Jordan's shoulder if that is taken as ref) similarly mostly voltage is 9v from ground potential or for a battery from -ve terminal if that is taken as a ref.
and the term Voltage ......Drop is self explanatory it is drop in voltage again has to be across two points. so if across smaller resistances of wire drop in potential is 1 volts it is drop from 9 volts to 8 volts across the whole wire. meaning the source lost that potential to do that work for sending that many electrons whose flow/rate is determinedagain by overall resistance. now since bulb resistance is more a substantial work must be done to overcome it for that flow of electrons and hence a substantial lost in potential to do work is lost that is the drop in potential say 7 volts and thus total 8 volts have been lost and now this point is at 1 volts again with respect to initial ref point (battery -ve terminal). the remaining potential is lost across the wire connecting to the -ve terminal.
now u might be confused if that one wire wasnt present and the bulb were somehow directly connected to the battery then where does the last one volt go?
if yes it will not be the case as that wud have led to a decrease in total resistance and the current wud have increased in such a fashion that the total voltage drop equals total voltage that is provided by the source.

 

The term "voltage drop" may be less than precise, but it's part of the electronic jargon, and so not really meant to convey a literal meaning.

 

There seems to be two conflicting things going on though...

Moving electrons through a resistance creates a voltage. However, voltage is required to move the electroncs in the first place!

If a battery provides voltage (eg. creates it) and voltage is created by moving current through a resistance, wouldn't they be added together? (eg. battery creates 9V and moving current through a lightbulb creates 1V so we now have 10V?)

 

the current thru the bulb drops voltage.
so it is 8volts.
if the direction of current was such that it opposed the current(opposite indirection) that wud be due to the battery then the voltage wud have added up. (this is possible in circuits with more than two sources.)

 

"Moving electrons through a resistance creates a voltage. However, voltage is required to move the electroncs in the first place!"

Wooaaah hold on there. You're confusing things. Electrons through a resistance creates a voltage IF AND ONLY IF there is a magnetic field moving them, like in an alternator or generator. A voltage on it's own will create a current through a resistance. Electrons don't spontaneously decide to move in unison across a conductor to create voltage. LOL.

"If a battery provides voltage (eg. creates it) and voltage is created by moving current through a resistance, wouldn't they be added together? (eg. battery creates 9V and moving current through a lightbulb creates 1V so we now have 10V?)"

Honestly, did you even read the 1st chapter of the guide? VOLTAGE = CURRENT X RESISTANCE.

If you have a voltage across a resistance you get a current. You don't get one without the other. Gravity doesn't exist because things fall, things fall because there is gravity. The sun doesn't shine because it's warm outside, it's warm outside because the sun shines.

 

Yes, I did read the first chapter. A couple of times actually. I've read quite a few chapters (in different books) and articles about it, which may be why I'm confused (eg. each author explains it differently).

I'm sorry that it's so frustrating trying to help me understand this

I think I am probably over-complicating this...

Is this really as simple as, "A 9V battery provides 9 volts and each resistance drops the voltage a little bit as the current passes through it?" That makes complete sense to me. (eg. it requires force [voltage] to push electrons through resistance... the more resistance the more force needed)

I know that voltage cannot be "used up", but it seems like it would work that way because if you have a lot of voltage drops and then something at the end of your circuit requiring a lot of voltage, there wouldn't be enough voltage left...?

Also, I was reading this line think that manifested meant created, but apparantly that is not the case:

Because it takes energy to force electrons to flow against the opposition of a resistance, there will be voltage manifested (or "dropped") between any points in a circuit with resistance between them.

 

Sofakng,

Get yourself a breadboard a few resistors and a 9V battery. Of course also your volt meter.

Built the circuit in the example (http://www.allaboutcircuits.com/vol_1/chpt_1/6.html)
and measure the voltage across each resistor, the battery and a section of wire.

Do as Dave indicates and look over Kirchoff's Voltage Law and Ohms law.

I think what you need is a bit of real life to plant all of the theory against.

Coming from a software background as well I have found many areas in electronics that are just down right confusing because I think as a programmer.

Everything in electronics can be viewed from several different view points. The sum of voltage drops is equal to total voltage. But you can look at just voltage drops from one view point and total voltage from another view point.

Take your time with this and get it well planted in your brain.

As a programmer I started out trying to flow chart what happens in a circuit over time as a piece of code would execute. This of course drove me nuts. Electrons are everywhere in the circuit, when current flows it starts to flow everywhere at once.

Not like a piece of code from one end to the other. You have to do a mind set change.

Keep with it. Build lots of circuits.

Thanks

Joe