# Why does voltage drop 100% across any component?

Discussion in 'General Electronics Chat' started by Mercfh, Feb 13, 2013.

1. ### Mercfh Thread Starter New Member

Apr 24, 2012
11
0
As the title says, lets say you have 5v coming from + and wire...then a resistor, or inductor, or LED or ANYTHING....how come the voltage drops COMPLETELY across it?
Like as in...how come there is zero volts on any wire after that component, all the way up till the - battery terminal?

I understand the water analogy, and resistors are like a smaller pipe limiting current flow...but it seems like pressure should be the same across the wire no matter what?

I can't seem to understand this basic principle!

Whats even more confusing is that say you have a Battery and an inductor or something.....then the voltage after the inductor is zero. But if you have a battery and an inductor then some wire then a resistor.....there is a +voltage after the inductor and then negative after the resistor? I don't get it?

Last edited: Feb 13, 2013
2. ### Biff383 Member

Jun 6, 2012
49
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Volume 1, Chapter 6.

3. ### crutschow Expert

Mar 14, 2008
13,498
3,374
Voltage is relative. To drop a voltage you need an impedance. Thus there may be voltage on a wire (relative to common) but that is not a drop.

For example in the circuit posted by Biff383, there is a 45V battery driving three resistors in series. Thus the voltage measured directly across the battery is 45V.

The sum of the voltage drops around a loop is always equal to the source (battery) voltage. The drop of each resistor around the loop is proportional to their relative resistance. Thus the drop across R1 is 10V, the drop across R2 is 20V and the drop across R3 is 15V for a total drop of 45V.

The voltage on the wire on any place in the loop is the sum of the voltage drops. Thus, for example, if point 1 is designated as common, then the wire voltage at point 3 to common is 20V +15V = 35V.

In an actual circuit there will also be a small drop across the wire resistance but typically that is so small that is ignored and assumed to be 0V. However that wire drop can be significant in power circuits. That's why high mains power circuits in a building have larger wire to minimize the wire resistance voltage drop (and consequent IR power loss).

That make sense?

4. ### Mercfh Thread Starter New Member

Apr 24, 2012
11
0
No it makes sense, Im asking why? Like the physics of it.

Voltage is The Potential Energy or "Pressure" so how come when we have multiple resistors or anything the voltage isn't completely taken up after the first resistor?

Since if 1 resistor was in the circuit it'd be 0v after, why if we have 3 resistors that after the first one it isn't 0v after?

5. ### SPQR Member

Nov 4, 2011
379
48

If you take a resistor and connect one lead to a 9V battery and measure the voltage at the unconnected end, you will see 9V.
What????? There is no voltage drop across the resistor??? What's up with that???

General rule: You cannot see a voltage drop unless there is current flowing.
In a resistor with only one lead connected, there is no current flowing, so there cannot be a voltage drop.

You will not see a voltage drop across a component unless that component "does" something.
A motor "turns", thus you see a voltage drop.
A resistor "heats", thus you see a voltage drop.
An LED "lights" thus you see a voltage drop.
A wire "does nothing" thus you do not see a voltage drop. (As mentioned by crutschow, that's not entirely accurate - it does "heat" but not very much).

You have a pipe connected to an enclosed waterwheel.
The pressure in the pipe is 100 psi before the waterwheel.
You turn on the flow, the waterwheel starts turning (doing something).
The pressure on the other side of the waterwheel falls to 50 psi.
The pressure MUST fall because the waterwheel is "taking" energy from the flow of water.
Put a cap over the other end of the pipe leaving the waterwheel.
The pressure in the pipe before the waterwheel is 100 psi - and is also 100 psi after the waterwheel.
If there is no flow of water, there is no energy being consumed, thus the pressure is the same.
Sounds just like the unconnected resistor.

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6. ### electron_prince Member

Sep 19, 2012
93
3
According to Ohms law

V = R.I

where R = Resistance of the resistor

I = current flowing through that resistor

and V = potential difference (voltage) across that resistor.

Now consider the circuit posted by Biff383

It has a three resistors (R1, R2 and R3) and ohms law is valid for each of these resistor. But if you put a box around that circuit, then the circuit itself is a resistor. In this case the circuit is a resistor of resistance (R1 + R2 + R2) ohms. Ohms law is also valid for this new "box" resistor.

so According to ohms law, the current that will flow through this "box" resistor is V/(R1 + R2 + R3).

Now look inside that "box" resistor. and think, what is the current flowing through R1, R2 and R3 resistors?

watch this video (http://www.youtube.com/watch?v=AfQxyVuLeCs). you will find that by studying circuit theory you agreed to adhere to the rules of lumped matter discipline. And according to lumped matter discipline, no charge is built or destroyed inside your resistor.

so the current that entered your "box" resistor must flow through resistor R1 and resistor R2 and resistor R3. It can't be destroyed.

so current through R1 = current through R2 = current through R3 = current through "box" resistor = V/(R1 + R2 + R3).

now apply ohms law on Resistor R1. calculate the potential difference required across the resistor R1 to let V/(R1 + R2 + R3) flow through it.

do it for resistor R2 and R3 as well.

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7. ### DC_Kid Distinguished Member

Feb 25, 2008
641
9
once the electrons make it from the "-" side (where the neg charge is) back to the "+" side (some other place that is charged more positive) their potential is gone. its like a big bucket of water held up 10ft, small 1/4 hose with valve coming out of the bottom. you open the valve and the water flows restricted until the water hits the ground (no pun intended). once on the ground that water has no more potential, etc.

for resistors in series, each junction point in between removes energy from the flowing electron, thus measuring the voltage (potential) at the junction points will show less voltage, etc.

are you wanting the physics and chemistry explanation, as in how electrons have EV and how that EV lowers as the eletron does work to cross over resistence??

8. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
From a logical perspective: At any point along the current path, as in the circuit posted above, the current flowing through that point is a constant, ie, no extra charge is being added or subtracted. Thus, current is the same at all points along the current path. That being the case, the voltage drops must be distributed amongst the resistances in the circuit. If all the voltage dropped in a sigle resistor, then no current would flow in the circuit past that resistor.

9. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
I'm going to toss my pennies in the pot:

Let's forget the water analogy for the time being, analogies are nice, but they are not the same.

Voltage is created by a separation of charge.

The electrons within a wire move according to the direction of an applied voltage. The electrons do not travel in straight lines, they bounce off of the lattice and hit atoms, with a net movement in the direction of the voltage. The electrons don't care what they move through, all they know is, "there is a force, and I'm going to move because of it." That force originates at the - terminal and ends at the + terminal. The amount of bouncing the electrons do is called resistance. The more impacts an electron has, the more resistance of a material.

The reason the voltage is dropped across anything attached to it is because the separation of charge is no longer maintained as electrons move to the point they were separated from (the + terminal).

For multiple resistors, the electrons are still separated after the first series resistor. The force is still present, and the electrons continue along their path toward the positive terminal.

Last edited: Feb 14, 2013
10. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
You're making the mistake of using the water analogy for voltage. You can't do that--in most cases it doesn't work. The water analogy really only applies to CURRENT, which flows through a conductor in a similar manner as water through pipes.

Voltage isn't an actual thing. You can't look under an extremely high-power microscope at a wire and say "Oh hey, that thing flowing through it is voltage". You can do that for electrons, but "voltage" is not an actual, physical thing. It's a reference, nothing more.

11. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
Voltage is a potential. You can have potential fields which move charge. In a circuit, the wire defines the potential field boundaries of interest.

12. ### tgil New Member

May 18, 2011
19
4
I am going to add my two cents as well, focusing on the physics side.

First, as mentioned, voltage is a potential energy (if you stand stationary on the top of a cliff, you have potential energy). When a circuit is complete, the potential energy converts to kinetic energy. Electrons literally start flowing on the wire. While the voltage potential travels at close to the speed of light through the circuit, physically the electrons flow at just a few centimeters per second and encounter friction along the way dissipating energy as heat. As energy is dissipated as heat, the potential energy (voltage) at that point in the circuit is less. In a resistor electrons encounter more friction and a relatively large amount of heat.

Current is measured in Amps which is Coulombs per second. A coulomb is a measure of charge. Current encompasses both the volume and speed of the electrons moving along the wire. Also, electrons (which have a negative charge) flow from negative to positive, but current (which has a positive charge) flows in the opposite direction from positive to negative.

Consider an ideal (able to source infinite current) battery of 9V. If you shorted the plus and minus with a wire that had a resistance of 0.1 ohms, the battery would output 90A (lots of electrons flowing from minus to positive). This would cause 810W (9V*90A) of heat to dissipate in the wire. The dissipation would be linear along the wire with each cm of wire dissipating about the same amount of heat. The voltage drop is also linear along the wire with each cm dropping about the same voltage. You can also visualize the electrons, being pushed by the voltage, flowing easily from plus to minus.

Now if we add a 1K resistor between the battery terminals, it creates a bottleneck in the electron flow. This decreases the flow of electrons by about 10,000 times to 8.999mA (9V/(1000+0.1)ohms). The wire and the resistor dissipate a total of 80.991mW. With just 8.999mA flowing through the wire, the electrons dissipate much less heat in each cm. They also drop much less voltage. The heat dissipation and the voltage drop are so little that we usually ignore them when analyzing circuits. Ignoring the wire resistance (which is assumed to be much higher at 0.1 ohms than it actually is) gives a current of 9mA and 81mW. So it is quite reasonable to ignore.

Hopefully, this gives you one more way to think about it.

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13. ### #12 Expert

Nov 30, 2010
16,693
7,334
Maybe my page on "Ohm's Law for Noobies" will help. It's at the top of the Chat page.
It goes into REALLY basic stuff.

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14. ### DC_Kid Distinguished Member

Feb 25, 2008
641
9
to get electron separation you have to do work. and its not just separation from the positive side, its not like how far away they are or how many points you can measure at. there is energy involved and that energy is diminished as the electron does work crossing through resistive paths to get home.

like a bowling ball rolling through 1,000 big sheets of tissue separated by 2ft each (zero friction surface, with Ei for initial energy of ball), passing through each sheet takes energy away from the ball, etc, and when it gets to the pins and stops all the energy that we put in is now gone into the sheets of tissue as deformation and heat. in circuits we see heat, in other things a moving mass of free electrons can cause deformation.

Last edited: Feb 14, 2013

Oct 18, 2012
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16. ### Mercfh Thread Starter New Member

Apr 24, 2012
11
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That makes sense....since Voltage is "Pushing" the current along. I guess just it's odd to me that when theirs one......voltage goes all the way down.

17. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Please read the E-Book at the top of the page, Vol I, DC.

Pay special attention to Ohms Law, Node analysis and Mesh Analysis.

Until you understand the above concepts, you'll be having a headache every time you try to make something work.

18. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
You know what--Instead of thinking of voltage as electrical pressure, think of it as the pressure difference. If something is in the circuit that holds back the current (a resistor, for example), then the pressure will be stronger before the resistor. If you check the pressure on ground and at the other side of the resistor, the pressure will be the same, meaning there's no pressure difference, and thus a 0 volt potential.

19. ### #12 Expert

Nov 30, 2010
16,693
7,334
That's the first thing I thought of. "Why is the measured voltage zero when both of my test probes are on the ground wire?"

Because every voltage is a voltage compared to someplace else. If you compare a voltage to the same place, it will always be zero difference...but this seems a bit advanced for this OP so I let other people have a go at it.