# A very basic question..causing a lot of confusion..

Thread Starter

#### Max Kreeger

Joined Oct 1, 2013
94
Hey guys,

This isn't a homework question, just something I've been thinking about..

Simple scenario... a battery in parallel with a resistor. If all the voltage drops across the resistor to satisfy KVL...then how does current continue to flow if there's no voltage to drive it? Shouldn't some voltage still be getting through?

#### MrAl

Joined Jun 17, 2014
8,259
Hi,

If not homework, then why post in the homework section Not sure what you mean with that question. The voltage drop is equal to the battery voltage right? So the voltage means there is a current flow if the resistance is finite.

The more unusual question i have heard in the past is, "Is the battery in parallel with the resistor or in series with the resistor" (ha ha ha)

There is a distinct answer however.

#### profbuxton

Joined Feb 21, 2014
419
Simple answer. Voltage does not "get" through. It is current that "gets" through. In your simple scenario, the battery voltage is applied to the resistor and this results in a current flow in the resistor.
The battery terminals(say positive) is applied to one end of the resistor and the battery terminal(say negative) is applied to the other end of the resistor.
With regard to KVL, if you take the negative terminal as starting point(0) then the positive terminal will be at 0+V(battery volts) then the loop is completed back to negative terminal (0) which will satisfy KVL in the simple scenario. Draw yourself a diagram and follow the loop.

#### ErnieM

Joined Apr 24, 2011
8,157
How do you define a "voltage drop"? The way I define it there are no drops in your simple circuit... Every micro volt supplied by the battery is across the resistor and hence driving the current.

KVL says the battery voltage is the same as the resistor voltage except for the sign*, so when you add A to B you get zero. But saying there is a sum of zero volts around the loop is NOT saying there is no voltage across any two points.

* a picture would probably be best to show the sign, but current leaves the battery at the + terminal and enters the resistor at its + terminal, so by the passive component convents one voltage is added and one is subtracted when summing them about the loop.

Thread Starter

#### Max Kreeger

Joined Oct 1, 2013
94
I posted it here since its a homework style question.

The confusion just gets me when you put two resistors in series, there is a slight voltage drop on R1 and then the rest on R2...thats what I mean when I said "voltage gets through" I'm thinking of it as a water analogy...a river. If all the pressure dissipates..then the water won't flow.

#### Jony130

Joined Feb 17, 2009
5,232

#### Kermit2

Joined Feb 5, 2010
4,162
Voltage is a measure of a type of force, called electromotive.
If you throw a stone you impart kinetic energy to the stone. When that energy is exhausted the stone stops moving and just lays where it landed.
When voltage returns to its source there is no EMF left to measure.
It is measured in reference to one of its own terminals.
A circuit is a closed path. References to something outside the closed loop will have no meaning to what happens in the circuit.

A true diagram of your battery would have its internal resistance shown, and there would be two resistances in series with the voltage source and not just the "load" resistance you are contemplating.

Thread Starter

#### Max Kreeger

Joined Oct 1, 2013
94
In water analogy the voltage is a water pressure, so how can all the pressure dissipates?
Voltage is a measure of a type of force, called electromotive.
A true diagram of your battery would have its internal resistance shown, and there would be two resistances in series with the voltage source and not just the "load" resistance you are contemplating.
In reference to Quote 1 Well that's exactly what confused me...according to the voltage drop formula Vin * (R1/Rtotal) when I only had the load resistor the entirety of the voltage source was dropped across it...and thinking of it as a river nothing would "come out the other end" if all the voltage would stop at the resistor....it was acting as an infinite resistance.

Quote 2: So in reality there is a voltage division between that internal resistance and the load? leaving a minimal current flow?

#### MrChips

Joined Oct 2, 2009
23,565
Voltage does not drop.

When we use the expression "voltage drop" we really mean "voltage difference".
If we have two nodes A and B in a circuit with voltages at VA and VB, the difference in voltage between nodes A and B is (VA-VB) or (VB - VA) depending on your perspective.
We call this difference a "voltage drop". No voltage is being lost or consumed anywhere.

#### Kermit2

Joined Feb 5, 2010
4,162
Yes the available voltage, or electromotive force, must push the electrons through both the load resistance AND the batteries internal resistance.

This is the reason a batteries voltage reads lower when it is loaded down.

Your car battery will show a voltage greater than 12.6 if you check it. It may go as low as 10 or 11 volts while it is starting a car. This is the effect of the batteries internal resistance.

#### BR-549

Joined Sep 22, 2013
4,938
A voltage gradient is GENERATED by CURRENT flowing thru the resistor.

Not from the supply, i.e. battery.

This voltage gradient is in opposition to the supply, and will increase until it equals the supply.

This is the point where current is set and in equilibrium.

#### WBahn

Joined Mar 31, 2012
26,398
The thing that you are losing sight of is that while the voltage is "dropped" across the resistor, that same voltage is "gained" across the battery. In your water analogy, voltage is equivalent to height and the resistor is like a stream flowing from the top of a hill to the bottom. The battery is like a pump that pumps water from the bottom of the stream back up to the top of the hill.

#### DGElder

Joined Apr 3, 2016
351
Good question Max. Sometimes naive questions are the best because they can force you to think very clearly about things you have taken for granted. Your confusion comes from thinking voltage at a point makes current flow at that point - it does not.

But first some definitions for the circuit and water analogy.

E-field is a measure of the force on a unit of charge, in SI units: N/C. Separate a positive and a negative charged particle and there will be an E-field between them. How strong is that field? You measure the force it exerts on a test particle of unit charge.

Volts is not a unit of force. It is a measure of potential energy per unit of charge (Joules/Coulomb), or the amount of energy it takes to move 1C of charge against a 1 Newton/Coulomb E-field. This can be understood by remembering that Work (energy) = force * distance, so for a coulomb of charge it is V = E-field * distance, or in SI units (N/C)*m = V = (N*m)/C = J/C

Note: EMF (electromotive force) is measured in Volts, but EMF is an unfortunate misnomer because it is not a force.

Pressure Gradient per unit volume of fluid (N/(m^3) would be analogous to E-field.

Pressure is not a unit of force, it is a measure of potential energy per unit volume of a fluid (Joules/m^3), or the amount of energy it takes to move 1 cubic meter of fluid against a pressure gradient of 1N/m^3. Again, Energy = Force * distance, so for a cubic meter of fluid it is P = Pressure gradient * distance, or in SI units (N/m^3)*m = N/m^2 = P = J/m^3. (in the second term you see the more familiar definition of pressure as force per unit area: a Pascal in SI units)

So to answer your question..... Strictly speaking Voltage does not drive the current. E-field is the force that drives the current. An E-field will exist between any two points with a different voltage potential. The voltage will drop as you move around the circuit and off course it would measure zero at any arbitrary point with respect to itself, such as the negative battery terminal. But there is still a voltage difference and E-field that exist between any two points along that path - and that's what makes current flow.

In the water analogy where the battery is replaced by a pump and resistor by a pipe: it is the pressure gradient in the pipe that makes water flow, not the pressure. You can measure pressures all along the route of a pipe circuit with respect to some arbitrary point in the circuit such as the pump inlet. It would jump up at the pump outlet and gradually drop down to the lower pressure at the pump inlet. The pressure at the inlet of the pump would of course be zero with respect to itself, but that pressure is lower than the pressure before the pump inlet and higher than immediately after the inlet - so water flows.

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#### shteii01

Joined Feb 19, 2010
4,644
Voltage is a field, it DOES NOT require a path.

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#### shteii01

Joined Feb 19, 2010
4,644
I posted it here since its a homework style question.

The confusion just gets me when you put two resistors in series, there is a slight voltage drop on R1 and then the rest on R2...thats what I mean when I said "voltage gets through" I'm thinking of it as a water analogy...a river. If all the pressure dissipates..then the water won't flow.
You are mistaken. The voltage drop across R1 IS NOT slight.

If R1 and R2 are identical, have same resistance, then voltage drop across R1 is equal to voltage drop across R2.
Example. Let us say that I have 1 Ohm resistor R1 in series with 1 Ohm resistor R2. This series combination of resistors is connected to 12 volt battery. What is the voltage drop across R1? 6 volts. What is the voltage drop across R2? 6 volts. Is the drop across R1 slight? NO.

#### WBahn

Joined Mar 31, 2012
26,398
Voltage is a field, it DOES NOT require a pass.
A "pass"?

What "pass"?

#### ErnieM

Joined Apr 24, 2011
8,157
A "pass"?

What "pass"?
Sounds like an autocorrect to me. #### MrAl

Joined Jun 17, 2014
8,259
I posted it here since its a homework style question.

The confusion just gets me when you put two resistors in series, there is a slight voltage drop on R1 and then the rest on R2...thats what I mean when I said "voltage gets through" I'm thinking of it as a water analogy...a river. If all the pressure dissipates..then the water won't flow.
Hi again,

Since your line of thinking for this circuit is a little different than usual (you are comparing a series circuit of two resistors with just one resistor) then all i can suggest is the following line of reasoning and experiment...

Start with your known and understood circuit, which is two resistors R1 and R2, setting them at maybe the same value say 1k, and some source voltage like 1v or 10v, then analyze the circuit in the way in which you prefer. Next, make R1 half as small as R2, like 500 ohms, then analyze the circuit again and see what changed. Later, make it smaller again, like 10 times smaller (100 ohms) then analyze again, try to figure out what is happening. Later still, make it 10 ohms, then 1 ohm, with the R2 still at 1k. Try to figure out what changed during these resistor changes, then try to figure out what should happen when R1 is set equal to some very very low value like 0.001 ohms, and then finally try to visualize what happens when R1 is set to exactly zero ohms (a direct short).

In this way we migrate from a two resistor circuit to a single resistor circuit all the while watching what happens to the current and voltage across R2. We can then gain a better understanding of what happens with just one resistor because the circuit with 1k and 0.001 ohms is very very close to the circuit with just one 1k resistor so any measurements will be almost identical.

Just to note, the analysis for the voltage across R2 which you probably already know is:
vR2=Vcc*R2/(R1+R2)

and the current through R2 (and also R1) is:
iR2=Vcc/(R1+R2)

Vcc is of course the battery or power supply voltage.

Try that experiment either on paper or in a circuit simulator and see if it all makes more sense.

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