# Simple Question About Sign of Vab w/ Example

#### bluetooth tamer

Joined Jan 16, 2015
23
Hi everybody,

I am preparing for a quiz I have on Thursday. It's about KCL and KVL, very basic.

The example goes like this, Va on top and Vb on bottom:

Find Vx with the answer being Vx = 2V.

I tried this problem, got -2V, and could not for the life of me figure out why my sign for Vx is wrong, unless the professor's answer is wrong.

I gave the resistors these terminals because if you follow the direction of current, current entering a positive and exiting a negative terminal, then you will get a voltage drop. After using the rightmost loop for KVL I found that going clockwise and keeping the same polarity for the resistor makes the current source a voltage drop and the resistor a voltage rise (current going into negative and exiting positive terminal).

Vab, I understand, is Vb-Va. So why is that when I take Vb (zero) and subtract Va (2V) I don't get the same answer as the professor provided me with?

#### Ramussons

Joined May 3, 2013
1,395
To be honest, I just don't get what you are talking about. Without some schematic, it will be impossible to even visualise the puzzle.
Can you ask the same question with some sketches?

#### bluetooth tamer

Joined Jan 16, 2015
23

#### bluetooth tamer

Joined Jan 16, 2015
23
I just uploaded sketches.

#### Ramussons

Joined May 3, 2013
1,395
Vx=2V is correct

Hi everybody,

I am preparing for a quiz I have on Thursday.

Find Vx with the answer being Vx = 2V.

I gave the resistors these terminals because if you follow the direction of current, current entering a positive and exiting a negative terminal, then you will get a voltage drop. After using the rightmost loop for KVL I found that going clockwise and keeping the same polarity for the resistor makes the current source a voltage drop and the resistor a voltage rise (current going into negative and exiting positive terminal).

Vab, I understand, is Vb-Va. So why is that when I take Vb (zero) and subtract Va (2V) I don't get the same answer as the professor provided me with?

#### bluetooth tamer

Joined Jan 16, 2015
23
I now know I'm looking at it incorrectly. Could you please explain how you got a positive 2V?

#### MrChips

Joined Oct 2, 2009
30,460
Moved thread to Homework Help.

#### MikeML

Joined Oct 2, 2009
5,444

#### bluetooth tamer

Joined Jan 16, 2015
23
Thank you, this is an effective way to communicate the answer. I'll have to learn how to use LTSpice. And it makes sense to me in a clear way other than the seemingly arbitrary decision that KCL is applied towards the top node.
If I use KCL at the bottom node I would get:

Current into node (bottom) = current out of node (bottom)
I2 + I(R1) + I(R2) = I1
I(R1) = V(bottom)/2
I(R2) = V(bottom)/2
3 + V(bottom)/2 + V(bottom)/2 = 5

V(bottom) = 2

What makes the above wrong and yours right? Nit picky? Yes, but I haven't been proven that Vx =/= -2V yet.

#### WBahn

Joined Mar 31, 2012
29,850
Vx is defined as being the voltage on the top node, Va, relative to the voltage on the bottom node, Vb. That is what the '+' and '-' labels tell you. That means that

Vx = Vab = Va - Vb

Where you are going wrong is two things. First, that you are forgetting that current is a directional quantity. If you define the current in R1 as being from Node A to Node B, then the voltage across the resistor has to be (Va-Vb) in order to match this definition. Second, you are forgetting that Ohm's Law relates the current THROUGH a resistor to the voltage ACROSS that same resistor. V(bottom) is NOT the voltage across R1 or R2 -- you can't ignore V(top)

So

I(R1) = [V(top)-V(bottom)]/(2Ω)
I(R2) = [V(top)-V(bottom)]/(2Ω)
Vx = [V(top)-V(bottom)]

I(R1) = Vx/2Ω
I(R2) = Vx/2Ω

3A + (Vx/2Ω) + (Vx/2Ω) = 5A

Multiply both sides by 1Ω

3A·1Ω + (Vx/2Ω)·1Ω + (Vx/2Ω)·1Ω = 5A·1Ω
3V + Vx/2 + Vx/2 = 5V
Vx = 2V

You also need to track your units through your work.

#### WBahn

Joined Mar 31, 2012
29,850
Thank you, this is an effective way to communicate the answer. I'll have to learn how to use LTSpice. And it makes sense to me in a clear way other than the seemingly arbitrary decision that KCL is applied towards the top node.
If I use KCL at the bottom node I would get:

Current into node (bottom) = current out of node (bottom)
I2 + I(R1) + I(R2) = I1
I(R1) = V(bottom)/2
I(R2) = V(bottom)/2
3 + V(bottom)/2 + V(bottom)/2 = 5

V(bottom) = 2

What makes the above wrong and yours right? Nit picky? Yes, but I haven't been proven that Vx =/= -2V yet.
Arguably it wasn't an effective way to communicate the answer since it left you not understanding why it is correct. But hopefully it moved you in the right direction.

#### MikeML

Joined Oct 2, 2009
5,444
LTSpice does not make mistakes about which direction current flows, though. If I had choosen the top node as the reference, then it would have correctly shown that the voltage on the bottom node is as shown:

Current into node bot = Current out of node bot

I(I2) + [0-V(bot)]/R1 + [0-V(bot)]/R2 = I(I5)

3 - V(bot)/2 -V(bot)/2 = 5

V(bot) = -(5-3) = -2

#### WBahn

Joined Mar 31, 2012
29,850
Agreed. But the point was about whether it communicated the material as effectively as the TS claimed. I maintain that it didn't. Not that it was wrong or that the information wasn't there, only that, in this case, it clearly didn't communicate the information effectively to the TS.

#### bluetooth tamer

Joined Jan 16, 2015
23
Thank you, this is an effective way to communicate the answer. I'll have to learn how to use LTSpice. And it makes sense to me in a clear way other than the seemingly arbitrary decision that KCL is applied towards the top node.
If I use KCL at the bottom node I would get:

Current into node (bottom) = current out of node (bottom)
I2 + I(R1) + I(R2) = I1
I(R1) = V(bottom)/2
I(R2) = V(bottom)/2
3 + V(bottom)/2 + V(bottom)/2 = 5

V(bottom) = 2

What makes the above wrong and yours right? Nit picky? Yes, but I haven't been proven that Vx =/= -2V yet.
Vx is defined as being the voltage on the top node, Va, relative to the voltage on the bottom node, Vb. That is what the '+' and '-' labels tell you. That means that

Vx = Vab = Va - Vb

Where you are going wrong is two things. First, that you are forgetting that current is a directional quantity. If you define the current in R1 as being from Node A to Node B, then the voltage across the resistor has to be (Va-Vb) in order to match this definition. Second, you are forgetting that Ohm's Law relates the current THROUGH a resistor to the voltage ACROSS that same resistor. V(bottom) is NOT the voltage across R1 or R2 -- you can't ignore V(top)

So

I(R1) = [V(top)-V(bottom)]/(2Ω)
I(R2) = [V(top)-V(bottom)]/(2Ω)
Vx = [V(top)-V(bottom)]

I(R1) = Vx/2Ω
I(R2) = Vx/2Ω

3A + (Vx/2Ω) + (Vx/2Ω) = 5A

Multiply both sides by 1Ω

3A·1Ω + (Vx/2Ω)·1Ω + (Vx/2Ω)·1Ω = 5A·1Ω
3V + Vx/2 + Vx/2 = 5V
Vx = 2V

You also need to track your units through your work.
Could it be that the key here is that the bottom node of a circuit is zero volts (maybe ground?). Therefore Va-Vb = Vx will always be positive.

and/or

According to your reasoning (and you can correct me if I'm wrong):

1)
Current has direction.
Because of this, when assigning direction to current through a resistor, the voltage across is defined as Va-Vb where Va is the origin of the current (current goes through Node A into Node B).

2)
Because ohm's law tells that current through a resistor is related to voltage across this resistor, voltage across is based upon the difference between two nodes -- meaning neither of the nodes can be ignored.

And as a follow up, will direction of current (for instance, going upward) translate into from the bottom to the top of a current source there is a voltage rise? Or can current source in some scenarios be a voltage drop or rise independent of the source's output direction?

#### WBahn

Joined Mar 31, 2012
29,850
Could it be that the key here is that the bottom node of a circuit is zero volts (maybe ground?). Therefore Va-Vb = Vx will always be positive.
Nope. The choice of reference node has nothing to do with the value of Vx. We don't even have to assign a reference node in order to find Vx.

1)
Current has direction.
Because of this, when assigning direction to current through a resistor, the voltage across is defined as Va-Vb where Va is the origin of the current (current goes through Node A into Node B).
Yes. This is critical. The polarity of the voltage across a device and the direction of current through a device must be properly coordinated.

2)
Because ohm's law tells that current through a resistor is related to voltage across this resistor, voltage across is based upon the difference between two nodes -- meaning neither of the nodes can be ignored.
Also critical. It is an extremely common mistake for people to take the voltage at one terminal of a resistor (which is therefore the voltage of that node relative to some other point in the circuit that has been arbitrarily chosen) and throw it at Ohm's Law.

And as a follow up, will direction of current (for instance, going upward) translate into from the bottom to the top of a current source there is a voltage rise? Or can current source in some scenarios be a voltage drop or rise independent of the source's output direction?
As with any device, a current flowing from one voltage to a less positive voltage corresponds to power being converted from electrical energy to some other form. Likewise, a current flowing from one voltage to a more positive voltage corresponds to power being converted from some other form to electrical energy. This is true whether it is a resistor, a capacitor, a voltage source, or a current source.