KVL Question

Hi all,

I'm currently reading about Kirchhoff’s voltage law (KVL) and come to a brick wall in understanding something. One of the diagram that was shown was below with the explanation about it.


View attachment 61861




I understand that to calculate the KVL it is going clockwise from 12V all the way to Vo but what I'm confused about the following

(1) The equation in the diagram states

-12 + 4i + 2vo - 4 + 6i = 0

my question is since we are going clockwise why it is calculating +6i and not -6i because from my understanding the flow is going from 4V and entering the (-) polarity of the 6ohms ?

(2) The equation

Vo = -6i

How to derive that Vo is -6i and not +6i ?


Thanks for the help

nicksydeny,

I'm currently reading about Kirchhoff’s voltage law (KVL) and come to a brick wall in understanding something. One of the diagram that was shown was below with the explanation about it.


Attachment 61861




I understand that to calculate the KVL it is going clockwise from 12V all the way to Vo but what I'm confused about the following

(1) The equation in the diagram states

-12 + 4i + 2vo - 4 + 6i = 0

my question is since we are going clockwise why it is calculating +6i and not -6i because from my understanding the flow is going from 4V and entering the (-) polarity of the 6ohms ?

Click to expand...

Both KVL and KCL apply in either direction. But you must be consistent. In the problem, it is assumed that the current exists in a CW direction. The calculation for current shows a minus value, so the assumption was wrong, and the current direction as shown by a ammeter will be CCW. If the assumption was the current existing is a CCW direction, then the calculated value would have been positive, confirming that the CCW direction assumption was correct.

You have no basis for assuming that the KVL was applied in a CW direction. In fact, the equation above proves that it was applied in a CCW direction. The results will be the same no matter which direction KVL is carried out.

(2) The equation

Vo = -6i

How to derive that Vo is -6i and not +6i ?

Click to expand...

The 6 ohm resistor has + and - signs showing the voltage sensing which applies to the 2*v dependent source. So we measure v by putting the negative lead of the voltmeter on the right side of the 6 ohm resistor and the positive lead on the left side. Since the current is assumed CW, we can assume a negative voltage from the voltmeter and relate it to v = -6*i.

To iterate, it does not matter what direction you assume the current, or the direction you measure the voltage, as long as it is consistent. Usually mesh analysis assumes a CW current direction.

Ratch

Hi Ratchit,

Thanks for the reply

Ratchit said:

You have no basis for assuming that the KVL was applied in a CW direction. In fact, the equation above proves that it was applied in a CCW direction. The results will be the same no matter which direction KVL is carried out.

Click to expand...

The reason why I 'assume' it's clockwise is because in the book, there was an explanation about how to determine the V variable is (+) or (-) based on the polarity and in the example is was explained and I quote

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Suppose we start with the voltage source and go clockwise around the loop as shown; then voltages would be −v1,+v2,+v3,−v4, and +v5, in that order. For example, as we reach branch 3, the positive terminal is met first; hence we have+v3. For branch 4, we reach the negative terminal first; hence, −v4. Thus, KVL yields

View attachment 61869
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

So you can see because of that explanation I'm taking into assumption that it is CW and not CCW. That is also why I was confused when I found another diagram (like in my first post) and notice that the equation is saying +6i and not -6i and this is still confuses me.

Ratchit said:

The 6 ohm resistor has + and - signs showing the voltage sensing which applies to the 2*v dependent source. So we measure v by putting the negative lead of the voltmeter on the right side of the 6 ohm resistor and the positive lead on the left side. Since the current is assumed CW, we can assume a negative voltage from the voltmeter and relate it to v = -6*i.

To iterate, it does not matter what direction you assume the current, or the direction you measure the voltage, as long as it is consistent. Usually mesh analysis assumes a CW current direction.

Click to expand...

So what you are saying if I 'assume' CW and go with my understanding of things the formula from


-12 + 4i + 2vo - 4 + 6i = 0


to


-12 + 4i + 2vo - 4 - 6i = 0

will come out with same result ?


Thanks for your help.

Forgot to add that the book that I'm currently reading is called "Fundamentals of Electric Circuits" by Charles. K. Alexander and Matthew N.O. Sadiku (3rd Edition)

nicksydney said:

So what you are saying if I 'assume' CW and go with my understanding of things the formula from


-12 + 4i + 2vo - 4 + 6i = 0


to


-12 + 4i + 2vo - 4 - 6i = 0

will come out with same result ?


Thanks for your help.

Click to expand...

No. Considering KVL:

-12 + 4i + 2vo - 4 + 6i = 0

would be the same as:

12 - 4i - 2vo + 4 - 6i = 0

if applied in the clockwise direction. The direction MUST be consistent, otherwise you will get the wrong answer. Looking at the diagram, seeing that it starts with -12 volts shows that they are assuming a counter-clockwise direction. That means that all resistors must be considered to have a voltage of +Ri, and voltage sources (with polarity against that direction of flow) would have to be -V. Any voltage sources with polarity following that flow would be +V.

Does that make sense?

DerStrom8 said:

No. Considering KVL:

-12 + 4i + 2vo - 4 + 6i = 0

would be the same as:

12 - 4i - 2vo + 4 - 6i = 0

if applied in the clockwise direction. The direction MUST be consistent, otherwise you will get the wrong answer. Looking at the diagram, seeing that it starts with -12 volts shows that they are assuming a counter-clockwise direction. That means that all resistors must be considered to have a voltage of +Ri, and voltage sources (with polarity against that direction of flow) would have to be -V. Any voltage sources with polarity following that flow would be +V.

Does that make sense?

Click to expand...

Hi DerStrom8,

I'm bit lost here from - let's take example of my first post in picture (a). If we use the equation

-12 + 4i + 2vo - 4 + 6i = 0

The -12 in the equation means the flow is "going out" from the 12V battery while flows "into" the 4ohms resistor to the (+) side it is therefore a positive in the equation, did I get this correct ?

nicksydney said:

Hi DerStrom8,

I'm bit lost here from - let's take example of my first post in picture (a). If we use the equation

-12 + 4i + 2vo - 4 + 6i = 0

The -12 in the equation means the flow is "going out" from the 12V battery while flows "into" the 4ohms resistor to the (+) side it is therefore a positive in the equation, did I get this correct ?

Click to expand...

You need to look at the polarity of the source. "Positive" voltage is depicted as current flowing from the - side of the source to the + side of the source (meaning out from the +, through the circuit, and back in through the -). Therefore, - voltage is shown by current flowing "out" from the -, through the circuit, and back in through +. Since it says the voltage is -12v, it means that current must be flowing from the + side of the source directly to - (not through the circuit--only the source itself). Considering that, the current is shown to flow in a CCW direction through the circuit ("out" the - and "in" the +). When moving that direction, and given the arrow that shows "i" in the CW direction, that means that everything must be opposite. That is why it is -12v and +6i--normally you'd say -6i, but since the given "i" is shown to be flowing in the opposite direction as the one shown by KVL, "i" is negative. Therefore, it is the same as the opposite of -6i, which is +6i.

DerStrom8 said:

You need to look at the polarity of the source. "Positive" voltage is depicted as current flowing from the - side of the source to the + side of the source (meaning out from the +, through the circuit, and back in through the -). Therefore, - voltage is shown by current flowing "out" from the -, through the circuit, and back in through +. Since it says the voltage is -12v, it means that current must be flowing from the + side of the source directly to - (not through the circuit--only the source itself). Considering that, the current is shown to flow in a CCW direction through the circuit ("out" the - and "in" the +). When moving that direction, and given the arrow that shows "i" in the CW direction, that means that everything must be opposite. That is why it is -12v and +6i--normally you'd say -6i, but since the given "i" is shown to be flowing in the opposite direction as the one shown by KVL, "i" is negative. Therefore, it is the same as the opposite of -6i, which is +6i.

Click to expand...

This is the diagram I draw based on your explanation, is this correct ?

View attachment 61877

Because the 'i' is pointing clock wise that's wise we have +4 + 2vo - 4v which is based on flow 1 in the diagram while the -12 + 6 is coming from flow 2 is this the correct assumption I'm making ?

Because the 'i' is pointing clock wise that's wise we have +4 + 2vo - 4v which is based on flow 1 in the diagram while the -12 + 6 is coming from flow 2 is this the correct assumption I'm making ?

Click to expand...

You are really getting wrapped around the axle because you are thinking of "current flow", right? Well, current does not flow. Charge flows. Current means charge flow and "current flow" means "charge flow flow". Current has a direction, but it does not flow twice.

Now, let's look at where you are going wrong. Assume the direction of the current is CW. That will determine the voltage polarity across the resistors, capacitors, and inductors if you had any. Voltage sources are unaffected by the direction of current. So don't try to determine the direction of the current by the polarities of the voltage sources or vice versa. In the diagram you posted, you show current going in both directions within a single loop. That never happens.

Now, pay attention to the way it should be done and read the following explanation carefully. Assuming the current is CW as given in the problem. Let's apply KVL CW. Starting with the independent source, we put the negative lead of the voltmeter on the bottom of the 12V source and measure 12 volts on the top of the 12 volt source. Then we put the neg lead on the left side of the 4 ohm resistor and measure -4*i on the right side of the resistor. Next, we put the neg lead of the left side of the 2*v voltage source and measure -2*v on the right side. Then +4 on the 4 volt source. Finally we put the neg lead of the voltmeter on the right side of the 6 ohm resistor and measure -6*i volts on the right side of the resistor. Adding up all the voltages we get 12 - 4*i - 2*v + 4 - 6*i = 0 . If you want to apply KVL CCW, then the voltages will be -12 + 4*i + 2*v - 4 + 6*i = 0 , which is the same equation with the signs reversed. If you assume a CCW current, then the voltages across the resistances will be opposite to the above two equations, but the voltages will stay the same. Do you understand how to apply KVL now?

Ratch

You can make this work using either direction for KVL. One will only have parts that are opposite from the other. Let's say you're setting up KVL in the clockwise direction, and we'll start with the 12v source. Since that is the direction the source is "pointing", we'll start with +12v. Now we come to a resistor, which gives you -4i. After that, you come to another source pointing in the opposite direction. This will give you a -2vo. Then there's another source, but it's pointing the same direction as the KVL "flow". Therefore, we'd add +4v. Finally, we come to another resistor, which is -6i. We set all that up and say it's equal to zero. You can solve for i using, say, Cramer's rule, and then use that to find any other desired values in the circuit.

Now let's do it the way the book did it. Instead of setting up KVL in a clockwise direction, we'll set it up to "flow" counter-clockwise. We start with the 12v source, and since it's pointing the opposite direction from the "flow" of KVL, we say it is -12v. Then we come to a resistor. Normally we'd say it's -6i, for a resistor, but the current flow is in the opposite direction. The sign is switched, and that gives us +6i. Then we move on to the 4v source, which is pointing against the KVL "flow". That means that it will be -4v. Next we move to the 2v source, which is pointing in the same direction as the flow, so we add +2vo. Then we finish with the last resistor. We use the same rule as before--you're going in the opposite direction, so the sign is switched from - to +4i.

Here is a diagram showing the problem using both directions for KVL:

**broken link removed**

The red arrows in the voltage sources are showing the polarity. If your flow goes with an arrow, you add +v. If it goes against an arrow, you use -v.

I hope this helps

DerStrom8 said:

Here is a diagram showing the problem using both directions for KVL:

**broken link removed**

Click to expand...

Hi DerStrom8,

I can't open the attachment it throws an error when clicking on the link.

Thanks

Ratchit said:

Now, pay attention to the way it should be done and read the following explanation carefully. Assuming the current is CW as given in the problem. Let's apply KVL CW. Starting with the independent source, we put the negative lead of the voltmeter on the bottom of the 12V source and measure 12 volts on the top of the 12 volt source. Then we put the neg lead on the left side of the 4 ohm resistor and measure -4*i on the right side of the resistor. Next, we put the neg lead of the left side of the 2*v voltage source and measure -2*v on the right side. Then +4 on the 4 volt source. Finally we put the neg lead of the voltmeter on the right side of the 6 ohm resistor and measure -6*i volts on the right side of the resistor. Adding up all the voltages we get 12 - 4*i - 2*v + 4 - 6*i = 0 . If you want to apply KVL CCW, then the voltages will be -12 + 4*i + 2*v - 4 + 6*i = 0 , which is the same equation with the signs reversed. If you assume a CCW current, then the voltages across the resistances will be opposite to the above two equations, but the voltages will stay the same. Do you understand how to apply KVL now?

Click to expand...

Hi Ratchit,

Thanks for the explanation I will sit down and try to draw it out so it's clear in my head, I'm a visual person more so it would be easier for me if I draw it

Cheers

nicksydney said:

Hi DerStrom8,

I can't open the attachment it throws an error when clicking on the link.

Thanks

Click to expand...

Hmm, you're right. Sorry about that. Let's try this one....

View attachment 61905

DerStrom8 said:

You can make this work using either direction for KVL. One will only have parts that are opposite from the other. Let's say you're setting up KVL in the clockwise direction, and we'll start with the 12v source. Since that is the direction the source is "pointing", we'll start with +12v. Now we come to a resistor, which gives you -4i. After that, you come to another source pointing in the opposite direction. This will give you a -2vo. Then there's another source, but it's pointing the same direction as the KVL "flow". Therefore, we'd add +4v. Finally, we come to another resistor, which is -6i. We set all that up and say it's equal to zero. You can solve for i using, say, Cramer's rule, and then use that to find any other desired values in the circuit.

Click to expand...

Hi DerStrom8,

Thanks for the diagram it was very helpful I'm able to grasp some of the things, but I have this question that is still bugging my head .

From your first diagram I understand how the arrow is drawn inside the +12V, 2vo and +4v and the direction of the flow, what I'm bit confused is since the 4ohms and 6ohms does not seem to be going against the flow how is it determined that it's -4 and -6 ?. I can understand why the 2vo become -2vo (because of the arrow direction going against the flow) ?

Thanks for your help.

Hello nicksydney
Assume a direction for current: clockwise in this case.
Assign the polarity or sign to the voltage source according to the terminal you encounter first in the direction you have assumed for current. Assign the voltage drop polarities across the resistors according to the rule: positive sign to the end that current enters the resistor.
So for the 12 volt source, the first terminal encountered in a CW direction is the negative terminal...therefore the term is -12
For the 4 ohm resistor the voltage is resistance x current; 4i, according to the rule, the term is +4i
For the dependant voltage source 2vo, the positive terminal is the first one encountered therefore theterm is +2vo
For the 4 volt source, the negative is the first terminal encountered so the term is -4
The 6 ohm voltage drop is 6i so according to the rule, the term is + 6i


Thus the equation becomes: -12 + 4i + 2vo - 4 + 6i = 0

But the diagram shows that the 6ohm resistor is connected across the output voltage. So vo has the same value as 6i, but with the opposite polarity.

vo=-6i

Substitute this value in the equation.

-12 + 4i + (2 x -6i) - 4 + 6i = 0

-16 + 10i - 12i = 0

-16 = 2i

i = -8A

Vo = -6i = -6 x -8 = 48v

Hope this helps as to the resistor polarities.

The assumptions as to current direction hold for the calculation process and may change when the result is known. That is the reason that the current we assumed as CW resolves to -8A, ie in the opposite direction to our assumption. But keep strictly to the assumptions throughout the calculation process.

Regards Phil

filbo said:

But the diagram shows that the 6ohm resistor is connected across the output voltage. So vo has the same value as 6i, but with the opposite polarity.

vo=-6i

Click to expand...

Hi Phil,

Thank you for your reply.

When I read your reply you mention the 6ohm resistor is connected across the output voltage are you referring to the 4V voltage source OR the 2vo ?

Thanks

No nick, the diagram in your post shows the 6 ohm resistor with + vo - shown below it.
Phil

filbo said:

No nick, the diagram in your post shows the 6 ohm resistor with + vo - shown below it.
Phil

Click to expand...

Ahh ok, so because of the + - symbol below the 6ohms that's how we can know that it is -6, so that means if in the diagram it does not show that symbol we use Vo as +6 correct ?

Maybe this PDF from steveb will help

nicksydney said:

Hi DerStrom8,

Thanks for the diagram it was very helpful I'm able to grasp some of the things, but I have this question that is still bugging my head .

From your first diagram I understand how the arrow is drawn inside the +12V, 2vo and +4v and the direction of the flow, what I'm bit confused is since the 4ohms and 6ohms does not seem to be going against the flow how is it determined that it's -4 and -6 ?.

Click to expand...

That is a rule for KVL. Resistors create a voltage drop when the KVL is set up in the same direction as the current flow (shown in your diagram as an arrow labeled 'i'). This will give you -4i and -6i. When going against the flow of current, you're going at it backwards, so it is a voltage rise (+4i and +6i).