Hi

Please have a look on the following linked diagram:
http://img32.imageshack.us/img32/1456/img0001vsi.jpg

While applying KVL it is not necessary to have the loop traversing direction and the current direction the same. e.g. the loop could be traversed CW and the current direction could be CCW. But in some situations having the directions for the both current and loop could make things easier such as in mesh analysis. Further, the direction of current defines the polarities of the resistors because the current flows from higher potential toward the lower. If a resistor is traversed in the same direction as the current, then the IR term would have -ve sign, otherwise it would be +ve.

In the linked diagram if we star traversing the loop from point "a", then the equation would be:

+E - IR1 -IR2 = 0

Please have a look on the link to notice another important point about the KVL: http://img97.imageshack.us/img97/9497/img0001xj.jpg

 

As mesh analysis is also an application of KVL, therefore what is said about KVL above is also applicable while using mesh analysis. For example, in the linked example, current directions and loop traversing directions were chosen arbitrarily. But answer was still correct. It should be noted that equations might appear simpler if there you use similar approach for all loops. e.g. If you traverse the loop in the same direction as the current in all the loops and all the loops have current running either CW or CCW, then it might make things simpler to solve.

Link: http://img442.imageshack.us/img442/3328/img0001ta.jpg

 

Hi

I don't know why I'm little confused about this. The sum of potential differences along a closed loop is zero. Here the sum of potential differences is "-10" instead. We started traversing the loop from the point "b" and ended up at "b" again and the potential at "b" is -10V. So, should we say that sum of potential differences along a closed loop is equal to potential difference of the point which is used starting and ending point. If you don't mind, please let me know this. Thank you.

Regards
PG

 

Hi

I don't know why I'm little confused about this. The sum of potential differences along a closed loop is zero. Here the sum of potential differences is "-10" instead. We started traversing the loop from the point "b" and ended up at "b" again and the potential at "b" is -10V. So, should we say that sum of potential differences along a closed loop is equal to potential difference of the point which is used starting and ending point. If you don't mind, please let me know this. Thank you.

Regards
PG

Potential difference and potential, are not quite the same thing. Potential can be referenced to any point, such as ground. In your circuit, you may want to use ground as the reference and say that the node b is at a potential of -10V. However, when you use KVL, you use potential differences; hence when you go from b to a, there is a potential difference of 10V.

 

Thank you.

Although I don't feel comfortable to make such queries about KVL at this stage, it's still better to clarify it.

In this picture we first talk about the second to last diagram. When I go from a to b, there is a rise of -10V ("a" is at 0V and "b" is at -10V). As we are traversing the loop in the direction of current, therefore there is potential drop across 2 ohm and 6 ohm resistors. We start from a and end at a.

Now come to the last diagram. When I go from b to a, there is change in potential of 0V because b is at -10V and a is at 0V, and again as we are traversing the loop in the direction of current therefore there is potential drop across 2 ohm and 6 ohm resistors. As we started at b and ended at b, therefore we will end up with the -10V.

Please excuse my verbosity. And let me know where I'm going wrong. Thank you.

Best wishes
PG

 

When I go from b to a, there is change in potential of 0V because b is at -10V and a is at 0V,


Best wishes
PG

Hmm,

Vab = Va - Vb = 0 V - (-10) = 10V

 

Yep, Jony130 has it right and shows the mathematical way to express what potential difference is.

As I said, potential difference is not quite the same as potential. The potential is a number we assign to every node in the circuit, with reference to another point. It is in fact the potential difference between the node and another "special" node. Any node can be called the "special" node because potential is itself physically meaningless. Only potential differences matter physically. So we typically label one special point as zero potential or ground (Vg=0), and then say that potential at any other point (say node a) is the difference from that point to ground. So whenever we are talking about the potential at the "a" node, we really mean Va-Vg, but this is of course equal to Va. The correct terminology is that Va is the potential difference relative to ground, but we make a shortcut and just say "potential" and all know what that really means.

As Jony130 showed you, potential difference is the difference in potential between one node and another "non-special" node. The potential difference of a relative to b is Vab=Va-Vb. The notation Vab, and its meaning as defined here, is an accepted convention.

 

Thank you, Jony, Steve.

I understand what Jony is saying. Actually I remember months ago Jony told me why we always need a reference point to measure potential differences.

Okay. Vab is applicable when the current is flowing from "a" to "b".

In a closed loop of a circuit algebraic sum of all the e.m.f.s is equal to sum of potential differences across all the resistors. Now I would like to repeat what I said earlier.

In this picture we first talk about the second to last diagram. When I go from a to b, there is a rise of -10V relative to the ground ("a" is at 0V and "b" is at -10V). As we are traversing the loop in the direction of current, therefore there is potential drop across 2 ohm and 6 ohm resistors. We start from a and end at a.

Now come to the last diagram. When I go from b (which is -ve terminal of battery) to a, there is change in potential of 0V relative to the ground because b is at -10V and a is at 0V, and again as we are traversing the loop in the direction of current therefore there is potential drop across 2 ohm and 6 ohm resistors. As we started at b and ended at b, therefore we will end up with the -10V.

I wrote 0 - 6I -2I = -10. I'm primarily confused about putting the expression equal to "-10". I say that I started the traversing the loop from point "b" which was at potential "-10V" so at the end I should also end up with -10V but that contracts the statement that sum of potential differences along a closed loop is zero.

Please note that I'm not trying to contradict you in any way but I'm just trying to come out of my own confusion and fill the gap in my own understanding, and please forgive my ignorance and don't get frustrated! Thanks.

Best wishes
PG

 

When we go from b to a we have potential difference equal to Vab = 10V
And when we go from a to b we have potential difference Vba = - 10V

 

When I go from a to b, there is a rise of -10V relative to the ground ("a" is at 0V and "b" is at -10V).

Note the mistake in terminology here. When you go from a to b, there is a rise of -10V. That is all you need to say. There is no "relative to ground" to speak of. I think this is where you are getting confused. Vab is the potential at node-a relative to the potential at node-b. We really don't care what Va and Vb are referenced to. Each can be relative to ground, or relative to potential at infinity, or relative to potential on the planet Mars. The only requirement is that each potential is referenced to the same point.

Let's try it with Va and Vb relative to that on planet Mars.

Vm is the potential on planet Mars
Vam is potential at node-a relative to that on Mars, so Vam=Va-Vm
In notation we say Va=Vam

Do the same for node-b and Vb=Vbm=Vb-Vm

Now what is Vab?

Vab=Va-Vb=Vam-Vbm=Va-Vm-(Va-Vm)=Va-Vm-Vb+Vm=Va-Vb

This is all very confusing looking, but see how it doesn't matter what Vm is?

So, when we talk about the potential difference Vab, it is not relative to ground or any other point. It has meaning only in reference to the node-a and node-b only.

 

Thank you very much, Jony, Steve.

I think I understand it now. But it's surprising my approach works the same too. Even here in the last KVL equation in blue I have applied KVL differently but ended up with the correct answer. I was looking at a problem I was working just now. Even in it I applied the KVL the same way. I understand that we can get the same form (which we will get by using the method you both described) by shifting the RHS member of the equation to the left but to me conceptually there is a huge difference. In other words, conceptually I was doing it wrongly though I always end up with the correct answer. Well, correct answer is all that matters!

Many thanks.

Best wishes
PG