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I want to know how to find the voltage difference between two points A and B which is $V_{ab}$ by Kirchhoff's law ... Also what does it mean if the value is negative? What is the difference between $V_{ab}$ and $V_{ba}$ ? So if $V_{ab}$ is positive 4 for example , does that mean that the voltage at A is +4 higher than the voltage at b with respect to the ground ? Does that mean that $V_{ab} = v_b -v_a$ ?

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  • $\begingroup$ When you say that the potential difference between point A and B is $x$ volts, it is inferred that $V_B-V_A=x$. $\endgroup$ – Sam Feb 2 at 15:28
  • $\begingroup$ The meaning of $V_{ab}$ is not universal, but depends on the author. But if $V_x$ is the potential at $x$, then $V_{ab}=-V_{ba}$. Which on is positive depends on the author, and it should be specifically stated in the text. $\endgroup$ – Bill N Feb 8 at 4:49
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Also what does it mean if the value is negative?

If $V_{ab}$ is negative, it means that $V_a-V_b$ is negative. In other words, it means that point b is at a higher potential than point a.

What is the difference between $V_{ab}$ and $V_{ba}$ ?

$V_{ab}$ is $V_a-V_b$. $V_b$ is $V_b-V_a$.

So $V_{ab}=-V_{ba}$.

So if $V_{ab}$ is positive 4 for example , does that mean that the voltage at A is +4 higher than the voltage at b with respect to the ground ?

$V_{ab}$ can't be "4". It can be 4 volts, also written as "4 V". But the number 4 by itself doesn't express a potential difference.

And it doesn't matter what you use as a reference voltage. If $V_{ab}$ is 4 V, it's 4 V whether you measure $V_a$ and $V_b$ with respect to ground or with respect to a certain battery terminal, or with respect to any other arbitrary point you like.

Let's choose any arbitrary point in the circuit as the reference and designate its potential as $V_r$. Now if we measure our other with respect to that point, we have

$$V_{ab} = (V_a-V_r) - (V_b-V_r)$$

But since we have a $V_r$ and a $-V_r$ being added together in this expression, the reference turns out not to matter at all and we just have

$$V_{ab} = V_a - V_b$$

Does that mean that $V_{ab} = v_b -v_a$ ?

The opposite.

$V_{ab} = V_a - V_b$

Now, back to your first question,

I want to know how to find the voltage difference between two points A and B which is $V_{ab}$ by Kirchhoff's law

You need to combine Kirchhoff's laws with the constitutive relations for each circuit element to be able to solve the circuit.

If you use Kirchhoff's Current Law, you can perform nodal analysis to get a complete set of equations from which you can solve the circuit.

If you use Kirchhoff's Voltage Law, you can perform mesh analysis to get a complete set of equations from which you can solve the circuit.

You can look online and find numerous tutorials on how to do either of these methods.

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  • $\begingroup$ Thank you for your help ❤️ $\endgroup$ – Anas Alaa Feb 7 at 15:52
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If you have a circuit and enough information about it then you may be able to deduce the voltage between $A$ and $B$ using Kirchoff's laws. There is no certain simple answer to this as it depends on the other components and what you know. E.g. if you had a battery connected to three resistors in series and you knew the battery's voltage and all of the resistances then you could deduce the voltage across each resistor. However, if you did not know the battery's voltage or all of the resistances then you could not.

Yes, if $A$ is $+4V$ with respect to $B$ then $B$ will be $-4V$ with respect to $A$. This means that conventional would flow from $A$ to $B$ if they were connected by a conductor. If the charge carriers were electrons (common bit not universal) then they would flow the other way.

There is no absolute zero voltage, you need to nominate a point that you will call zero. Often, but not always, this is ground. An example when this might be so is an old TV from the valve / tube age. Some of these rectified the mains without a transformer so the chassis was not at the same voltage as the neutral or ground; it was not safe to touch. For the purposes of the circuits inside the TV, the chassis was $0V$ even though this was not ground.

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  • $\begingroup$ Thanks for your time and help $\endgroup$ – Anas Alaa Feb 7 at 15:53
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To understand the reason for the difference in signs, it may be helpful to consider the following definition of potential difference, or voltage $V$.

The potential difference, $V$, between two points is the work required per unit charge to move the charge between the two points.

By convention, the direction of the electric field is the direction of the force that a positive charge would experience if placed in the field. This means that in order to move a positive charge against the direction of the electric field an external agent has to do positive work on the charge. Moving positive charge against the direction of the field increases its electrical potential and potential energy. It is analogous to the increase in gravitational potential energy when raising a mass against the gravitational field. When the charge is moved in the direction of the electric field, the electric field does work and there is a decrease in electrical potential and potential energy.

Now, in terms of Kirchoff's law consider the resistor below. The current flows from A to B, in the direction of the electric field. The field does work moving charge from A to B. This results in a drop in electrical potential. Therefore $V_{AB}$ = $-IR$. That means $V_{BA}$ = $-V_{AB}$ = $+IR$. External work is required to move positive charge from B to A against the electric field.

In applying KVL going around a loop in the direction of current flow, the voltage across R (A to B) is $-IR$, a voltage drop.

Hope this helps.

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  • $\begingroup$ I really appreciate your help thanks $\endgroup$ – Anas Alaa Feb 7 at 15:52

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