6.13 Potential difference between two points in a circuit from Office of Academic Technologies on Vimeo.

**6.13 Potential difference between two points in a circuit**

Let’s consider a simple circuit which consists of a power supply, an electromagnetic force, let’s say a battery, such that the potential difference between its terminals is equal to 10 volts. Let’s connect this to a resistor with resistance of 5 ohms. And let’s also use an ammeter, which measures the current to the circuit. As soon as we turn the switch on, of course from Ohm’s law, since *R* is equal to *ε* over *i*, we will expect a current of 10 volts divided by 5 ohms, which is equal to 2 amperes, to flow the circuit.

But, in reality, from the ammeter, the value that we’re going to be reading will be slightly less than 2 amperes, indicating an existence of another resistance in the circuit. That resistance is basically coming through what we call the internal resistance of the battery. And it is, in general, represented with this symbol. And therefore, our EMF unit will be consisting with this internal resistance.

Now, whenever the current is not flowing through the power supply, in other words, whenever the power supply EMF is not pumping the charges, then the potential difference between its terminals is 10 volts. But, whenever we introduce the battery to the circuit by connecting to the ends of the resistance, *R*, then we will be introducing the internal resistance of the battery connected in series form to the resistance, *R*. In other words, *R* equivalent (*R**eq*) in this circuit is going to be equal to little *r* plus the big *R*.

So, in the ideal circuit, or in the case of the ideal battery, whenever we turn the switch on, we are going to be having a current of *ε* over *R* flowing through the circuit. But in this case, *i′* is going to be equal to, again, the potential difference across the resistor, which is *ε*, divided by the total resistance of the circuit, and that’ll be *R* plus *r*. Since the denominator is getting larger, therefore the current drawn from the power supply, or from the electromotive force, will be relatively smaller in comparing to the ideal battery case. So, if the current *i* is flowing through this system, then *i′* is going to be flowing through this system.

From here, let’s say we’d like to determine the potential difference between any two points in our given electric circuit. Let’s say we’re interested in the potential difference between these two points. To be able to do that, we’re going to start from the first point and trace the change in potential as we move to the other point along any one of the possible branches, in other words, either going through this way or through this branch.

Let’s say we start at point *a*, so it has the potential of *V**a*. And then, first go through this path, and the current is flowing from positive towards the negative end. So, as we go through this branch, we will cross the resistance *R* in the same direction with the flow of direction of current. Therefore, the potential will decrease by minus *i* times *R*. Then, we will eventually end up with the other point that we’re interested.

Therefore, this total potential should be equal to the potential at point *b*. So from here, we can say that the potential difference point *a* and point *b*, *V**a* minus *V**b*, will be equal to *i* times *R*. Or, we can express this as *V**a* minus *V**b* is equal to *i* in explicit form. Actually, in this case, we call this one as *i*′, *ε* over *R* plus *r* times *R*.

Now, if we choose the other path, let’s say this is path 1, and that is following through this branch, and if we choose the other path going from *a* to *b* along the second available branch, then we can write down our equations as — let’s say this is the path 2. For the path 2, starting from the first point, we have *V**a* and now we’re moving in an opposite direction to the direction of flow of current. Therefore, the potential will increase by *i* times the internal resistance *r*.

And now, we will cross the EMF in opposite direction to the direction of EMF arrow, from positive to negative. Therefore, the potential will decrease by *ε* volts. Now, we ended up with the other point. Therefore, some of the changes in potential should be equal to the potential at point *b*. Rearranging this expression, we will have *V**a* minus *V**b* will be equal to *ε* minus *i* times *r*.

Again, if you write down the current in its explicit form, we will have *ε* minus *ε* over *R* plus *r* times little *r*. Having common denominator on the right hand side will give us *ε* times *R* plus *ε* times little *r* minus *ε* times little *r* divided by *R* plus *r*. *ε* times little *r* will cancel, and that’s going to give us *ε* times *R* over *R* plus *r*. So, you can easily see that we have exactly the same result as the potential difference between point *a* and point *b* by going from point *a* to point *b* through these two different ways. And this is the general procedure that we apply to be able to calculate the potential difference between any two points in a given electric circuit.

So, if we state the procedure over here, we can say to find the potential difference between any two points in a circuit, start at one point and traverse the circuit to the other point, following any path and add, algebraically, the changes in potential that you encounter.

Let’s go back to our circuit and cut it open just right before the battery over here at point *b*. If we do that, it will look like this. We start from the point *b* and here is the battery, seat of EMF, or the electromotive force. Its internal resistance, *r*, and then moving on, resistance *R* and here is the point *a*. And eventually, again, we’re going to end up with the point *b* right here.

Well, if we look at the change in potential as we go as we start from this end towards the other end of this circuit, I’m marking each component along this branch. At point *b*, we have the potential of *V**b*, and that potential does not change until we end up with the electromotive force. And that electromotive as it pumps its charges from the negative terminal to the positive terminal, it will cause the potential to increase to *ε* volts. And then, potential will remain at that level until the charges go through the internal resistance of the electromotive force. That internal resistance will cause the potential to decrease by –*i* times *r*.

**So you know that the current is flowing from positive end towards the negative end along this branch. Once the charges emerge from the internal resistance part, then the potential remains at that level, which has the same potential as this point a over here. Then, whenever the charges enter into this resistance, R, then the potential decreases further by –i times R, i times big R, down to the potential Vb. That’s how the potential will change as we move in the direction of current flow along that circuit. Again, the units over here represent the actual battery. And, if we neglect the internal resistance of the battery, then we call that battery as “ideal battery”.**