After some in depth research, I have come down to an understanding of the ECG produced by RBBB as well as that of LBBB. However, I might be mistaken and please do correct me if I am wrong.
First, we must understand the fact that an ECG is measuring the movement of current at an instant of time, also known as the instantaneous vector of current. As soon as the current is moving exactly the opposite way to the positive electrode, the wave crosses the zero potential and moves to the negative side of the graph. We must also know that from the thickest to the thinnest wall we have the left ventricle, the septum, and then the right ventricle. This helps in understanding which wall depolarizes first, second, and last, or possibly during the same time, which is crucial to figuring out the instantaneous vector for the current.
Bear in mind, when I say "approaching to becoming perpendicular/parallel to the axis of V1", I do not mean that the movement of current at some point actually becomes perpendicular/parallel to the axis of V1, I just mean that it is APPROACHING towards becoming perpendicular/parallel with the axis of V1, it does not necessarily actually become perpendicular/parallel to the axis of V1. Remember also that if the movement of current is exactly parallel to the axis of V1, we get the maximum voltage recorded. On the other hand, if the movement is perpendicular to the axis of V1, the recording is zero voltage.
Let's first look at LBBB:
The septum first depolarizes from right to left, however, not fully just yet. As a result, on the ECG, we get an instantaneous vector of the current to be away from V1, hence we get a downward slope below the zero potential. There is a downward slope, indicating that the instantaneous vector is each time getting bigger with time (until the depolarization of the right ventricle begins). That is because the instantaneous vector is approaching to becoming less perpendicular to the axis of V1.
As the septum is depolarizing, the right ventricle begins to depolarize as well. Now we still have current moving in the septum from right to left but we also have current moving from left to right in the right ventricle. However, the instantaneous vector in the right ventricle is smaller than that of the septum, because, remember, the septum is thicker. As a result, the net instantaneous vector of current is still away from V1, but smaller than that of when only the septum was depolarizing because the current moving towards V1 in the right ventricle subtracts some of that magnitude of the vector moving away from V1 (the current in the septum). As a result, we have an upward slope (since the instantaneous vector is now getting smaller), but still below the zero potential (since the current is still moving away from V1). The upward slope is indicating that the instantaneous vector is getting smaller with time (until the left ventricle begins to depolarize). That is because the instantaneous vector is approaching to be perpendicular to the axis of V1.
The depolarization of the right ventricle and that of the septum finishes during the same time. Next, the left ventricle begins to depolarize from right to left, hence we get a downward slope on the ECG, because the current is moving away from V1. It is a downward slope because the movement of current is approaching to be parallel to the axis of V1. Lastly, we get an upward slope because the current is moving in a direction towards becoming perpendicular to the axis of V1. At some point, there is no current, hence V1 records zero potential.
Now, let's look at RBBB:
First, the septum depolarizes as normally from left to right. This is towards V1 and hence produces an upward slope above the zero potential on the ECG. We have an upward slope because the instanataneous vector of current is getting bigger with time (it is APPROACHING towards becoming parallel with the axis of V1).
Then the left ventricle begins to depolarize. The septum finishes depolarizing before the left ventricle finishes because the left ventricle is much thicker, even though the septum began earlier to depolarize. While the septum is depolarizing from left to right, the left ventricle is depolarizing from right to left, hence the instantaneous vector of current movement is still towards V1 but has a smaller voltage than the voltage that we had when only the septum was depolarizing because the net instantaneous vector is approaching towards becoming perpendicular to the axis of V1; the vector of current moving away from V1 subtracts from the vector moving towards V1 (depolarization of the septum). (Although I am confused here because shouldn't the net instantaneous vector be away from V1 because the left ventricle is thicker than the septum?)
Edit: As you can see, above, in brackets, in the end, I was confused about one thing but now I have gotten it clarified. So, let's get into it. First, the septum depolarizes from left to right and produces an upward slope above the zero potential line AND another downward slope above the zero potential line. I stated earlier that the downward slope above the zero potential is related to the depolarization of the LV and of the septum. However, that is not the case, and the downward slope is still related to only the septum. It is a downward slope because the instantaneous magnitude of the vector of the current is getting smaller since it is approaching to move less parallel to the axis of V1. Thereafter, the depolarization of the left ventricle begins, the septum has probably not completely finished depolarizing yet, but this does not affect the graph in the ECG since it does not affect the direction of the instantaneous vector of current because the left ventricle is thicker than the septum, hence the net vector is from right to left, and thus, the graph crosses the zero potential line into the negative side, producing a downward slope. Then we get an upward slope, which represents the movement of current in the left ventricle approaching to become less parallel to the axis of V1. Thereafter, the depolarization of the RV begins, which depolarizes from left to right, hence the graph crosses the zero potential line into the positive upper half, producing a positive upward slope. The movement of current in the RV is approaching to becoming perpendicular to the axis of V1 and at some point there is no current at all in the heart since depolarization at some point finishes, hence we get a downward slope reaching zero potential (still in the positive side of the graph). The upcoming part of my text does not have to be read since this edit corrects it, but just keeping it here to show what kind of a misunderstanding I had.
Then at some point in time the septum finishes depolarizing but we still have some left of the left ventricle that has not finished depolarizing, hence the net instantaneous vector of current movement is away from V1, giving us a downward slope below the zero potential. We have a downward slope because the movement of current is approaching towards becoming parallel to the axis of V1. As the left ventricle is depolarizing, the movement of current starts to approach towards becoming perpendicular to the axis of V1 and then reaches zero potential, either because it became fully perpendicular to the axis of V1, or because the movemet of current stopped for a moment. As a result, we get an upwards slope on the ECG, but still below the zero potential. After this, the right ventricle begins to depolarize, that is, from left to right, so towards V1, producing an upwards slope above the zero potential line. Then we start to get a downward slope above the zero potential line because the instantaneous vector of current movement is approaching to be perpendicular to the axis of V1, and finally there is no more curreny, reaching zero potential.
Phew! That was long, but worth it. I hope that this can help others and I really hope that my explanation is not wrong.
Used sources:
Guyton & Hall Medical Physiology
Acadoodle.com
