Saturation current in photoelectric effect

While studying photoelectric in my school, my teacher drew a graph of current versus the potential difference across the two electrodes:

I am not able to understand why do we get saturation current. I know that at saturation current all the electrons emitted from the metal surface are able to reach the other electrode. My question is suppose we increase the potential difference between the two electrodes further. I think that one of the following two things should happen:

1. 
   

   current may increase as now the electrons in the wire will be under greater potential difference thereby reducing the time in which the electrons complete one round to reach the same point again and hence increasing current. ($V=IR$)

   
   


2. 
   

   current may not be continuous (steady current might not be achieved) as the time taken by the photoelectrons to reach the positive electrode may be more than the time taken by the electrons inside the wire to go from end of the wire nearer to the positive electrode to the other end.

   
   

So in no case should we get a graph as shown in the link.

It may be that I have not understood the photoelectric effect completely. If that is the case please tell me where am i wrong?

Answer

In the photoelectric effect, photons incident on the cathode cause the emission of electrons. Assuming there is a sufficient electric field, these electrons will make their way across to the anode, contributing current.

For simplicity, let's assume every photon generates a photo-electron. Then if $N$ photons per second hit the cathode, the current will be carried by a total of $N$ electrons per second. We always assume there are "infinitely many" electrons waiting for their turn, and the thing limiting the current is how many electrons get "released" from the cathode (i.e. how many photons hit the cathode).

Current is charge per unit time. If the electron has a charge $q_e$, then $N$ electrons per second carry a current

$$I = N\; q_e$$

There is nothing here about the velocity of the electrons... not about the time it takes them to cross the gap. If they went 100 times faster, it would not change the number of electrons crossing the gap per second. That number is determined by "how many start the trip per second" and "how many don't make it". The second of these explains that the curve starts out not completely flat: very slow electrons may not make it, especially with a small retarding potential. But once they go fast enough to fully escape, their final speed really doesn't matter. And neither does the transit time in the wire.

Did you ever calculate how slowly electrons move in a current carrying wire (for example, copper wire)? While the electrical signal is very fast, the drift velocity of the electrons themselves is very very slow... because there are so many electrons per unit volume. But that is only tangentially relevant here.