Photo-Current vs Potential Graph What I don't understand is that when we increase the value of potential why does the photocurrent reach a saturation point? Because when we increase the potential the velocity of the electrons increases so they reach the collector plate faster, that is the number of electrons reaching the collector plate increases in a smaller time period so shouldn't that increase the photocurrent?
 A: In a photoelectric tube the number of electrons leaving the surface of the emitter plate depends on the brightness of the incoming light (the photon density).  Many of these are not directed toward the collector and end on the inner surface of the vacuum tube. Increasing a voltage which accelerates them toward the collector increases the fraction which reach the collector (until you are getting them all).  Applying a reverse voltages pushes the electrons back toward the plate. The reverse voltage which gives zero current determines the maximum energy of the emitted electrons.
A: Yes, you are correct that they will reach faster at the other end but it's doesn't affect the photo current. Imagine a cross sectional area.Let's assume that we have already reached saturation point.
Imagine 100 electrons are the maximum electrons that can pass through that imagined cross sectional area.
Now if you increase the potential at any instant still 100 electrons cross that same cross sectional area but now the difference is they just have  more velocity at that instant.
Now current is defined as $I=Ne$
Where:
e: Charge on an electron
N is the number of electrons passing per second.
Now we can see it's independent of velocity. That's why photocurrent remains same.
I hope it makes sense.
