If the frequency of the incident wave is increased, then the kinetic energy of the photoelectrons increases. If so, why doesn't the photoelectric current increase?
If the kinetic energy of electrons increases then their velocity also increases, if their velocity increases then the number of electrons passing an area in 1 second should also increase, i.e. the current should increase.
In the context of the photoelectric effect, the key thing to remember is that you get one electron per photon. Current is a measure of how much charge flows per unit time, which is proportional in this case to the number of electrons per unit time, and therefore the number of photons per unit time. The energy of the electrons doesn't come into it at all.
Your intuition may be that faster-moving electrons must imply more flow of charge, much like a river that starts moving faster will result in a greater flow of water. However, this is only true if the density of the fluid is held constant. In the case of a flowing river, this happens more or less automatically: if there were gaps in the river, water would immediately flow in to fill those gaps. But that is not the case with the electrons ejected by the photoelectric effect. The faster the electrons are moving, the more distance each one covers before the next is ejected, which means that the electrons are more spread out, i.e., the density is lower. The lower density and the higher speed combine to give a constant rate of flow, as it must do, by the logic in the first paragraph
I think it's worth describing a couple examples where the original poster's logic does apply. Then I'll answer the question by (a) distinguishing those earlier scenarios from the photoelectric effect, (b) explaining why that earlier logic doesn't apply to the photoelectric current, and (c) deriving the equation for photoelectric current.
There are many contexts where increasing the speed $v$ causes the current $I$ to increase. As a first example, imagine that there are $n$ charge carriers per cubic meter traveling through a wire of cross-sectional area $A$. The charge of the charge carriers is $q$, and their drift speed is $v$. The current is given by $I = nAqv$, and thus the current $I$ is directly proportional to speed $v$ as long as $n$, $A$, and $q$ are constant. As a second example, if you send a proton around a particle accelerator $5,000$ times per second, the current is $I=(5000/\textrm{sec})(1.6\times10^{-19}\textrm{C})\approx8 \textrm{ atto amperes}$. But if you send the proton around a particle accelerator $10,000$ times per second, the current is twice as big, $I\approx16 \textrm{ atto amperes}$.
Here's why that drift speed equation and that logic don't apply to the photoelectric current: the photoelectrons are halted every time they completely one lap. Consider an electron that is ejected with high $KE$ from the metal plate, then moves around the circuit really fast, and finally returns back to the metal plate in a very short time. That electron won't be ejected again until another photon strikes it. So the photon density regulates the current, and electrons are ejected only as frequently as the photons arrive at the metal plate. If, somehow, the electrons were able to continue whizzing around the circuit as fast as they were originally ejected (without ever slowing down or being stopped), then higher speeds $v$ would produce higher currents. However, the ejected photoelectrons do not move with constant speed around the circuit, and thus the high initial ejection speed does not indicate how many laps the electrons make per second.
Similarly, the reason we can't look at the equation $I=nAqv$ and say "high speed $v$ implies high current $I$" is that the values in this equation are not constant. In the vacuum tube, the speed $v$ would be higher and the density $n$ of the charge carriers would be lower. In the wire where the electrons get "backed up," the speed $v$ would be lower and the density $n$ would be higher.
$$\textrm{power}=\frac{\textrm{energy}}{t}$$
But the total energy carried by the light is equal to the number of photons multiplied by the energy per photon:
$$\textrm{energy}=nhf$$
Substituting gives:
$$\textrm{power}=\frac{nhf}{t}$$
$$\frac{P}{hf}=\frac{n}{t}\textrm{ }\textrm{ }\textrm{ }\textrm{ Eq. 1}$$
The quantity $\frac{n}{t}$ is the number of photons striking the metal plate per second. Only some of the incident photons eject electrons. Let's say the fraction of incident photons that actually produce photoemission is $k$. The photoelectric current $I$ is:
$$I=\left(\textrm{charge of electron}\right)\times\left( \textrm{number of electrons ejected per second}\right)$$
$$I=\left(e\right)\left(k\frac{n}{t}\right)\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ Eq. 2}$$
Substituting Eq. 1 into Eq. 2 gives:
$$I=\frac{ekP}{hf}$$
where $I$ is the photoelectric current, $e$ is the elementary charge, $k$ is the fraction of photons that produce photoemission, $P$ is the power of the incident light, $f$ is the frequency of the incident light, and $h$ is Planck's constant.
It's the number of released photoelectrons per second that determines the current. So if you increase the frequency, but leave the number of incident photons constant, the velocity of the electrons increase, but the number of released electrons per second stays the same. If you increase the intensity of your light wave, that is you increase the number of incident photons, you will see a rise in the photocurrent.
