A

Step 1: Understand the photoelectric effect, which states that when light of a certain frequency hits a metal surface, it can cause the emission of electrons. The energy of the incoming photons is given by the equation: $$E = h u$$, where $E$ is the energy of the photon, $h$ is Planck's constant, and $ u$ is the frequency of the light.

Step 2: The intensity of light is related to the number of photons incident on the surface per unit time. If the intensity increases while keeping the frequency constant, it means that more photons are hitting the surface.

Step 3: As the number of incident photons increases, more electrons are emitted from the surface, leading to an increase in the photo-current (the current due to the flow of emitted photoelectrons). The frequency of the light determines the kinetic energy of the emitted photoelectrons, but since the frequency remains constant, the kinetic energy of individual photoelectrons will not change.

Conclusion: Since the number of emitted photoelectrons increases with the intensity, the photo-current also increases.
Therefore, the correct answer is A: photo-current increases.