How to draw constant current independent of voltage for capacitor testing

The current when charging a capacitor is not based on voltage (like with a resistive load); instead it's based on the rate of change in voltage over time, or ΔV/Δt (or dV/dt).

The formula for finding the current while charging a capacitor is:

$$I = C\frac{dV}{dt}$$

The problem is this doesn't take into account internal resistance (or a series current-limiting resistor if you include one) or if the capacitor already has some charge.

You have to account for the continually changing charge being applied to the capacitor. In other words, at the very beginning, it looks like a short circuit to your power supply (barring resistance, again). Thus, whatever maximum current your power supply can handle is the theoretical max current. As the capacitor charges, this current decreases exponentially, until the capacitor reaches max charge Q.

The formula for this is:

$$I = \frac{V_b}{R}e^{-t/RC}$$

Where \$V_b\$ is the source voltage, R is resistance, t is time and RC is the time constant (product of resistance and capacitance).

Let's say you don't use a current-limiting resistor and your power supply has an internal resistance of 4Ω:

$$I = \frac{12}{4}e^{-0/0.0132}$$

At time 0 s, the current is 3A. If we figure for, say, 1 ms later:

$$I = \frac{12}{4}e^{-0.001/0.0132}$$

Now the current is ~1 A.

So, how long will it take to charge the capacitor? If you take the time constant, RC (the 0.0132 in the exponent) as a value in seconds, there's a rule of thumb that a capacitor will be charged in 5 times this duration:

$$5\cdot0.0132 = 0.066s$$

The initial current (or the current during some portion of this duration) is referred to as the inrush current. You may want to reduce it by adding a series current-limiting resistor to protect your power supply.

The picture in your question assumes that the voltage waveform started some time earlier and that the transient of it beginning is no longer affecting things.

Basically Q=CV and this translates to I = C dv/dt and, if you applied a sinewave the differential of that sinewave voltage gives rise to the cosine wave of current but, of course at t=0 things are a little different; For a start you can't suddenly start a sinewave from rest - that would imply infinite bandwidth. Given this fact, there is a small finite time which the current rapidly ramps up to the starting value in your picture. From thereon it pretty much follows the equation given above.

EDIT section, mechanical analogy

A mechancial analogy could be regarded as a flywheel i.e. a rotating mass. The force applied to the end of the flywheel will accelerate the speed at which the flywheel rotates but when the flywheel (lossless assumed) is at constant speed, no force is needed. You can imagine the flywheel speed like voltage; the flywheel has charged up to speed n and there is no longer any force needed to keep it charged at that speed. Just like a capacitor, once charged to a constant voltage there is no current needed to keep a perfect capacitor at that voltage.

However, if you applied a constant force to decelerate the flywheel, the speed decelerates linearly and if the constant force is a true constant force, the flywheel speed will decelerate through n=0 and start rotating in the opposite direction after a little while. Force is -X and speed ramps down linearly. Ditto with the capacitor, if you take a constant current from the capacitor the voltage falls linearly and eventually becomes negative and charges up to a negative voltage.