I'm having some trouble understanding what a voltage-independent current source is. How can you have a current without a voltage? As I understand it, voltage or EMF is the force that drives the electrons to move, thus making a current.
[Physics] How does a voltage independent current source work
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What flows is not the voltage but the charge, and that flow is called current. There can be voltage without a current; for instance if you have a single charge, that charge induces a voltage in space, even if it's empty. Voltage, in the most physical way, is a scalar field that determines the potential energy per unit charge at every point in space.
Now, you can't have currents without voltages because if there's a current there's a charge moving, and every charge produces a voltage, but you can have currents without voltage differences in space. For example, if you have a charged sphere, and you make it rotate, the charge will be on the surface and by rotating the sphere you will have a current on the surface, but the voltage is the same at every point of the surface. Also magnetization of materials can induce currents by the same way.
If you introduce a charge in a circuit without a voltage it just doesn't move?
That's true, it won't move, unless you have some changing magnetic field that may introduce "voltage differences" between the same point, making $\nabla\times E\not =0$, although that wouldn't be electrostatic voltage the way you're seeing it.
The answer to your question lies in the fact that you are dealing with two different types of electric field (conservative and non-conservative) and that the non-conservative electric field owes its existence to a changing magnetic flux produced by a changing current.
The definition of self-inductance is $L=\dfrac {\Phi}{I}$ where $\Phi$ is the magnetic flux and $I$ is the current.
Differentiating the defining equation with respect to time and then rearranging the equation gives $$\dfrac{d\Phi}{dt} = L\dfrac{dI}{dt} \Rightarrow \mathcal E_{\rm L} = - L\dfrac{dI}{dt} $$ after applying Faraday's law where $\mathcal E_{\rm L}$ is the induced emf produced by a changing current.
The electric field associated with the changing magnetic flux is non-conservative.
Consider a circuit consisting of an ideal cell of emf $V{\rm s}$, a switch and an ideal inductor all in series with one another.
At time $t=0$ the switch is closed.
The initial current must be zero which you can understand with an appreciation of the fact that mobile charge carriers have inertia and thus cannot undergo an infinite acceleration.
The conservative field produced by the cell is trying to increase the current from zero but the non-conservative field produced by the inductor is trying to stop the current changing.
Which field wins?
At $t=0$ there is no current so it would appear that it is a draw between the two fields, but the non-conservative field can only stop a current flowing at $t=0$ on condition that the current changes.
So the current has to increase despite the opposition of the non-conservative field and so it continues with the current increasing due to the conservative field despite the opposition of the non-conservative field.
All that the non-conservative field can do is slow down the rate at which the current changes; it can never stop the current changing as then it (the non-conservative field) would no longer exist.
For this example the current $I = \dfrac{V_{\rm s}}{L}\,t$ and the energy delivered by the battery $\dfrac 12 V_{\rm s} I t = \dfrac 12 \dfrac{V_{\rm s}^2t^2}{L}$, is equal to the energy stored in the magnetic field produced by the inductor, $\dfrac 12 L I^2 = \dfrac 12 \dfrac{V_{\rm s}^2t^2}{L}$, and is the area under the power against time graph (shaded green).
Best Answer
A current source is a device that will (ideally) provide a constant pre-specified current to the connected load no matter what the load's resistance is, so that it will adjust the voltage across its terminals to get the current right (instead of the other way around as in a(n ideal) voltage source).
Of course, this is only ever approximately possible, and the quality of that approximation will depend (as with voltage sources!) on how much is known or expected about the load resistance and how fancy the device is -- see for example wikipedia.