How can a voltage follower's output have any current through it?

Question

Voltage follower circuit is used to create isolation between two different kind of circuits. Due to high input impedance,, so the input current is much lower than the output current while the output voltage follows the input voltage. So the voltage follower provides large power gain across its output. Due to this behavior, Voltage follower used as a buffer circuit and can be used to isolate stages while building multistage filters or some other multistage circuit.

This quote is from this webpage.
I think I understand clearly the purpose of a voltage follower circuit, and this quote explains it pretty well.
There is one key aspect that I can't seem to grasp and I can't find the info anywhere:
The part of the circuit which is connected to the output consumes current.
Where is this current coming from? Since the battery is connected to the input of the op amp, it is hi-z and there is no current possibly flowing out of the battery.
How can a current form from the battery to the output stage of the circuit?
To my understanding, the output load would have the battery's voltage, but should not be able to operate because of the high impedance input being the only connected component to the battery.
It doesn't seem to make any sense.

vangelo · Accepted Answer

The output current is provided by the voltage source, which is connected to the pins 4 and 7 of the op. amp.
Ignoring the negligible input currents, these are the relevant ones:

Notice that B1 provides current to the potentiometer RV1, which acts like a voltage divider, and B2 provides power to the op. amp. (and its output load).
As noted, this is a real op. amp., so there is negligible currents at the inputs. There are also other non-ideal characteristics, like, e. g., the model in your post is not a rail-to-rail component, so the inputs and the output do not work using the full power source voltage range.

How can a voltage follower's output have any current through it?

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