question about amps

[shane_95]

I have a 9V 2A transformer, i hooked it up to my 12v 34mA fan. my question will than only draw the 34mA and leave me with 1.966A?

[Classicmacintosh]

If the fan is a brushless DC type then current draw will most likely be reduced to some extent, probably to roughly 25-30mA.

[vk6zgo]

I have a 9V 2A transformer, i hooked it up to my 12v 34mA fan. my question will than only draw the 34mA and leave me with 1.966A?

Sorry,Shane,but you have a strange way of phrasing questions.

As Classicmacintosh said,the motor will probably draw a bit less at the reduced voltage,however,for the sake of argument,we will go with 34mA.
With no other load connected to the transformer, that will be the only current in the circuit.
The transformer can now,if a load drawing 1.966A is connected in parallel with the motor,deliver that current as well,for a total draw of 2A,
On that basis,you can say that your transformer after supplying the motor,still has the capability of supplying a further 1.966A.

[T4P]

I have a 9V 2A transformer, i hooked it up to my 12v 34mA fan. my question will than only draw the 34mA and leave me with 1.966A?

lol no.
Unless it's a DC-DC Conv then will the V=I/R law apply.
So it will draw less than 34mA at 9V, because the literal resistance at 12V is 352ohms
so at 9v it draws only 25.5mA
And so the transformer is loaded at 25.5mA leaving the rest possible for another device
BUT it only applies if it's at 9V ... if it's a linear unregulated transformer you see more than just 9V

[digsys]

I have a 9V 2A transformer, i hooked it up to my 12v 34mA fan. my question will than only draw the 34mA and leave me with 1.966A?

Another way to explain this is -
P=VxI so the transformer can supply 18VA of Energy (AC). If you add a bridge to make DC, you still have the SAME ENERGY available,
(- the diode loss of the bridge, which we'll ignore for now) BUT VDC= 1.4X VAC and IDC = IDC / 1.4 (again, forget losses for now)
So for simplicity, you have 18VA (or WATTS) of power available.
The fan draws 9VDC x 0.04A = 0.35W (close enough), which leaves you with 17.65W available BEFORE the Transformer OUTPUT starts
to drop, and heat up. You CAN do this with just AMPS, as the others have suggested, both methods are fine.
SOMETIMES, it is easier to visualise what you have available using POWER. You need to be fluent with the primary relationships -
V=IR, P=VI and the variations, you'll use these ALL the time !!