# Does Clarke transformation convert scalar quantity to vector quantity?

Current is a scalar quantity, is it converted to vector quantity using Clarke's transformation? If it is correct, I would like to know how transformation converts from scalar to vector?

Thanks and best regards

• No. Current is a scalar, but current in a coil in a motor is a vector dependent on the coil's orientation. Jul 31 '21 at 13:09

Leblanc's theorem: A coil supplied with a current (ti ) = I 2 cos(ωt) creates a field $$\H(vector) = Hm*cos(w*t)*ux\$$ (vector). which is equivalent to the sum of two fields of modulus Hm/2 rotating in opposite directions at the angular velocity ω.

Ferraris theorem: three windings evenly spaced in the plane and powered by sinusoidal currents of pulsation ω and forming a balanced system make it possible to create a field rotating at the speed ω.

We can consider that the current is a scalar quantity originally, even if in time, we can have a qualification of "complex" number and therefore also "vector" in time.

If we speak of a current of 1 A, we do not generally attribute to it a "geometric sense" (a direction in space).

However, if we consider that the magnetic field created in a coil by a current (mechanical element) is "axial" in the coil, one thus finds oneself in a certain "mechanical configuration" according to the position of the assembly of this one in space and one can then consider that the current has a "geometric" qualification ... therefore "vector" (x, y, z, or angular ...) in the broad sense of the term. This is the case with three-phase machines where the 3 coils create a "rotating" magnetic field in "space" and in "time".

This "rotating" field can be described in many different ways ... hence the various types of "vector transformations".