how is the voltage drop across a single resistor calculated? In a simple E Mosfet circuit from where are the measurements for Vg and Vd taken?

 

how is the voltage drop across a single resistor calculated?

Ohm's Law:
E = sqrt(PR), or Voltage = the square root of (Power in Watts x Resistance in Ohms)
E = P/I, or Voltage = Power in Watts/Current in Amperes
E = IR, or Voltage = Current in Amperes x Resistance in Ohms

In a simple E Mosfet circuit from where are the measurements for Vg and Vd taken?

The reference is the source. It's usually specified (respectively) as Vgs and Vds or Vdss.

 

I have 15v Vdd and one resistor 2.7 Kohms gong into a E mosfet and I measure 3.544v for Vd but have not been able to calculate it.

 

I have 15v Vdd and one resistor 2.7 Kohms gong into a E mosfet and I measure 3.544v for Vd but have not been able to calculate it.

Why don't you post a schematic of your circuit?
That's much easier than trying to describe it.
As of this moment, I have no clue as to what your Vgs is, what MOSFET you're using, or even if it's N-ch or P-ch.