I (Current) Phasor and whether it is a compound number 1

[hyboxis79]

I hope anyone can help me with this. When calculating voltage drops with a cos (Phi) lets say 0.8[**see note at hte bottom]. I am not sure whether I am supposed to calculate the voltage drop like


delta V = I (R cos (Phi) + X sin (Phi))

R being the real impedance and X being the reactive impedance
from what I understand it's agreed upon that "V" will always stay real while "I" can be complex.

does this mean that when I calculate the voltage drop on the reactive part I will write

delta V = I(R sin(Phi) + X cos(Phi))





[**]I hope that people understand this notation. I mean by it that it's the cos of the angle between the real P power(measured in watts) and the complex power S (measured in [VA]).

 

[Marmite]

In complex number terms Resistance is real, Reactance is imaginary. Impedance Z is the complex sum of the real resistance and the imaginary reactance.
Z=(Rcos(phi)+ jXsin(phi))
Ohms law applies so V=IZ.
Don't forget this is in rectangular form. To get the same result as you would get from a multimeter you would need to convert to polar form.
Regards
Marmite

 

[jghrist]

delta V = I (R cos (Phi) + X sin (Phi))
is an approximation of the difference in voltage magnitudes. In most cases the approximation is good. See the attached calculations.

 

[hyboxis79]

thanks for you reply

I've encountered a few calculation from a course I took a few years ago where a voltage drop is calculated like this


delta V = I (R cos (Phi) + X sin (Phi))
the answer recieved was a scalar value of the this addition. although obviousely vector addition does allows it.
I thought they were refering to the fact that "I" is a vectorwhich is in an angle Phi to "V". And still it doesn't make too much sense mathematically. maybe they were refering to "V" in R.M.S value?

 

[hyboxis79]

thank you for your posting jghrist. it's very helpful