I (Current) Phasor and whether it is a compound number
I (Current) Phasor and whether it is a compound number
(OP)
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]).
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]).






RE: I (Current) Phasor and whether it is a compound number
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
RE: I (Current) Phasor and whether it is a compound number
is an approximation of the difference in voltage magnitudes. In most cases the approximation is good. See the attached calculations.
RE: I (Current) Phasor and whether it is a compound number
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?
RE: I (Current) Phasor and whether it is a compound number