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

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]).  

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

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

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

(OP)
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?  

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

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

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