I measured a 33mH inductor and I have 30.7mH. I do my calculation for reactance @ 300Hz and I have 57.839Ω. I also measured 76.1Ω across the inductor. Do I add the resistance to the reactance to use for my Z Impedance calculation? I would assume yes? Or do I add it to the resistance part of the equation?

Sam

After thinking about it the inductor's resistance gets added to R not Xl? It's strictly resistance not a product of inductance. I think that is where some of the error came in on the last exercise because I had added it to Xl instead of R.

 

Do I add the resistance to the reactance to use for my Z Impedance calculation?

You add it as a complex number to get the complex impedance, i.e. 76.1Ω+ j57.839Ω

 

Since there is also a 50Ω resistor in series with the coil then it is (50Ω+j0)+(76.1Ω+j57.839Ω)=(126.1Ω+j57.839Ω). And ELI wins the series. Thx Crutschow! It's starting to gel a bit. Finally getting intuitive as to LEAD/LAG, C/L up or down.

Sam

 

Since there is also a 50Ω resistor in series with the coil then it is (50Ω+j0)+(76.1Ω+j57.839Ω)=(126.1Ω+j57.839Ω). And ELI wins the series. Thx Crutschow! It's starting to gel a bit. Finally getting intuitive as to LEAD/LAG, C/L up or down.

Sam

Hi,

Pure resistance is always the real part, the reactive part is the imaginary part.
For an inductor with series resistor this would be R+jwL
and for a capacitor iwth series resistance this would be R+1/(jwC)
where jw is j*w and w=2*pi*f with f being frequency in Hertz.

Just so you know, you can always get the phase shift easily by using the two argument inverse tangent function usually depicted as:
ph=atan2(imag,real)

Note the imaginary part comes first inside the parens then teh real part. So to get the phase angle of an inductor with series resistor you would do:
ph=atan2(w*L,R)
and note here the "w*L" is the imaginary part not "j*w*L" because "j" is just an operator.

For an inductor with two series resistances it would be:
ph=atan(w*L,R1+R2)

and with three resistors:
ph=atan(w*L,R1+R2+R3)

because the total resistance is always the 'real' part.

Now if you have a resistor in series R1 and a resistor in parallel R2 with the inductor, then you first have to calculate the parallel part then the series part as:
first the parallel resistor with the inductor:
Z1=(j*w*L*R2)/(R2+j*w*L)

then add the series resistor R1:
Z=Z1+R1=(j*w*L*R2)/(R2+j*w*L)+R1

we can simplify that or first simplify Z1:
Z1=(j*w*L*R2^2)/(R2^2+w^2*L^2)+(w^2*L^2*R2)/(R2^2+w^2*L^2)

then add the series resistance R1:
Z=j*(w*L*R2^2)/(R2^2+w^2*L^2)+(w^2*L^2*R2)/(R2^2+w^2*L^2)+R1

and note the imaginary part is the first term above and the rest is the real part.

However, once you know the imaginary part and real parts you can always use the abvove:
ph=atan2(imagpart,realpart)
where from above we have:
imagpart=(w*L*R2^2)/(R2^2+w^2*L^2)
realpart=(w^2*L^2*R2)/(R2^2+w^2*L^2)+R1

and once we have the values and frequency this becomes very easy to calculate using a program that has the function atan2() in it.

The above is with the resistor R2 in parallel with the inductor, then the resistor R1 in series with that parallel combination. If the second resistor is in parallel with the series combination of R1 and L though then it's different. We just use general circuit analysis of combining passive components into one lumped network.