Inductive / capacitive reactance in parallel

[amateur_24]

Hi
I want to finish off my hw. We've only covered reactance in series. How would I calculate the following question

A coil of 0.159H is connected with 100ohm resistor accross 230v 50hz supply. What's it's impedance?

It would be great if you show all the calculations you've done.

 

[alec_t]

1) The above title mentions capacitive reactance but your circuit has no capacitor.
2) Presumably the resistor is in parallel with the coil?
3) When you say 'its' impedance, do you mean the impedance of the coil alone, or of the parallel combination of coil/resistor ?

 

[RCinFLA]

Calculate the reactance of the 0.159 H coil at 50 Hz. Assuming you mean 100 ohm is in series with it then impedance would be 100 + jXinductor.

If you want to converter between series and parallel equivalents use your polar and rectangular functions on your calculator. Convert to polar form, invert Mag, change sign of angle, revert back to rect. form and take the inverts of each component (change sign of reactive part) to get parallel equivalent. Reason to work in polar form is inverting a polar form is much easier then inverting in rect. form.

 

[amateur_24]

Yeah sorry There isn't a cap in the circuit. I wanted the impedance of the resistor and coil together.

 

[panic mode]

XL= sL = j*(2*pi*f)*L
or if you only care about magnitude:
|XL|=49.95 Ohm


Z=XL||R=XL*R/(XL+R)=100*50j/(100+50j)
Z=5000j*(100-50j)/(100^2+50^2)
Z=(250000+500000j)/(12500)
Z=20+40j
|Z|=sqrt(20^2+40^2)=89.44 Ohm

 

[Ratchit]

panic mode,

XL= sL = j*(2*pi*f)*L
or if you only care about magnitude:
|XL|=49.95 Ohm


Z=XL||R=XL*R/(XL+R)=100*50j/(100+50j)
Z=5000j*(100-50j)/(100^2+50^2)
Z=(250000+500000j)/(12500)
Z=20+40j
|Z|=√(20^2+40^2) = 89.44 Ohm

Wrong answer. 1/sqrt((1/49.05)²+(1/100)²) = 44.69 ohms

Ratch

 

[RCinFLA]

0.159H in parallel with 100 ohms @ 50 Hz = 44.686 /_ 63.457 degs = 19.969 + j39.977 ohms.

0.159H @ 50Hz = +j49.95

100 || +j49.95 -> take 1/100 = 0.01 , take 1/+j49.95 = -J0.02 -> convert 0.01 - j0.02 to polar = 0.0224 /_ -63.457 degs

Invert mag -> 1/0.0224 = 44.687, change sign of angle = +63.457 degs : you have impedance in polar form = 44.687/_ 63.457 degs.

Convert polar form to rect. form -> 44.687/_ 63.457 degs = 19.969 + j39.977 ohms.

 

[tazman171]

From the sounds of his question I doubt he really needs a answer dealing with polar/rectangular coordinates, etc., as it is just an inductive load with no capacitive reactance and there will be no phase shift between current and voltage XL will be 90 Deg out of phase with the resistive Ohms. The total is called impedance and you need to do a vector diagram.

The current "lags" the voltage in a purely inductive circuit and therefore vector, VL is drawn 90o in front of the current.

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So 2x(3.14)x(50 Hz)= 49.926 Ohms

So the square root of (49.926² + 100²)= 111.77 Ohms impedance

230V/ 111.77 Ohms(Z) = 2.0578 Amps

That's the easy version with out polar, scalar or using your calcs complex mode as it is jus t and inductive load.


To show this easily I linked to the above pictures.

 

[Ratchit]

tazman171,

From the sounds of his question I doubt he really needs a answer dealing with polar/rectangular coordinates, etc.,

From the sound of his question, he wants the parallel impedance or admittance. If so, then polar/rectangular coordinates cannot be avoided.

it is just an inductive load with no capacitive reactance and there will be no phase shift between current and voltage

Of course there will be, unless the capacitive and inductive reactances are equal and cancel out.

The total is called impedance and you need to do a vector diagram.

Impedance and admittance are complex quantities, but they are not vectors or phasors.

So 2x(3.14)x(50 Hz)= 49.926 Ohms

So the square root of (49.926² + 100²)= 111.77 Ohms impedance

You are calculating the series impedance, the OP wants the parallel impedance.

That's the easy version with out polar, scalar or using your calcs complex mode as it is jus t and inductive load.

That is the incomplete version, as resistance and inductance cause a phase change between voltage and current in both series and parallel circuits.

Ratch

 

[panic mode]

>> |Z|=√(20^2+40^2) = 89.44 Ohm
>Wrong answer. 1/sqrt((1/49.05)²+(1/100)²) = 44.69 ohms

thank you for fixing it... i guess i must have entered it wrong; √(20^2+40^2)=44.7

 

[tazman171]

tazman171,



From the sound of his question, he wants the parallel impedance or admittance. If so, then polar/rectangular coordinates cannot be avoided.



Of course there will be, unless the capacitive and inductive reactances are equal and cancel out.



Impedance and admittance are complex quantities, but they are not vectors or phasors.



You are calculating the series impedance, the OP wants the parallel impedance.


That is the incomplete version, as resistance and inductance cause a phase change between voltage and current in both series and parallel circuits.

Ratch

MY bad, I mis interpreted his statement. What I gave though does get you the answer with out showing the phase relationship, simple like I said.

I meant no phase shift between IL and IC as there was XL and did say that voltage 90 deg out of phase with current, with current leading by 90 deg for XL. The phase angles of resistive and inductive impedance are always 0 deg and +90 deg, respectively, regardless of the given phase angles for voltage or current. So the IL will be -37.016 deg out while the VL is 52.984deg, 90 deg leading for the inductor.

Just trying to find total impedance though right?

Purely a simple method that's damn close without messing with any phase relationship, Square root of (1/(1/100 squared + 1/50 squared))= 44.721 Ohms.

Saw no mention of needing more that that, but I also missed it being parallel.

You can run out the polar and rectangular values (and will need to, complex mode on your calc will be your friend) but seeing as he said that they had only done series I was assuming that they weren't to that level of instruction yet as we had covered the basics of both parallel and series XL and XC separate and combined before we ever got into polar and rectanguler methods of doing the calculations. Didn't want to throw all the stuff at him that you guys did, so I threw the "simplified" way of finding the impedance.

So what ever, I did it for series and just showed the value for parallell using a simple method with out worrying about the phase angles and got just about the same answer as the other method..

In the real world though, the phase angle of your load are extremly important on everything from sizing your wire to circuit protection to what enclosures you'll need to buy to put it all in.