Inductance

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linguist

New Member
I have a basic circuit that has an inductor which has a resistance of 10Ω, it is connected to a 110v 50Hz supply that draws 8A.

I want to calculate the Inductance of the Coil but I seem to be having difficulty understanding if the Resistance of the coil has any bearing on the calculations.

I have attempted the Math below with an explanation of my dilema.

I first calculated XL

XL= V/I

XL= 110/8 = 13.8Ω

L = XL/2(pi)f

= 13.8/314

L = .044H

Now to verify, if I calculate 2(pi)fL to get XL I get

6.28 X 50 X .044

XL=13.8 Which corresponds to above.

My misunderstanding is that in the First equation XL= V/I.
Doesn't this give the Total Impedance Z @ 50Hz & not just the XL of the coil, it has the 10Ω resistance included giving ZT & not XL to be used in the equations to find L.

Are my equations correct to find L?

Cheers

ericgibbs

Well-Known Member
hi,
You must include the series resistance, this link explains fairly well.

linguist

New Member
ericgibbs,

Thanks for the reply & the link, however I am not sure I understand to find L of the coil from L = XL /2(pi)f

If you just ADD the 10Ω resistance to the total given by

XL= V/I

XL= 110/8 = 13.8Ω

This gives ZT

If you subtract the 10Ω resistance, this means that XL does not equal V/I as this would be ZT

I have this all upside down somewhere?

Cheers

ericgibbs

Well-Known Member
ericgibbs,

Thanks for the reply & the link, however I am not sure I understand to find L of the coil from L = XL /2(pi)f

If you just ADD the 10Ω resistance to the total given by

XL= V/I

XL= 110/8 = 13.8Ω

This gives ZT

If you subtract the 10Ω resistance, this means that XL does not equal V/I as this would be ZT

I have this all upside down somewhere?

Cheers

hi linguist,

There was an error in my maths regarding the formula, please accept my apologises if this has caused you a problem.

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JimB

Super Moderator
Ztot = 10R + 2*pi*50*0.0119 = 10R + 3.73R = 13.75R
Err...
don't forget that Z = √(R² + X²)
Your recent morphing (avatar) must have affected your grasp of fundamentals, most out of character for you!

JimB

Last edited:

ericgibbs

Well-Known Member
Err...
don't forget that Z = √(R² + X²)

JimB

Thanks Jim for the heads up , I should know better , a very careless error......

ericgibbs

Well-Known Member
hi linguist,
I will redo the calculations:

The inductor draws a current of 8 amps from a 110v at 50Hz and the 'DC' resistance of the coil is 10R

So the Total impedance is 110/8 = 13.75R

Let XL = inductive reactance
This is Ztot = sqrt[10^2 + XL^2] ,where the 3.75R is the inductive reactance part.

So,
13.75 = sqrt[10^2 + XL^2]
189 = 100 +XL^2
89 = XL^2
So, XL= 9.434

Zind= 2*pi*f*L, thats: 9.434 = 2 * 3.14 * 50 * L, where L is the unknown, so transpose the formula.

L = 9.434/[ 6.28 * 50] = 9.434/314 = 0.03H
Inductance = 0.03H

Checking the maths:

Ztot = sqrt(10^2 + (2*pi*50*0.03)^2
Ztot = sqrt[100 + 88.73]
Ztot = sqrt[188.73]

So, Ztot = 13.74R

JimB

Super Moderator
Thats better Eric, back to your old self again!

JimB

ericgibbs

Well-Known Member
Thats better Eric, back to your old self again!

JimB

Thanks Jim,

JimB

Super Moderator
Thanks Jim,

A good question?
I had from time to time wondered whether I should have one, but I could never decide on something suitable.

JimB

linguist

New Member
Hi Eric,

Thanks for your posts, I am actually glad to see that we can all make mistakes, I am sure that you don't make many,
Your posts have helped me understand, many thanks!

JimB,

No problems, I didn't regard it as thread hijacking, I could see your ears prick forward at the opportunity to correct Eric, I don't think it would happen that often so pounce when the timing is right.

Cheers

MrAl

Well-Known Member
Hi there,

The way i like to do these problems is to transform all the components into the frequency domain and then solve using complex numbers.

For example, for the resistor R stays the same but the inductor L transforms to s*L, and the series circuit impedance is made up of the sum of these two:
Z=R+s*L

and then of course the current is:
I=E/Z

where E is the source voltage (AC).

That makes the current:
I=E/(R+s*L)

and then substituting s=j*w we get:
I=E/(R+j*w*L)

and that's the complex current.
To get the amplitude, we can multiply top and bottom by the conjugate of the denominator which is R-j*w*L and so we get:
I=E*(R-j*w*L)/((R+j*w*L)*(R-j*w*L))

and after simplifying the denominator we get:
I=E*(R-j*w*L)/(R^2+w^2*L^2)

and now expanding the numerator we get:
I=(E*R-E*j*w*L)/(R^2+w^2*L^2)

and so the real part is:
REAL=E*R/(R^2+w^2*L^2)

and the imaginary part is:
IMAG=-E*w*L/(R^2+w^2*L^2)

and the amplitude is:
AMPL=sqrt(REAL^2+IMAG^2)

and if we were to work this out we would get:
AMPL=E/sqrt(R^2+w^2*L^2)

and so that is the amplitude of I:
I(AC)=E/sqrt(R^2+w^2*L^2)

If we use rms values for E then I is also in rms, if we use peak units for E then I is also in peak units.

Although this seems like quite a rigamarole to go through the nice thing about this technique is that we can use this for circuits with any number of resistors, capacitors, inductors, and AC voltage or current sources. For example, if we had two resistors and two inductors all in series we would calculate the impedance as:
Z=R1+R2+s*L1+s*L2

and we would proceed the same as before.

If we had a third resistor R3 in parallel with L2 we would first have to write the parallel combo for those two:
R3*s*L2/(R3+s*L2)

and use that in the impedance equation instead of just L2:
Z=R1+R2+s*L1+R3*s*L2/(R3+s*L2)

but again we would proceed to solve this the same basic way.

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linguist

New Member
MrAl,

Thanks for the reply, sorry to seem ignorant but could you explain what (s) is please.

Cheers

MrAl

Well-Known Member
Hi,

Oh sorry about that, that's my fault for not stating why we were using the variable 's' (lower case 19th letter of the English alphabet).
You probably already know that the letter 'F' (lower or upper case) is used many times over again to represent frequency, such as 50Hz, 60Hz, etc, well that 's' variable is sometimes called the "complex frequency" rather than just "frequency". As frequency F is a variable that usually takes on simple values like 50, 60, 100, etc., the complex frequency variable 's' can take on complex values like 5+20j where the '5' there is called the 'real' part and the 20 there is called the 'imaginary' part, and the lower case 'j' is used to show that the 20 is the imaginary part more or less, and is considered mathematically to be multiplied by the imaginary part. That means it may be written as:
5+20*j
to be more exact about it.
To put it another way, the complex frequency 's' is a two dimensional quantity, where the usual frequency 'F' is just a one dimensional quantity. That means the complex frequency can be shown on a plane where the real part is along the 'x' axis and the imaginary part is along the 'y' axis. This plane is called the complex frequency plane.

But no need to concern yourself too much about all this just yet, because when we solve problems such as those i have presented in my previous thread we can simplify the complex frequency to contain only an imaginary part, and so we are back to one dimensional again. So instead of 5+20*j we will be using only 20*j with the real part equal to zero, which means it can be written as 0+20*j or just 20*j or as some people like to write it 20j.
Of course this brings us to the new variable, lower case 'j', which is the 10th letter of the English alphabet, and that is sometimes called the imaginary operator. The value of j is equal to the square root of minus 1 such as: sqrt(-1), but you will note that we dont actually take the square root like we do with other numbers like sqrt(4) which equals 2. The square root of minus 1 is only used to show the imaginary part, and in the computations that follow there are some simplifications that come up when j is multiplied by other numbers. We dont really calculate the square root of minus 1, but we do calculate the square of the square root of minus 1, which may be written as (sqrt(-1))^2, and you will note that if we square a number after taking the square root we end up with the same number back again, so the square of the square root of minus 1 is simply minus 1, written of course as -1. This helps us resolve the final solution to our circuits into just two parts, the real part and the imaginary part, and then we have ways to deal with that that actually gives us a real world number like 1,2 3, 5.5, etc., that indicates some level of current or voltage in our circuit, as well as a phase angle like 50 degrees, 20 degrees, etc., which tells us the phase angle between some currents or voltages.
There is also one more little thing to know about though, and that is called the 'conjugate' of a number. The conjugate of a complex number is simply the same number only with the imaginary part made the opposite sign. For example, for the complex number 5+20*j the conjugate is simply 5-20*j, and for the complex number 6-34*j, the conjugate is simply 6+34*j. So you see, this is all rather easy after learning just a few simple rules. The main thing is that a complex number has two parts unlike regular numbers that have only one part.

Lets look at a simple example where we have a solution we came up with for our circuit and we want to find the final numerical solution...
Say we have the equation:
I=2/(s+3)
where we have one 's' in the denominator.
The first thing we do is replace the 's' with "j*w", because that is the imaginary part of 's' that we use to evaluate AC circuits. The equation then becomes:
I=2/(j*w+3)
and writing the denominator with the real part first and imaginary part second:
I=2/(3+j*w)
Now to solve this we need to simplify the denominator so that it becomes completely real, ie with no imaginary part. It just so happens that when we multiply a complex number (the denominator) by it's conjugate, we get a real number only. Of course since we multiply the denominator by the conjugate that means we must also multiply the numerator by the conjugate to keep the equation equal to the original value, so that gives us:
I=2*(3-j*w)/((3+j*w)*(3-j*w))
and now to simplify the denominator we just multiply it out following the rules for algebra, keeping in mind that when we multiply j by itself we get minus 1 (-1). After we multiply out the denominator we end up with:
((3+j*w)*(3-j*w))=9+w^2
so we can rewrite the equation now to:
I=2*(3-j*w)/(9+w^2)
Now we can multiply out the numerator and we get:
I=(6-2*w*j)/(9+w^2)
and now we have an equation with a real denominator and a complex numerator. That was the main goal here.

Now that we have what we were after, we can separate the real and imaginary parts of this equation by expanding, and that means we simply break apart the numerator and use the same denominator for both parts. We end up with:
I=6/(9+w^2)-2*w*j/(9+w^2)
and now the left term here is the real part and the right term is the imaginary part. The imaginary part has a 'j' in it same as before.
In this form we can immediately identify the real and imaginary parts of the current I:
real=6/(9+w^2)
imag=-2*w/(9+w^2)
and you will note that we carried the negative sign into the imag part, but we got rid of the multiplication of the imag part by j. We drop the 'j' when we identify the two parts this way.
Now we are ready to calculate the amplitude from:
Ampl=sqrt(real^2+imag^2)
and note again all we are doing here is taking the square root of the sum of the real and imag parts squared. This is the quantity you would measure in an AC circuit if you were to connect a meter to the circuit.
The phase angle is also of interest in many AC circuit analysis problems, and that is calculated from:
Angle=taninv(imag/real)
where 'taninv' is just the inverse tangent, and (imag/real) is just the division as indicated. There is one little catch here however, and that is that the angle calculation has to be checked to make sure the resulting angle is in the correct quadrant because the single argument taninv function returns only the principle value. What we are after with the phase angle is the phase angle of the point (imag,real) in the complex plane, so all we have to do is check to see that this is correct, and if not we correct it.
For example, with imag=-2 and real=-2 if we divide we get +1, but if imag=2 and real=2 when we divide we also get +1, yet the first set is in the third quadrant and the second set is in the first quadrant, so we would have to adjust the first set after taking the inverse tangent. It's a little tricky there but not too hard once you do a couple of these problems.
Alternately, if your calculator or program has the two argument inverse tangent function you can easily use that:
Angle=taninv(imag,real)
and note this last taninv function takes two separate arguments rather than one, so the function does recognize and adjust the final result for the correct quadrant.

Ok, now that we can calculate the amplitude and phase angle, what we usually want to do is calculate this for one single value of w, where w=2*pi*F (and there is our single value frequency F, were F is the frequency of the AC source in our circuit), or calculate over and over again using a frequency that increments so we can see the response for many different frequencies. We can do this in at least two different ways:
1. Calculate a formula for any w
2. Keep inserting w into:
real=6/(9+w^2)
imag=-2*w/(9+w^2)
and keep calculating the amplitude and phase angle.

Calculating a formula as #1 requires knowledge of algebra and a little work, but doing it as #2 above it's quite simple especially if we use a small program in a language that is easy to use, or a programmable calculator. Either way, we usually sweep the frequency F from some low value to some high value and plot the amplitude and phase angle for these various frequencies, or else we use only one value for F (as in power circuits) and simply calculate that one answer, and this gives us the resulting current or voltage that we were solving for in the first place.

Again, this might seem like a bit of a task to go through, but the beauty of it is that this very same technique can be used for any circuit no matter now many inductors, capacitors, resistors, and voltage and current sources, with a sine wave excitation of a single frequency F (perhaps swept through many values).

Lets work out that one example for the amplitude with say a frequency of 10Hz. Recalling these three:
real=6/(9+w^2)
imag=-2*w/(9+w^2)
Ampl=sqrt(real^2+imag^2)

First we calculate w from:
w=2*pi*F
and this comes out to approximately:
w=2*pi*10=62.83
so now we can calculate the real and imag parts:
real=6/(9+w^2)=6/(9+62.83^2)=0.001516
imag=-2*w/(9+w^2)=-2*62.83/(9+62.83^2)=-0.03176
and now the amplitude:
Ampl=sqrt(real^2+imag^2)=sqrt(0.001516^2+(-0.03176)^2)=0.0318
and so the resulting current is about 31.8 milliamps AC. Note this is a peak measurement if our source was specified as peak, or an rms value if the source was specified as rms.

As a slightly more advanced topic, if we sweep F through all the possible (or relevant) frequencies with the right amplitudes and add the results we can get a solution to the circuit for other types of sources like square waves for example. We dont usually have to do this, but it is interesting that this also works and i believe this theory is due to Fourier.

For single frequency sine waves though, it's a very fast technique. It's also very very useful when we want to sweep the frequency in order to obtain the plot for a filter circuit.

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linguist

New Member
MrAl,

Thank you for taking the time to explain things in your last post.

Maybe I should have said, that was excellent, I will have to sit down & read it through several times for it to sink in properly.

I can't thank you enough!

Cheers

MrAl

Well-Known Member
MrAl,

Thank you for taking the time to explain things in your last post.

Maybe I should have said, that was excellent, I will have to sit down & read it through several times for it to sink in properly.

I can't thank you enough!

Cheers

Hello again,

You're welcome, and if you have any questions about that procedure just give me a yell with a PM here.

RCinFLA

Well-Known Member
If you have 10 ohm of measured resistance and 8 amps at 110v you may be driving inductor so hard you are saturating the core, driving its inductance down.

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