Admittances

[mike_bhoy]

Hi all, newbie here, been searching the net for help with my assignment, hoping you can help me

basically i'm stuck. i have an rlc series circuit. the supply voltage is 10v, and the current at 1khz is 5ma 53o leading, then at 2 khz its 7 ma 33.5o leading.

basically i have worked out that z = 1203 + j1597, using z = v/i, but i can't figure out how to work out the values of Xl and Xc with just the info provided, how do you solve 2 unkowns?

this is supposed to work using admittances.

Any help welcome, my head is busting.

thanks
Mike.

 

[Roff]

It looks to me like you have two equations and 3 unknowns. You have equations for I1 and I2, but you have R, L, and C as unknowns. Were you given a value for one of these?

 

[ljcox]

It looks to me like you have two equations and 3 unknowns. You have equations for I1 and I2, but you have R, L, and C as unknowns. Were you given a value for one of these?

No, he has only 2 unknowns since the resistor must be 1203 hm: in both cases (assuming that his maths is correct - I have not checked it).

So now you can find L and C from Z at the 2 frequencies by solving the 2 simultaneous equations.

In fact, you can use either the modulus or the angle.

 

[Roff]

No, he has only 2 unknowns since the resistor must be 1203 hm: in both cases (assuming that his maths is correct - I have not checked it).

So now you can find L and C from Z at the 2 frequencies by solving the 2 simultaneous equations.

In fact, you can use either the modulus or the angle.

I hadn't assumed his impedance was correct. Since you posted, I calculated Z1 and Z2. He had Z1 almost right - the reactance term should be negative. Z2 does give almost the same resistance. I got R=1191 (and a reactance term, which I will leave to the OP). If you take the arithmetic mean of 1203 and 1191, and assume that is the resistance, you can, as you said, easily calculate L and C by solving the two simultaneous equations.

 

[mike_bhoy]

Thanks for the help, sorry i know reactance should have been negative, at least i know what to do now
thanks again.
Mike

 

[mike_bhoy]

Sorry still struggling, can you post the simultaneous equations, i don't need the answer, just how to start, thanks.

 

[Roff]

Sorry still struggling, can you post the simultaneous equations, i don't need the answer, just how to start, thanks.

OK, you already got Z1=1203-j1597. Call that R1+jX1. You can get Z2=R2+jX2 the same way, right? Assuming you get R2≈R1 (as I did), you now have X1 and X2.
You should know that, for a series RLC circuit, Z=R+jwL+1/jwC.
This means that X=jwL+1/jwC. Keep in mind that you will actually have 2 equations, one for X1 which includes w1, and one for X2 which includes w2. Since you only have 2 unknowns (L and C), you should be able to solve for them.
Can you take it from there? If not, tell us where you're stuck. This is just complex algebra from here on out.

 

[mike_bhoy]

ok so i get 2 equations

1597 = (2pie x 1000 x l) + (1 / (2 pie x 1000 x c)
788.5 = (2pie x 2000 x l) + (1 / (2 pie x 2000 x c)

but how do i multiply to cancel on out,

sorry to seem so dim, but i have no notes on how to do these, i have only covered admittances with known values.

i really appreciate the help.

 

[Roff]

ok so i get 2 equations

1597 = (2pie x 1000 x l) + (1 / (2 pie x 1000 x c)
788.5 = (2pie x 2000 x l) + (1 / (2 pie x 2000 x c)

but how do i multiply to cancel on out,

sorry to seem so dim, but i have no notes on how to do these, i have only covered admittances with known values.

i really appreciate the help.

First: X1=-1597 and X2=-788.5. They are not positive as you stated (you said in your second post that you knew the reactances were negative).

Next, X=jwL+1/jwC, right? We have to get j in the numerator before we can factor it out. I'm assuming you know 1/j=-j. Therefore,
X=j(wL-1/wC), not j(wL+1/wC) as you show in your equations above.
Anyhow, |X|=wL-1/wC. Let's just call it X.
You now need to simplify it. Multiply both sides of the equation by wC:
wCX=(LC*w^2)-1

So, (LC*w1^2)-w1CX1-1=0
and (LC*w2^2)-w2CX2-1=0 Remember X is negative in both equations!

You now have two simultaneous equations. The only unknowns are L and C. Do you know how to solve for them?

 

[mike_bhoy]

ok thanks, sorry for any silly mistakes but i'm still learning it, i understand it better now you have explained, i'll give the equation a go, if i get stuck i'll post up again.

Again thanks for your help Ron H, i really appreciate it

 

[mike_bhoy]

ok i multiplied the first equation by w2^2 and the second by w1^2, to cancel the first part out in each giving me the equations

w2^2*w1*c*x-w2^2=0
w1^2*w2*c*x-w1^2=0
does this seem right?

 

[Roff]

ok i multiplied the first equation by w2^2 and the second by w1^2, to cancel the first part out in each giving me the equations

w2^2*w1*c*x-w2^2=0
w1^2*w2*c*x-w1^2=0
does this seem right?

EDIT: hang on, let me think about that.
EDIT: I forgot to put the "subscripts" on X1 and X2 in my previous post. They have been fixed.
If you use X1 and X2 in your equations, subtract one from the other, and solve for C, you'll get the same answer I did.
I sure think you did it the hard way, though.
I did it by recognizing that w2=2*w1, and substituted 2*w1 for w2 in the second equation. Then I multiplied the 1st equation by 4, and subtracted one from the other. It seemed easier to me. Whatever works is fine.
Post your answers when you think they are correct.
I simulated the final circuit to make sure I hadn't made any errors - which are sure easy to make, as you know.

 

[mike_bhoy]

Yeah its a home assignment, as i said before i haven't covered admittances using simultaneous equations, thats why i am stuck, i have only learnt them with known values of resistance and reactance. which is why i can work out the first part but not the individual values of xl and xc, i'm first year in electronic engineering. If you don't want to help i understand, even if you know a website that might help, i do want to learn myself, but i didn't know were to start, thats why i'm on here asking.

 

[Roff]

Yeah its a home assignment, as i said before i haven't covered admittances using simultaneous equations, thats why i am stuck, i have only learnt them with known values of resistance and reactance. which is why i can work out the first part but not the individual values of xl and xc, i'm first year in electronic engineering. If you don't want to help i understand, even if you know a website that might help, i do want to learn myself, but i didn't know were to start, thats why i'm on here asking.

I think you responded to my previous post before I edited it. Sorry.

 

[ljcox]

Yeah its a home assignment, as i said before i haven't covered admittances using simultaneous equations, thats why i am stuck, i have only learnt them with known values of resistance and reactance. which is why i can work out the first part but not the individual values of xl and xc, i'm first year in electronic engineering. If you don't want to help i understand, even if you know a website that might help, i do want to learn myself, but i didn't know were to start, thats why i'm on here asking.

once you set up the equations, it's just a problem in maths.

See that attachment for an alternative method.

BTW, they are not admittances. It is an impedance and hence the j part is reactance.