No, he has only 2 unknowns since the resistor must be 1203Ron H said: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?
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.ljcox said:No, he has only 2 unknowns since the resistor must be 1203hm: 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.
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.mike_bhoy said:Sorry still struggling, can you post the simultaneous equations, i don't need the answer, just how to start, thanks.
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).mike_bhoy said: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.
EDIT: hang on, let me think about that.mike_bhoy said: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?
I think you responded to my previous post before I edited it. Sorry.mike_bhoy said: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.mike_bhoy said: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.
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