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I was solving a problem of single phase eletrical circuits where I had to find the inductor $L$ and resistance $R$. I managed to get two equations containing the two unknowns. $$\frac{R}{R^2+(w*L)^2}=c_1$$ and $$\frac{wL}{R^2+(w*L)^2}=c_2$$ where $w,c_1 \text { and } c_2$ are known.How do I solve this?

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  • $\begingroup$ sorry it is resistance and inductor. $\endgroup$ – GTX OC Nov 14 '13 at 6:24
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Squaring both equations and adding them you get

$$\frac{R^2}{(R^2+(w*L)^2)^2}+\frac{(w*L)^2}{(R^2+(w*L)^2)^2}=c_1^2+c_2^2$$

or

$$\frac{1}{R^2+(w*L)^2}=c_1^2+c_2^2$$

This yields:

$$R^2+(w*L)^2=\frac{1}{c_1^2+c_2^2}$$

Now replace the denominators in both equations.

Alternate solution

Dividing the two equations you get

$$\frac{wL}{R}=\frac{c_2}{c_1}$$

Thus

$$wL=\frac{c_2R}{c_1} \,.$$

Replacing in either equation you get an equation in $R$.

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Presuming that by w$*$L you mean the product of w and L, and presuming w is a constant:

Let u(R,L) = $R^2 + (wL)^2$ We have

R = c$_1$u and wL = c$_2$u giving

c$_2$R = c$_1$c$_2$u and

c$_1$wL = c$_1$c$_2$u so that

c$_2$R - c$_1$wL = 0.

L = c$_2$R/c$_1$w

Now substitute L back into your original equations

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