Desirae Washington

Answered

2022-07-09

Technique to solve this equation of 2 unkowns

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.

$\frac{R}{{R}^{2}+(w\ast L{)}^{2}}={c}_{1}$

and

$\frac{wL}{{R}^{2}+(w\ast L{)}^{2}}={c}_{2}$

where $w,{c}_{1}\text{and}{c}_{2}$ are known.How do I solve this?

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.

$\frac{R}{{R}^{2}+(w\ast L{)}^{2}}={c}_{1}$

and

$\frac{wL}{{R}^{2}+(w\ast L{)}^{2}}={c}_{2}$

where $w,{c}_{1}\text{and}{c}_{2}$ are known.How do I solve this?

Answer & Explanation

Ronald Hickman

Expert

2022-07-10Added 18 answers

Squaring both equations and adding them you get

$\frac{{R}^{2}}{({R}^{2}+(w\ast L{)}^{2}{)}^{2}}+\frac{(w\ast L{)}^{2}}{({R}^{2}+(w\ast L{)}^{2}{)}^{2}}={c}_{1}^{2}+{c}_{2}^{2}$

or

$\frac{1}{{R}^{2}+(w\ast L{)}^{2}}={c}_{1}^{2}+{c}_{2}^{2}$

This yields:

${R}^{2}+(w\ast L{)}^{2}=\frac{1}{{c}_{1}^{2}+{c}_{2}^{2}}$

Now replace the denominators in both equations.

$\frac{{R}^{2}}{({R}^{2}+(w\ast L{)}^{2}{)}^{2}}+\frac{(w\ast L{)}^{2}}{({R}^{2}+(w\ast L{)}^{2}{)}^{2}}={c}_{1}^{2}+{c}_{2}^{2}$

or

$\frac{1}{{R}^{2}+(w\ast L{)}^{2}}={c}_{1}^{2}+{c}_{2}^{2}$

This yields:

${R}^{2}+(w\ast L{)}^{2}=\frac{1}{{c}_{1}^{2}+{c}_{2}^{2}}$

Now replace the denominators in both equations.

Holetaug

Expert

2022-07-11Added 8 answers

Alternate solution

Dividing the two equations you get

$\frac{wL}{R}=\frac{{c}_{2}}{{c}_{1}}$

Thus

$wL=\frac{{c}_{2}R}{{c}_{1}}\phantom{\rule{thinmathspace}{0ex}}.$

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

Dividing the two equations you get

$\frac{wL}{R}=\frac{{c}_{2}}{{c}_{1}}$

Thus

$wL=\frac{{c}_{2}R}{{c}_{1}}\phantom{\rule{thinmathspace}{0ex}}.$

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

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