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Given a very simple circuit such as this.

enter image description here

If I where given the RMS values for the three currents, how could I go about solving R and Xl? (the resistor and inductor in paralell)

I know I can not apply KCL using the RMS values, even if I tried the sum of the three currents is not zero (as in total current is not equal to the sum of the currents of the two branches).

I tried getting the peak value from the RMS values and then apply KCL but I'm stuck because the total current and the current that goes thru the inductor will have some unknown angles.

What would be a good strategy to solve this?.

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  • \$\begingroup\$ Hint 1: the current through an inductor is in phase withers the voltage across it. Hint 2: the complex current phasors do add to zero. \$\endgroup\$ – The Photon Nov 4 '18 at 21:02
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Well, we know that:

$$\underline{\text{Z}}_{\space\text{in}}=\frac{1}{\frac{1}{4}+\frac{1}{\text{R}}+\frac{1}{\text{j}\omega\text{L}}}=\frac{\frac{1}{4}+\frac{1}{\text{R}}}{\left(\frac{1}{4}+\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}+\frac{1}{\left(\frac{1}{4}+\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\cdot\frac{1}{\omega\text{L}}\cdot\text{j}\tag1$$

So, know you know that:

$$\left|\underline{\text{I}}_{\space\text{in}}\right|=\frac{\left|\underline{\text{U}}_{\space\text{in}}\right|}{\left|\underline{\text{Z}}_{\space\text{in}}\right|}=\frac{\left|\underline{\text{U}}_{\space\text{in}}\right|}{\sqrt{\left(\frac{\frac{1}{4}+\frac{1}{\text{R}}}{\left(\frac{1}{4}+\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\right)^2+\left(\frac{1}{\left(\frac{1}{4}+\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\cdot\frac{1}{\omega\text{L}}\right)^2}}\tag2$$

And:

$$\left|\underline{\text{I}}_{\space2}\right|=\frac{\left|\underline{\text{U}}_{\space\text{in}}\right|}{4}\space\Longleftrightarrow\space\left|\underline{\text{U}}_{\space\text{in}}\right|=4\cdot\left|\underline{\text{I}}_{\space2}\right|\tag3$$

And we can write:

$$\left|\underline{\text{I}}_{\space1}\right|=\frac{\left|\underline{\text{U}}_{\space\text{in}}\right|}{\sqrt{\left(\frac{1}{\text{R}}\cdot\frac{1}{\left(\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\right)^2+\left(\frac{1}{\omega\text{L}}\cdot\frac{1}{\left(\frac{1}{\text{R}}\right)^2+\left(\frac{1}{\omega\text{L}}\right)^2}\right)^2}}\tag4$$

And remember that:

$$\left|\underline{\text{I}}_{\space1}\right|=\sqrt{2}\cdot\text{I}_{\space1\space\text{rms}}\tag5$$

And that:

$$\omega\text{L}=\text{X}_{\space\text{L}}\tag6$$

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  • \$\begingroup\$ @JoaquinBrandan Using \$(3)\$ you can get the voltage, I edited that into my answer just a minute ago. \$\endgroup\$ – Jan Nov 4 '18 at 21:20
  • \$\begingroup\$ If you gave me the numerical or exact values you got for the currents I can solve it for you using Mathematica?! \$\endgroup\$ – Jan Nov 4 '18 at 21:21
  • \$\begingroup\$ I see, You can compute the RMS voltage across the 4Ohm resistor and since they are all in paralell that is the overall RMS voltage, then solve for the remaining components. brillant. It feels like this way of solving it is a concecuence of the shape of the circuit and not the general method I was picturing in my head. in \$(5)\$ you state that the \$I_1\$ is not RMS, however all currents are given in RMS so that should not be necessary, right?. I am given the three currents in RMS already. How did you use mathjax here? the dollar signs dont seem to work for me :c \$\endgroup\$ – Joaquin Brandan Nov 4 '18 at 21:29
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    \$\begingroup\$ @JoaquinBrandan On stackexchange you need to use a backslash before the dollar sign. \$\endgroup\$ – Hearth Nov 4 '18 at 21:31
  • \$\begingroup\$ @Felthry, only on EE, actually. \$\endgroup\$ – The Photon Nov 4 '18 at 21:37

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