I've a parallel RL circuit:


simulate this circuit – Schematic created using CircuitLab

And I know that (the RMS input-current) \$\overline{\text{I}}_{\space\text{in}}=2\$ A, (the RMS input-voltage) \$\overline{\text{V}}_{\space\text{in}}=200\$ V, the input frequency is \$50\$ Hz and \$\overline{\text{P}}_{\space\text{in}}=69.44\$ W now I need to find \$\text{R}\$ and \$\text{L}\$:

$$ \begin{cases} \overline{\text{I}}_{\space\text{in}}=\frac{\overline{\text{V}}_{\space\text{in}}}{\left|\underline{\text{Z}}_{\space\text{in}}\right|}=\frac{\overline{\text{V}}_{\space\text{in}}}{\sqrt{\frac{R\omega L}{R^2+(\omega L)^2}}}\\ \\ P=\overline{\text{V}}_{\space\text{in}}\cdot\overline{\text{I}}_{\space\text{in}}\cdot\cos\left(\varphi_{\space\text{in}}\right)=\overline{\text{V}}_{\space\text{in}}\cdot\overline{\text{I}}_{\space\text{in}}\cdot\cos\left(\frac{\pi}{2}-\arctan\left(\frac{\omega L}{R}\right)\right) \end{cases}\tag1 $$

Using the given values:

$$ \begin{cases} 2=\frac{200}{\sqrt{\frac{R\cdot2\pi\cdot50L}{R^2+(2\pi\cdot50L)^2}}}\\ \\ 69.44=200\cdot2\cdot\cos\left(\frac{\pi}{2}-\arctan\left(\frac{2\pi\cdot50L}{R}\right)\right) \end{cases}\tag2 $$

But when I tried to solve the system I get imaginary numbers, what is my mistake?


I can write:


So, I get:

$$ \begin{cases} 2=\frac{200}{\sqrt{\frac{R\cdot2\pi\cdot50L}{R^2+(2\pi\cdot50L)^2}}}\\ \\ 69.44=200\cdot2\cdot\frac{2\pi\cdot50L}{R}\cdot\frac{1}{\sqrt{1+\left(\frac{2\pi\cdot50L}{R}\right)^2}} \end{cases}\tag4 $$

Now, for example we get:

$$2=\frac{200}{\sqrt{\frac{R\cdot2\pi\cdot50L}{R^2+(2\pi\cdot50L)^2}}}\space\Longleftrightarrow\space L=\frac{R\pm Ri\sqrt{3999999999999}}{200000000\pi}\tag5$$

  • \$\begingroup\$ How could you get imaginary numbers from this equation?? Other than by having L or R negative... \$\endgroup\$
    – Eugene Sh.
    Feb 26, 2018 at 15:26
  • \$\begingroup\$ Why do you think power factor angle is Pi/2 - arctan(wL/R)? Why is there Pi/2? \$\endgroup\$
    – Deep
    Feb 26, 2018 at 15:30
  • \$\begingroup\$ +1 for perfect formatting and embedded Mathjax - and on the OP's first post. \$\endgroup\$ Feb 26, 2018 at 15:56
  • \$\begingroup\$ @Deep That is the input impedance \$\endgroup\$
    – asd33
    Feb 26, 2018 at 16:02

1 Answer 1


Try and find R first.

You know the applied voltage and you know the power. All that power is dissipated in the resistor so, 69.44 = \$\frac{200^2}{R}\$ hence R = 576.04 ohms.

Can you solve it from here? Hints are not needed any more because the OP spotted his mistake so: -

enter image description here

Knowing apparent power (400 VI) and active power (69.44 watts) you can calculate reactive power as \$\sqrt{400^2-69.44^2}\$ = 393.926 VIr.

Reactive power is the reactive watts taken by the inductor hence \$X_L\$ is \$\frac{200^2}{393.926}\$ = 101.542 ohms. Hence L = 323 mH.

  • \$\begingroup\$ Ok, but what is the mistake I made? \$\endgroup\$
    – asd33
    Feb 26, 2018 at 16:03
  • \$\begingroup\$ @asd33 From my standpoint, the mistake you made is not thinking about the problem and getting too early into the maths. R is so easily solvable and you are two steps away from finding L= 0.324 H using my approach. Apart from anything else you have only gone half way without giving us any clue about the path you took to get "imaginary numbers"? \$\endgroup\$
    – Andy aka
    Feb 26, 2018 at 16:06
  • \$\begingroup\$ Yes true I also find that value, and I also agree that my method is (maybe) to mathy but it is important so see what is my mistake in the mathy method I used :(? \$\endgroup\$
    – asd33
    Feb 26, 2018 at 16:08
  • \$\begingroup\$ I was still adding to my comment LOL! \$\endgroup\$
    – Andy aka
    Feb 26, 2018 at 16:09
  • \$\begingroup\$ Can you see my edit?! \$\endgroup\$
    – asd33
    Feb 26, 2018 at 16:14

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