# Solving a parallel RL circuit for the r and l values

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?

EDIT:

I can write:

$$\cos\left(\frac{\pi}{2}-\arctan\left(\frac{2\pi\cdot50L}{R}\right)\right)=\frac{2\pi\cdot50L}{R}\cdot\frac{1}{\sqrt{1+\left(\frac{2\pi\cdot50L}{R}\right)^2}}\tag3$$

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:

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

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

## 1 Answer

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: -

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.

• Ok, but what is the mistake I made? Feb 26, 2018 at 16:03
• @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"? Feb 26, 2018 at 16:06
• 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 :(? Feb 26, 2018 at 16:08
• I was still adding to my comment LOL! Feb 26, 2018 at 16:09
• Can you see my edit?! Feb 26, 2018 at 16:14