Reducing reactive power by using a resistor

Question

I have a school exercise that goes like this:

Voltage is 230V. Frequency is 50Hz. Inductor is 400mH.

How big of a resistor should be added with the inductor (in a series) to reduce reactive power by 75%? (The correct answer is apparently 218 ohms.)

How do you calculate this? Is this even possible? I know only of capacitors reducing reactive power.

I can only figure out that 25% of the reactive power is be 105,24 var. Also sorry, I don't study in English, so the terms might be a bit strange. Thank you.

Answer 1

The resistor in series with the inductor just serves to reduce the overall current draw through the inductor. Thus, the resulting reactive drop (Q) in the inductor goes down.

The reactive drop in the inductor is,

$$ Q = I^2X_L$$

You can calculate the current with the formula I give above.

Answer 2

Well, the real power is given by:

$$\text{P}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\left(\varphi\right)\tag1$$

Where \$\varphi=\left|\arg\left(\underline{\text{V}}\right)-\arg\left(\underline{\text{I}}\right)\right|\$.

The complex power is given by:

$$\text{Q}=\text{V}_\text{rms}\text{I}_\text{rms}\sin\left(\varphi\right)\tag2$$

And the apparent power is given by:

$$\text{S}=\sqrt{\text{P}^2+\text{Q}^2}=\text{V}_\text{rms}\text{I}_\text{rms}\tag3$$

So, for your circuit we get:

• Real power: $$\text{P}=\text{V}_\text{rms}\cdot\underbrace{\frac{\text{V}_\text{rms}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}}_{=\space\text{I}_\text{rms}}\cdot\underbrace{\frac{1}{\sqrt{1+\left(\frac{\omega\text{L}}{\text{R}}\right)^2}}}_{=\space\cos\left(\varphi\right)}=\frac{\text{R}\text{V}_\text{rms}^2}{\text{R}^2+\left(\omega\text{L}\right)^2}\tag4$$
• Complex power: $$\text{Q}=\text{V}_\text{rms}\cdot\underbrace{\frac{\text{V}_\text{rms}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}}_{=\space\text{I}_\text{rms}}\cdot\underbrace{\frac{\omega\text{L}}{\text{R}}\cdot\frac{1}{\sqrt{1+\left(\frac{\omega\text{L}}{\text{R}}\right)^2}}}_{=\space\sin\left(\varphi\right)}=\frac{\omega\text{L}\text{V}_\text{rms}^2}{\text{R}^2+\left(\omega\text{L}\right)^2}\tag5$$
• Apparent power: $$\text{S}=\text{V}_\text{rms}\cdot\underbrace{\frac{\text{V}_\text{rms}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}}_{=\space\text{I}_\text{rms}}=\frac{\text{V}_\text{rms}^2}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}\tag6$$

Now, we want to solve:

$$\frac{\text{V}_\text{rms}^2}{\omega\text{L}}\left(1-\frac{\eta\space\text{%}}{100}\right)=\frac{\omega\text{L}\text{V}_\text{rms}^2}{\text{R}^2+\left(\omega\text{L}\right)^2}\tag7$$

Using your values, we get:

$$\frac{230^2}{2\pi\cdot50\cdot400\cdot10^{-3}}\cdot\left(1-\frac{75}{100}\right)=\frac{2\pi\cdot50\cdot400\cdot10^{-3}\cdot230^2}{\text{R}^2+\left(2\pi\cdot50\cdot400\cdot10^{-3}\right)^2}\space\Longrightarrow\space$$ $$\text{R}=40 \sqrt{3} \pi\approx217.656\space\Omega\tag8$$