# Can a purely reactive element exhibit resistive properties?

This question arises from a power factor correction example, consider the circuit below: simulate this circuit – Schematic created using CircuitLab

For the specified frequency $$Z_R=36\Omega$$ $$Z_L=48j\Omega$$ $$Z_C=-75j\Omega$$

Combining yields: $$Z_{eq}=\frac{(Z_R+Z_L)*Z_C}{Z_R+Z_L＋Z_C}=100\Omega$$

Now I kind of understand why the reactive components disappeared, since the inductor and the capacitor worked to cancel each other out. I don't however understand where the extra 64ohms of resistance came from.

And since we are on the subject, I'd also like to know what happens to the voltage drop across the inductor and capacitor in the power factor corrected situation. Do the impedances disappear and therefore there wouldn't be a voltage drop?

• The do not "cancel each other", in my way of seeing it. It's a peculiarity of the way complex numbers work. You can better see what happens in the complex plane by considering that the parallel of those two branches is the reciprocal of their reciprocal impedances. And the reciprocal of a complex number has reciprocal absolute value and conjugate angle. If you build 1/ztot = 1/z1 + 1/z2, you can see how it can happen for the resulting impedance to fall again on the real axis (despite the fact that in the last step you are composing a proper complex value with a purely imaginary one). Feb 11, 2017 at 23:09
• It makes mathematical sense that multiplying complex numbers together can generate a purely real number. But what does it mean in the circuit? is the capacitor and inductor creating resistances? But how is that so if they are purely reactive? Feb 11, 2017 at 23:25
• The voltage across the capacitor is exactly the same as the source voltage. This means that the current flowing through the capacitor is exactly the same as if the rest of the load was not there. BUT, the inductor supplies current at just the right time so that the source sees a purely real load. This is a resonance phenomenon of the capacitor and inductor. If the frequency is changed, the impedance seen by the load will become complex rather than real. Feb 11, 2017 at 23:29
• The power factor is unity, but the useful power supplied is less than would be the case if the R=36 ohm load were connected directly to the supply. This is because the voltage across R is less than V1 due to the voltage divider formed by R and L.
– Chu
Feb 11, 2017 at 23:51

By way of illustration, let $\small R=1; X_L=1$, hence the series connection is: $\small Z=1+j$. Now, this series connection of $\small 1$ and $\small j1$ is equivalent to a parallel connection of $\small 2$ and $\small j2$, $\small \left(=\frac{j4}{2+j2}=1+j\right)$, hence place an additional capacitive reactance of $\small -j2$ in parallel and the overall load across the source (i.e. the load that the source sees) is $\small 2\Omega$ resistive. Thus, the power factor is unity but the effective resistance seen by the source is $\small 2\Omega$, hence the power [W] transferred to the load is less than would be the case if the true resistive load were connected directly to the source.

• Your answer makes perfect mathematical sense, but I would still like to know what causes the source to see the extra resistance that isn't there? Is the source actually seeing the voltage drop at the inductor and taking it as resistance since the the phase shift has been cancelled out? Feb 12, 2017 at 0:41
• The capacitor and inductor take energy from the source, then - while the souce is at a low voltage - they burn this energy through the resistor Feb 12, 2017 at 1:15
• The source does not "see extra resistance". The source generates a voltage. The load decides how much current will flow. Because of the way current flows back and forth between the inductor and capacitor, the overall load ends up taking current in phase with its voltage. And the current is less than would be the case if the load was the resistor only. Feb 12, 2017 at 1:39
• @mkeith, the voltage and current supplied by the source are the same as would be supplied to a 2 ohm resistor, so I think it's accurate to say that the source 'sees' 2 ohms
– Chu
Feb 12, 2017 at 10:42
• Sorry, Chu. I guess I was referring to the OP's circuit. Of course you are right. Feb 12, 2017 at 18:04