I'm trying to measure the impedance (\$R_x\$) of C1 in the RC circuit shown below, but I'm getting some results I can't explain.


simulate this circuit – Schematic created using CircuitLab Measurement:
On VM1 and VM2 I measure the voltage by consecutively taking a sample of \$10^4\$ points over 4 ms on each channel then I calculate the RMS.
(I'm using a multichannel DAQ card for output and input. I can't find the symbol, hence the analog VMs).
Using Ohm's law I calculate \$R_x\$:

\$R_x = R1 \frac{VM_2-VM_1}{VM_1}\$

The applied current is a sine curve of 0.5V where I varied the frequency between 1, 5, 10, 50 and 100 kHz. It is turned on for about 2-3 seconds during the consecutive reading of the two channels.

For each frequency I make 10 measurements and take the mean of those.

I would expect the values to go like: $$R_x=\frac{1}{2\pi f C}$$ where f is the frequency and C the capacity. Fx at 1 kHz for a \$0.1 \mu F\$ capacitor I would get \$ 1591.59 \Omega \$. But my measurement at that frequency is about \$ 500 \Omega \$

These are my measurements for different capacitors: enter image description here

Why are my numbers this far of?

If I let something out please let me know and I will add it to the post.
Any tips, remarks or comments are appreciated.

I've done the calculations again thanks for the helpful answers. It fits a lot better now: enter image description here

There seems to be some increasing deviation though, is there an apparent reason for this?

  • 1
    \$\begingroup\$ That's usually written as \$X_C=\frac{1}{2\pi\:f\:C}\$. Note that R isn't used? Do you know why? \$\endgroup\$
    – jonk
    Oct 22, 2017 at 18:12
  • \$\begingroup\$ @jonk Is it to underline the frequency dependence, which isn't the case for a simple resistor? Is it to distinguish impedance from resistance? \$\endgroup\$
    – Alex
    Oct 22, 2017 at 22:18
  • \$\begingroup\$ There's already plenty written on the topic and already an answer here. But I'll add a different approach for you that avoids the fancy stuff and see if it helps. \$\endgroup\$
    – jonk
    Oct 22, 2017 at 22:39

3 Answers 3


Let's take your case of the \$X_C=1591.\overline{591}\:\Omega\$ computation that assumed \$f=1\:\textrm{kHz}\$ and \$C=100\:\textrm{nF}\$. (I'm assuming you didn't actually measure the \$C\$ value but just assumed it... so we'll assume it here, too.) Your resistor, I take it, is actually measured with some meter. Again, I'll assume that your meter is perfectly accurate (it isn't, but who cares?) I'm also going to assume your "DAQ" board was used properly and that you interpreted the results correctly. No reason not to.

Let's see if we can work out what should be done and work out what you did.

If you know a fixed frequency, then you can consider resistance (\$R\$) to be the x-axis (positive-only because I don't want to drag this out into never never land) and inductance and capacitance will be on the y-axis. By convention, capacitance (\$X_C\$) is on the negative y-axis and inductance (\$X_L\$) is on the positive y-axis. If you want to know what the total series impedance will look like (and you are using a voltage divider, so it's 'series' here) to the power supply, then you mark out \$R\$ on the x-axis, mark out \$X_C\$ on the negative-going side of the y-axis, and this forms the two sides of a right triangle. The length of the hypotenuse is the magnitude of the "complex impedance."

I'm stealing the following image from here:

enter image description here

The above image gives you a picture of what I'm suggesting.

So, with this in mind you should expect to see a magnitude value of \$\sqrt{\left(1797\:\Omega\right)^2+\left(1591.59\:\Omega\right)^2}\approx 2400\:\Omega\$. That's the magnitude.

Now. Let's see. You probably worked out your equation so that it subtracts your nearly \$1800\:\Omega\$ resistor from this, directly. (Not as a vector.) So that would yield about \$600\:\Omega\$. Not far from what you wrote as the value you figured for \$X_C\$.

But the problem is that you did a direct subtraction.

You don't say what you measured in this case, but let me haul out a couple of numbers. You write that your source voltage is set to \$500\:\textrm{mV}\$ peak. Let's say you measured (using your DAQ board) a voltage peak of \$380\:\textrm{mV}\$ across \$R_1\$. Then you would have computed \$1797\:\Omega\cdot\frac{500\:\textrm{mV}-380\:\textrm{mV}}{400\:\textrm{mV}}\approx 567\:\Omega\$ for \$X_C\$ (using your equation.)

So let's do this differently.

You should have realized that the equation is derived this way:

$$\begin{align*} Z &= \sqrt{R_1^2+X_C^2}\tag{1}\\\\ I&=\frac{V}{Z}\tag{2}\\\\ V_{R_1}&= I\cdot R_1= \frac{V}{\sqrt{R_1^2+X_C^2}}\cdot R_1\tag{3} \end{align*}$$

From the above, you can solve (3) to get:

$$ X_C = R_1\cdot\sqrt{\left(\frac{V}{V_{R_1}}-1\right)\left(\frac{V}{V_{R_1}}+1\right)}$$

Plugging in my figures of \$V=500\:\textrm{mV}\$ and \$V_{R_1}=380\:\textrm{mV}\$ I find \$X_C\approx 1537\:\Omega\$.

Which is more like it.


You need to take into account that the voltages across the capacitor and the resistor are \$90^{\circ}\$ out of phase. The impedance of a capacitor is

$$Z = \frac{1}{j\cdot\omega \cdot C}$$

where \$j{\equiv}\sqrt{-1}\$ is the imaginary unit. This makes all the difference. You need to use phasors and complex math.

Your circuit is simple enough, that you can solve it with a trick. Since the voltages are \$90^{\circ}\$ out of phase you can use the property $$\left | V_C \right |^2=\left | V_{\text{M}2} \right |^2 - \left | V_{\text{M}1} \right |^2$$

  • 1
    \$\begingroup\$ Absolute function isn't necessary since the terms are squared. \$\endgroup\$
    – jonk
    Oct 22, 2017 at 18:26
  • 1
    \$\begingroup\$ jonk: these \$|\cdot|\$ were done at least for educational purposes, because short of explaining the whole complex phasor business, OP might do things like \$\left((j100+0.02)\text{V}\right)^2\$ and get results in the range of \$-10\;000\$. \$\endgroup\$ Oct 22, 2017 at 19:46
  • 1
    \$\begingroup\$ @MarcusMüller I suppose. But I also think the OP is a long way away from worrying \$ V \in \mathscr{C}\$. Almost sure, still stuck with \$ V \in \mathscr{R}\$. But point is taken. \$\endgroup\$
    – jonk
    Oct 22, 2017 at 20:46
  • 2
    \$\begingroup\$ @jonk agreed; Alex, if you're reading this, don't be confused. I swear, learning complex phasors is worth it; it opens up a whole world. \$\endgroup\$ Oct 22, 2017 at 21:02
  • 2
    \$\begingroup\$ @Nat Precisely that, lowercase i already had a meaning in the field, so to avoid retroactive confusion j is used instead. Which is better to those who don't need to switch fields too often. \$\endgroup\$
    – Kroltan
    Oct 22, 2017 at 22:55

It seems that part of the problem is that you are confusing reactance with resistance. This led you to derive the wrong equation for Xc, which results in the wrong calculation for Xc. The correct equation is: $$X_c =R_1 \sqrt{ \frac{V_2^2 - V_1^2}{V_1^2}}$$

Use this equation and see if you get better results.

Another thing you need to keep in mind, is that this equation applies to "ideal" circuits. In real life, you will find that capacitors, do in deed, have resistance in addition to reactance.


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