# How to determine an equivalent RC circuit of a black box?

I've been given information about a "black box" device that I know is made of a simple combination of resistors and capacitors, and I've been asked to determine an equivalent circuit. I took an "Electrical Engineering for Non-EE Engineering Majors" class many years ago, but the skills and knowledge that I learned back then are quite rusty, and I could use some help.

Here's what I know: when an RLC bridge is connected to the black box and set to measure the equivalent parallel circuit, then the measurement is 140 pF. The dissipation factor is 0.16. The measurement frequency is 1 kHz. I don't have access to the device, I just have specifications. I've shared all that I know, although I strongly suspect that the device uses a minimum number of components.

What has me confused is that when I looked up Dissipation Factor, the Wikipedia article talks about an equivalent circuit of a capacitor in series with a resistor. That seems simple enough; I could easily calculate the component values if I knew that the black box were such a series circuit. But when I looked up "RLC bridge" I quickly ended up at the Wien bridge Wikipedia article, which talks about an equivalent circuit of a capacitor in parallel with a resistor.

Let me guess that a black box circuit made from resistors and capacitors can have an equivalent circuit of either a resistor in series with a capacitor, or a resistor in parallel with a capacitor. Is that right? And would someone please point me in the right direction to understand how to calculate the component values? If they can't be directly calculated, or if my learning curve would be too steep in the time I have available, could I arrive at the component values by simulating a circuit in LTspice and adjusting component values until the calculated results are sufficiently close to the numbers I was given? I'd be grateful for any help.

By the way, this isn't a homework question, although it does sound like one. This is a project my boss assigned to me.

• Is this a homework question? Commented Nov 30, 2021 at 20:22
• The method of measuring depends on the number of capacitors and AC impedance vs frequency but if only 1 cap , then there are many ways. wikiwand.com/en/Dissipation_factor Commented Nov 30, 2021 at 20:22
• Since there is only one frequency involved and you only have resistors and capacitors and no non-linear devices like diodes, the black box circuit can be driven by a voltage and you can observe the current and its relative phase (or driven by a current and observe the voltage and its phase.) From that you can determine an equivalent simple circuit. I gather that while you refer to RLC, your question is only about some number of resistors and capacitors. Is that right? Homework, it appears. That also right?
– jonk
Commented Nov 30, 2021 at 20:44
• As jonk says apply an ac voltage with an external small series resistance (10Ω). Measure applied voltage and voltage across resistor (current will be in phase with resistor) with a scope and you can get phase angle and leading or lagging . Commented Nov 30, 2021 at 20:52
• I tried to simulate the device in LTspice as one leg of a Wien bridge. The parallel form of the device balances in the Wien bridge, but the serial form of the device doesn't, so I asked a follow-up question, Why doesn't my Wien bridge simulation work? Commented Dec 6, 2021 at 20:13

a black box circuit made from resistors and capacitors can have an equivalent circuit of either a resistor in series with a capacitor, or a resistor in parallel with a capacitor.

At a given frequency, a resistor in parallel with capacitor has the same impedance as a different resistor in series with a different capacitor.

Call $$\-jX_p\$$ the impedance of the parallel capacitor, $$\R_p\$$ the resistance of the parallel resistor, $$\-jX_s\$$ the impedance of the series capacitor, $$\R_s\$$ the resistance of the series resistor,

then

$$R_s -jX_s = R_p || - jX_p$$

$$R_s - jX_s = \frac{-jX_pR_p}{R_p-jX_p} = \frac{(-jX_pR_p)(R_p+jX_p)}{(R_p-jX_p)(R_p+jX_p)}$$

$$R_s - jX_s =\frac{-jX_pR_p^2-j^2X_p^2R_p}{R_p^2-j^2X_p^2}$$

$$R_s - jX_s =\frac{-jX_pR_p^2+X_p^2R_p}{R_p^2+X_p^2}$$

$$\\therefore\$$

$$R_s = \frac{X_p^2R_p}{R_p^2+X_p^2}$$

$$X_s = \frac{X_pR_p^2}{R_p^2+X_p^2}$$

• Just the help I needed! I was able to solve the problem, thanks! Commented Dec 1, 2021 at 20:56
• I came up with $R_p$ = 7.10 MΩ, $R_s$ = 177 kΩ, and $C_s$ = 144 pF, for those of you following along. Commented Dec 6, 2021 at 19:04

Dissipation factor is 1/Q and for any reactance, $$\Q = \frac{|X|}{r_{series}}=\frac{R_{parallel}}{|X|}\$$

You say you are measuring the equivalent parallel circuit, so you can easily work out the parallel resistance. The equivalent series circuit will have a slightly different capacitance, which you can work out using MathKeepsMeBusy's answer, but the dissipation factor / Q will be the same for both.