# Measuring R in AC circuit - Ideas / Maths ?

#### FredM

Joined Dec 27, 2005
124
Hi -

I have an unknown value of R in parrallell with an unknown value of C - I need to measure R through an AC coupled circuit (there are High DC potentials I must isolate). C is likely to be in the order of a few nF, but MIGHT be a lot larger (100nF?).. And the circuit must produce 2 alarm levels, one when R is less than 200k, and one when R is less than 40k.

My plan was to inject (via known resistance, and large capacitance of say 10uF, so that the Z of this capacitance did not affect result) sine signals switching between (say) 100Hz and 1kHz, and to measure the resulting attenuation for each frequency

(F) --[== 10k ==] --(M)--| 10uF|---[ZZ Unknown R || Unknown C]---(GND)

F = Frq switching between 100Hz and 1kHz, through (say) 10k. Monitor amplitude at (M) with respect to Gnd. Decouple through 10uF, Unknown R||C connect to Gnd.

Total Z of R||C = 1/((1/R)+(1/(1/(6.28*F*C)))). (correct?)

There should be a simple way to compute R from the above, having Z for two known frequencies, and I feel sure someone must have done this before and be able to give me an answer.. But annoyingly, I cannot seem to find it !

Any help would be appreciated!

#### bloguetronica

Joined Apr 27, 2007
1,541
You can measure the AC voltage at the terminals of the unknown device and measure the AC current passing trought the whole series. Then you can calculate the impedance using the equations:
R = Vrms / Irms
R = Vp / Ip

These eq. are valid under a sinusoidal waveform regime.

#### recca02

Joined Apr 2, 2007
1,212
since r and c are in,
current thru r and voltage across it will give the value of r.
then find out equivalent impedance for resistance and capacitance parallel combination.
Total Z of R||C = 1/((1/R)+(1/(1/(6.28*F*C)))). (correct?)
i think that wud equal 1/z not z.

#### FredM

Joined Dec 27, 2005
124
Hi folks - thanks for your replies..
I am a bit foggy about this.. I have an R and a C in parrallell (R||C) and both are unknowns..
I can get Z for this network for a known frequency easily..
I can get a Z for 100Hz, and (assuming the C is reasonable size) a different Z for 1000Hz.
.. BUT, I want to get the value of R (effectively I want to get rid of ZC, although, I do not think it matters which I "get rid of" - If I know ZC, I can find R.. And if I know R I can find ZC and therefore C).

R should remain constant regardless of frequency, whereas ZC will change as a function of frequency.. So I figured that by passing 2 different frequencies through the network, and measuring the Z of the network at each frequency I should be able to determine the value of R (and also C).

I am hoping for a simple formula or whatever so that I can easily seperate the R and C 'components' of the Z found for each frequency.

Resistance || = 1/((1/R1)+(1/R2)) (?)
ZC = 1/(2*PI*f*C) (?)
therefore (?)
R||C = 1/ ((1/R)+ (1/(1/(2*PI*f*C)))) (?)

#### beenthere

Joined Apr 20, 2004
15,819
Is there some compelling reason you can't just look at the resistor? Possibly take power off and measure it with an ohmmeter? Solving for two unknowns in a systen where you can't determine one by itself is going to be very difficult.

The next most obvious way to go about it is to put DC across the RC. The C becomes an open, and the current is all up to the resistor. If the cap is in the nF region, even AC down around a couple hundred Hz ought to give results pretty much dependant on the resistance.

Theory is fine up to a point - sometimes it's just easier to open the horse's mouth and count the teeth. This should tell that I am a technician by background.

#### FredM

Joined Dec 27, 2005
124
Hi again.. Oh, it would be real nice to just measure the DC.. If only!
I am attaching a rough schematic showing what I am trying to do.. I have tried loads of DC schemes (using isolated DC-DC converters and connecting these to battery terminals to get local current loops I can measure) - but they all fail because current flow cannot be directly monitored, and is swamped by the battery current.

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#### nanovate

Joined May 7, 2007
666
Why can't you measure the current directly through the 10M resistors? It is in parallel with the unknown R and you can detect if the R drops below 200K or 40K. Or am I misunderstanding what you want... do you want to measure both the unknown R and C?

#### beenthere

Joined Apr 20, 2004
15,819
One more point - if your chassis is properly grounded, then detecting leakage current in the ground line is quite simple. That is how GFCI outlets are protected. Looking for leakage is lots easier - and at non-lethal voltages.

#### FredM

Joined Dec 27, 2005
124
Why can't you measure the current directly through the 10M resistors? It is in parallel with the unknown R and you can detect if the R drops below 200K or 40K. Or am I misunderstanding what you want... do you want to measure both the unknown R and C?
Current through the 10M resistors is entirely a function of the voltage across them.. If the leakage resistances are equal, the voltage between each battery terminal and chassis will be 200V, so, say for example both leakage resistances are 10k (well below the required thresholds) the current through the 10M resistors would be the same as if both leakage resistors were 1M (well above the threshold) or there were no leakage resistors at all.

One more point - if your chassis is properly grounded, then detecting leakage current in the ground line is quite simple. That is how GFCI outlets are protected. Looking for leakage is lots easier - and at non-lethal voltages.
This circuit is for electric vehicles – there is no actual ‘ground’ return path which can be monitored as far as I can tell.. I have placed the 10M resistors between the battery terminals and chassis as I was worried that, with the battery completely isolated, electrostatics could cause the battery to float at a high potential WRT chassis, and pose an ESD risk.. Adding the 10M’s ‘ties’ the battery centre voltage to the same potential as the chassis, and with a fully charged (in fact this is 350V, but I say 400) battery this means its +Ve terminal is +200V and its –Ve terminal -200V WRT Chassis (provided there are balanced resistances between chassis and terminals)

A fault where both terminals are connected to chassis via < 1M is probably extremely unlikely.. but I would prefer not to depend on voltage imbalance as a means of detecting a fault state..

If I have a measured total Z for the network with f = 100Hz, and call this A, and I have a measured total Z for the network with f = 1000Hz, and call this B.. What I want (I think) is a formula which gives me R from A and B. I have always had difficulty with this kind of equation manipulation, and am hoping someone says “oh, that’s easy!”

#### recca02

Joined Apr 2, 2007
1,212
if u got the value of Z at two different frequencies.
i think u can solve that using simultaneous eqn.
B1= A + 628C
B2=A + 6280C
WHERE
A=1/R
C= capacitance
B1= 1/Z1
B2= 1/Z2
or did i get something wrong?

#### Ron H

Joined Apr 14, 2005
7,014
Why can't you do it as below? It's a simple attenuator, with no frequency switching required. Just measure the peak-to-peak voltage and calculate the resistance. You can, of course, scale the input voltage to anything convenient.
The 3 traces are for Rleak=40k, 200k, and 10Meg.

#### FredM

Joined Dec 27, 2005
124
Thank you everyone - Ron, your method looks simplest.. I will do a simulation which includes the battery etc and see if this works.

I am still interested in a mathematical solution to derive R.. More now because I have a problem solving simultaneous equations, and this is a problem I want to rectify.. I often find that seeing a solution to a 'real' problem teaches me more than hours getting lost in taxt books.

Once again.. Thank you all. I have not posted here before, and am stunned by the fantastic level of interest and help.

#### recca02

Joined Apr 2, 2007
1,212
maybe you can post the values of z for the two frequencies.
and we can have a go at finding values of r and c.
BTW some calculators can solve simultaneous eqn for u.