# Calculate Voltage across resistor of rlc circuit

#### darkknight

Joined Oct 7, 2009
41
knowing that the circuit is critically damped:
l = 33mh
c = .001µf
r = 9.15 kohms

i want to calculate vr at certain times frames but i am doing some wrong

i am using the the formula

i(t) = (V/L)*t*e^(-R/2L)

R=2√(.033*.000000001) = 11489.1

R/2L = 174077

E^174077 gives me 0 so that throws the solution off, can anyone help me on what i am doing wrong?

i(t) = (V/L)*t*e^(-R/2L)

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#### t_n_k

Joined Mar 6, 2009
5,455
The exponential term should include time 't' as well. You show a constant (-R/2L) as the exponent. That should be multiplied by 't'. At 1kHz the half cycle intervals are 500usec.

Presumably you are trying to predict the resultant resistor voltage waveform with respect to time....?

#### darkknight

Joined Oct 7, 2009
41
yes that is what i trying to do, know the voltage across the resistor at several time points but

even if i multiply the exponent by t, i get

at time at 1 microsecond
(R/2L)*t = 174077*.000001 = .011489

e^ans = 1.012

(4/.033)*.00001*1.012 = .0000123

is this is way off what i got measuring it, i can figure out what is wrong

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#### t_n_k

Joined Mar 6, 2009
5,455
That's just telling you the value of the current at 1ms when you really want the current waveform over the entire interval from 0 to time t.

At 1kHz the charge and discharge half cycles will only be 500us - not the full 1ms interval.

So you would need to calculate & plot the current and hence resistor voltage values for sufficient points to give a true representation of the output waveform.

You could do this in EXCEL series table say and do the plot as part of the same worksheet.

#### darkknight

Joined Oct 7, 2009
41
i have a table with values measured and now i have to calculate and compare the measured and calculated but i am having difficulty calculating it.

i am struggling with the formula

#### t_n_k

Joined Mar 6, 2009
5,455
You'll probably need to show some working - what you've done so far.

Also note that your assumed solution is not exactly correct. You have a repetitive source waveform whose period is much longer than the circuit time constant. This means the capacitor will be charged to ±4V (one or the other) at the start of a source half cycle. So the effective source voltage would be ±8V (again, one or the other) at the start of any half cycle.

It would also be good if you showed your measured waveforms as a cross check.

#### darkknight

Joined Oct 7, 2009
41
values i got were for example

at 1 microsec i measured .75V
at 2 microsec i measured 1.6V
at 3 microsec i measured 2.2V

#### t_n_k

Joined Mar 6, 2009
5,455

Attached my graph ...

#### darkknight

Joined Oct 7, 2009
41
but my difficulty is proving that through written form, i am having trouble with the formula

#### t_n_k

Joined Mar 6, 2009
5,455
O.K. - let's take the 2usec (2e-6) case

i(t)=4/(.033)*(2e-6)*exp(-(2e-6)*11,489/(2*.033))=0.000171A

So Vr(t)=11,489*.000171=1.96V

Which is somewhere near what you read at 2us. Measurement errors could be significant and we don't know the tolerances on the values you gave - how did you determine the critically damped condition from your observations? R looks to be adjustable according to some experimental condition.

I would have expected about twice that value for the circuit configuration you posted (and as indicated in my graph) but let's forget that for the moment.

Was your actual source voltage 4V peak-to-peak? What was the source value at 2us on the other scope channel.

• darkknight

#### darkknight

Joined Oct 7, 2009
41
yes it was 4 p-p and we only used one scope and i see that your values are close to the measured, i have to relook at my calculations

#### t_n_k

Joined Mar 6, 2009
5,455
Comparing with your 1us calculation I have i(t)=0.000102 A which would give a resistor voltage of 1.17V. Again this is somewhat different from your observed reading.

Keep in mind the point that the inductor will also have some winding resistance which may or may not be significant in relation to the calculated critically damped resistance value of 11,489Ω.

#### darkknight

Joined Oct 7, 2009
41
something should be wrong because if its 4 p-p it should go over 2V correct?

#### t_n_k

Joined Mar 6, 2009
5,455
So what was your maximum reading for the resistor voltage & at what time?

#### darkknight

Joined Oct 7, 2009
41
at 6 now that i am checking my notes i measured 2.75

#### t_n_k

Joined Mar 6, 2009
5,455
If you have access to or can download a Spice based circuit simulation program you would be able to better appreciate what is happening in the circuit. For example you could download TINA-TI. There are several available for free - albeit with certain restrictions on simulation capability.

#### darkknight

Joined Oct 7, 2009
41
i dont have that program and not familiar with the software and time if not on my side i(t)=4/(.033)*(2e-6)*exp(-(2e-6)*11,489/(2*.033))=0.000171A

So Vr(t)=11,489*.000171=1.96V

its supposed to be

Vr(t)=9150*.000171=1.56V

correct?

we want vr which is the voltage against the resistor

#### t_n_k

Joined Mar 6, 2009
5,455
Yes - sorry about that. The R value is what you want. I was distracted by the critical damping value of 11,489Ω. Forgot to check R was different.

That's a good outcome since the answers will be much closer to your observed values.

Also, I withdraw my comments about the factor of 2 in the theoretical value. Since it is 4V peak-to-peak the capacitor only charges to ±2V. So the original formula using 4V is correct.

#### t_n_k

Joined Mar 6, 2009
5,455
So I finally obtain this waveform using Excel