RC Time Constant Problem Solving for Capacitive Charge @ 2TC

Thread Starter

SamR

Joined Mar 19, 2019
5,031
Here is the problem: What is the instantaneous value of charge current exactly after two time constants?

For:
RC.PNG

TC=RC=0.01uF * 15kΩ = 150us 2TC= 300us t=elapsed time = 2TC = 300us ic = instantaneous current @ Cap
Imax = V/R = 40V/1500Ω = 26.67mA
ic = Imax (1 - e^-t/(RC)) = 26.67mA (1 - e^-300us/150us) = 26.67mA(1 - 0.135335) = 23.061mA

And 2TC = ~86% .: 0,86 * 26.27mA = 22.592mA

So those 2 approaches seem to agree but the answer key gives 360.69uA. And that is my quandary...

The -t/(RC) = -2 and e^-2 = 0.135335

I don't see the error?
 
Last edited:

crutschow

Joined Mar 14, 2008
34,280
I don't see the error?
I see two. ;)
One is that you used 1500Ω in your calculation, not the 15kΩ ohms in the schematic.

Second, you are using the formula for voltage.
The formula for current is Imax (e^-t(RC)).
Thus the current at 2 time constants is 2.6667mA * 0.13535 = 360.94μA.
 

Thread Starter

SamR

Joined Mar 19, 2019
5,031
One is that you used 1500Ω in your calculation, not the 15kΩ ohms in the schematic.

Second, you are using the formula for voltage.
1.5kΩ is correct, my bad on the ckt diagram Oh good lord I am blind, V is 90 not 40. so V/R is = 60mA

K then the ic = Imax(1-e^-t/(L/R) doesn't transpose to -t(RC)... That is the "charging" curve for L/R The "discharge" is Imax * e^-t(L/R)
iC and vR are the same curve, AHHH and that is not the iL curve, it is the opposite. K I see it now.

I found the i for LC but didn't find if for RC and that is my error. Then why is the .83 * V/R wrong? Wouldn't that be 0.17 * V/R = 4.533mA Where did the 2.6667mA come from? Isn't that for 4V 2.6667mA * 1500Ω = 4V

Problem 3
IMG_0507.JPG

So Imax (e^-t(RC)) = 0.06A* e^-2 = 8.12mA?
 

crutschow

Joined Mar 14, 2008
34,280
So Imax (e^-t(RC)) = 0.06A* e^-2 = 8.12mA?
Yup.
I determine which equation to use by noting whether the exponential is going up (1-e^t/τ) or going down ( e^t/τ).
Thus if you suddenly apply a voltage to an RC circuit, the voltage is going up and the current is going down.
The opposite is true for an RL circuit.
 

RBR1317

Joined Nov 13, 2010
713
Sorry no idea... I have avoided Calculus since college
Calculus is not required. My linear circuits textbook had an entire chapter on first-order transient response where the circuit has a resistor (or resistor network), a switch, and a capacitor or inductor in series or parallel with the resistor. In all cases the voltage and/or current followed one of those exponential curves as it went from the initial value to the final value. Your job as the circuit analyst is to determine the initial value & final value of interest (thereby giving the magnitude of the transient) and to determine the time constant. Then determine whether the transient is rising or falling in value, so you know which exponential equation to use (there's only two to choose from). Calculating first-order transients is an intuitive process, but very simple and always the same.
 

Thread Starter

SamR

Joined Mar 19, 2019
5,031
I know the curves and CrutSchow did a pretty good job describing the difference between RL and RC which are the inverse of each other. The term 1st order transient is new to me though and when I looked it up the description was in calculus. Have worked thru the phasors and angles so now getting into the timing. Still, the calc I did after making my corrections does not match up with what the answer key gives...
 

RBR1317

Joined Nov 13, 2010
713
The term 1st order transient is new to me...
The term 1st order means there is only one energy storage element (L or C) and a resistor or resistor network where all voltage & current transients follow either a rising or falling exponential curve. It's when there are more than a single energy storage element, or a varying source that calculus becomes involved. Actually, Laplace transforms (nobody uses calculus to solve circuit problems.)
 

Thread Starter

SamR

Joined Mar 19, 2019
5,031
Something is amiss... After 5 time constants, the current should be 0

RC = 1500Ω * 0.01uF = 15us =1TC AHA I was using 150us
Imax = V/R = 90V/1500Ω = 60mA
5TC = 75us = t

i = Imax(e^-t/RC) = .06A(e^-(75us / 15us) = .06A(e^-5) = .06A * 6.738m) = 404uA which is not 0

and after 2 TC = 30us = t
i = Imax(e^-(30us/15us) = 0.06A(e^-2) = 8.12mA which is not the 360.9uA given by the answer key..
 

MrChips

Joined Oct 2, 2009
30,706
The current is never zero, even after infinite time-constants.

After 5 TC the current will fall to 1/e^5 from the initial value, i.e. to 0.7%


capacitor charging voltage.jpg



capacitor discharging voltage.jpg
 

Thread Starter

SamR

Joined Mar 19, 2019
5,031
OK I knew it wouldn't be exactly 0 @ 5TC and 0.007 * 60mA = 420uA so (420 - 360.9) / 420 = 0.14*100 = 14% Still not close.
 

Thread Starter

SamR

Joined Mar 19, 2019
5,031
7% of 60mA is 420uA and my number was Ahhh not 360 but 404uA

so 420-404/420=0.038 * 100% = 3.8% error which is within the 5% margin for experimental error and therefore valid.

For 2TC 0.137 * 60mA = 8.2mA which is not even close to the 360.69uA which is the answer that the key gives me. Which now that I can validate the equation giving the correct results for 5TC I can consider valid.

So... i = Imax(e^-(30us/15us) = 0.06A(e^-2) = 8.12mA for 2TC is therefore correct and the key's answer is an errata...

OK mystery solved! Thanks Guys!
 
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