Engineering RC Circuit Transient Response Analysis

[dudforreal]

for t=24μs I get Vo as 6 x 0.1236 = 0.742V.

 

[gneill]

for t=24μs I get Vo as 6 x 0.1236 = 0.742V.

Where'd the 6 come from?

 

[dudforreal]

Isn't that Vth?

 

[gneill]

Isn't that Vth?

Perhaps we're looking at different problems? I thought you'd stated that Vo goes from 0V to 3.322V?

 

[dudforreal]

oh sorry...so it is 3.322 x 0.1236 = 0.411V.

 

[gneill]

oh sorry...so it is 3.322 x 0.1236 = 0.411V.

That looks good. The numbers and calculation are good.

 

[gneill]

Gotta go catch some zZ's now. I'll check back in the morning (my morning, that is).

:zzz:

 

[dudforreal]

I think that my answer is wrong because when I used this equation on a different question with the same circuit with the voltage source being 4V, R1 = 1600, R = 2000 and C = 10n the answer I got was 0.798V but the correct answer is 1.4239V.

 

[gneill]

Good morning. (*YAWN*)

I took another look at the problem. A small correction is required to the time response formula.

This time the exponential is starting at zero and heading towards a final value that's equal to the Thevenin voltage of the resistor network. So the form is:
V_o(t) = V_{th} \left(1 - e^{-t/\tau} \right)

Regarding your "different question with the same circuit", apply the above suggested formula and recheck your result -- also check to see if the time t1 to be plugged in is the same for both problems.

 

[dudforreal]

ok cool i got the answer. Now for the second part of the question with the switch between the resistor and the voltage source, my answer for the time before and after 0 as 0V and it is going to -5(7500/17500) = -2.143V.

 

[gneill]

ok cool i got the answer. Now for the second part of the question with the switch between the resistor and the voltage source, my answer for the time before and after 0 as 0V and it is going to -5(7500/17500) = -2.143V.

Yes, that's what I see for that circuit.

 

[dudforreal]

ok thanks i got the answer