# RC Circuit Transient Response

• Engineering
gneill
Mentor
ok I got the answer but is the equation the same for the question shown below:

In the circuits shown below the switch has been opened for a very long time and closes at
t = 0. Calculate the output voltage VO at:
1: t = 0−; 2: t = 0+; 3: t = t1 = 5.6 µs; 4: t → ∞.
View attachment 39459 View attachment 39460

No, the equation won't be the same for them. They will exhibit slightly different behavior. Pick one to analyze and describe what you think will happen when the switch closes.

For the first one, Vo before t=0 would be 3.35V and after t=0 it would be 0V and would be going to 3.35V.

gneill
Mentor
For the first one, Vo before t=0 would be 3.35V and after t=0 it would be 0V and would be going to 3.35V.

I think that should be 3.322V, but yes, that's the behavior (Assuming that the capacitor is initially uncharged at t = 0. The circuit doesn't provide any mechanism for discharging the capacitor when the switch is open, so who knows what voltage it may be carrying from its "prehistory"?).

What do you figure for the time constant?

Yes it should be 3.322V, my mistake. I think the time constant should 2942.37 x 3.9n = 1.1475*10^-5.

gneill
Mentor
Yes it should be 3.322V, my mistake. I think the time constant should 2942.37 x 3.9n = 1.1475*10^-5.

Yes, that's right, about 11.48 μs. And since its a simple exponential rise from 0V to 3.322V the equation is the simple form. What result do you get for t=24μs?

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

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

Where'd the 6 come from?

Isn't that Vth?

gneill
Mentor
Isn't that Vth?

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

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

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

That looks good. The numbers and calculation are good.

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

:zzz:

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
Mentor
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.

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
Mentor
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.

ok thanks i got the answer