RC Circuit Transient Response Analysis

[dudforreal]

Homework Statement

In the circuit 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 = 24 µs; 4: t → ∞.

In the circuit shown below the switch has been closed for a very long time and opens at
t = 0. Calculate the output voltage VO at:
1: t = 0−; 2: t = 0+; 3: t = t1 = 24 µs; 4: t → ∞.

The attempt at a solution

The only part I have trouble on is the getting the output voltage at t1 because I don't know how to obtain the equation in order to get the answer.

 

[gneill]

The only part I have trouble on is the getting the output voltage at t1 because I don't know how to obtain the equation in order to get the answer.

You need to determine the time constant for the circuit and write the expression for the voltage as a function of time. Replacing the resistor network and voltage source with a Thevenin equivalent may help with that.

 

[dudforreal]

I calculated Vo.e^(-t/RC) using the advice you gave but I got the wrong answer

 

[gneill]

I calculated Vo.e^(-t/RC) using the advice you gave but I got the wrong answer

You don't provide enough information to see where you went wrong; You'll have to post your calculations.

 

[dudforreal]

for a question similar to the last one only but t1=3.6µs the correct answer is -11.2066V. I calculated the equivalent resistance of the two parallel resistors being 750 ohms. Therefore
RC = 2.25µs and -t/RC = -1.6 and finally using V0.e^(-t/RC) I get -1.6404V which is incorrect.

 

[gneill]

for a question similar to the last one only but t1=3.6µs the correct answer is -11.2066V. I calculated the equivalent resistance of the two parallel resistors being 750 ohms. Therefore
RC = 2.25µs and -t/RC = -1.6 and finally using V0.e^(-t/RC) I get -1.6404V which is incorrect.

This doesn't help! I can't see the 'similar question'. Besides, you've got three different circuits in this thread. How are we to know which circuit your similar problem resembles?

Why don't you take one of your circuits that you're having problems with and detail your attempt at solution? Working though one may make the others clear.

 

[dudforreal]

Sorry about that, the 'similar question' is:

In the circuit shown below the switch has been closed for a very long time and opens at
t = 0. Calculate the output voltage VO at:
1: t = 0−; 2: t = 0+; 3: t = t1 = 3.6 µs; 4: t → ∞.

View attachment 39461

which the answer is -11.2066V for t1.

 

[gneill]

Sorry about that, the 'similar question' is:

In the circuit shown below the switch has been closed for a very long time and opens at
t = 0. Calculate the output voltage VO at:
1: t = 0−; 2: t = 0+; 3: t = t1 = 3.6 µs; 4: t → ∞.

View attachment 39461

which the answer is -11.2066V for t1.

How is that possible? The for the given circuit the voltage source is only 3V. When the switch opens the output can only rise to that.

 

[dudforreal]

Oh sorry there must be something wrong it should read -13V for the voltage source and 1200 for R1 and 2000 for R

 

[gneill]

Oh sorry there must be something wrong it should read -13V for the voltage source and 1200 for R1 and 2000 for R

Okay. Let's take those values. What value do you calculate for the output voltage before the switch is opened?

 

[dudforreal]

i get -13*(2000/3200) = -8.125V

 

[gneill]

i get -13*(2000/3200) = -8.125V

Okay, that looks good. Now, what happens when the switch is opened? What is the initial value of Vo, and where will it head?

 

[dudforreal]

It will be the same.

 

[gneill]

It will be the same.

and... where will it head? What value will it decay towards as time goes on?

 

[dudforreal]

It will head to -13V I think.

 

[gneill]

It will head to -13V I think.

Yes, that's right. So it will start at -8.125V and head to -13V. What's the time constant?

 

[dudforreal]

RC = 750*3n

 

[gneill]

RC = 750*3n

Okay, looks good. So how would you write an equation to describe the voltage transitioning from -8.125V to -13V with that time constant?

EDIT: My bad. When the switch opens, only R1 = 1200Ω is in series with the capacitor. So RC is 1200Ω * 3nF.

 

[dudforreal]

Vo*e^(-t/RC) = -8.125e^(-3.6/2.25)

 

[gneill]

Vo*e^(-t/RC) = -8.125e^(-3.6/2.25)

Vo is the output voltage. It's what the equation should be describing. I think you meant to use another variable name there. Let's call it Vth, which will have the value Vth = -8.125. So then you've provided:
$$Vo(t) = V_{th} e^{-\frac{t}{RC}}$$
This equation would describe an exponential decay from -8.125V up to zero volts. Not quite what you want. You should be ending at -13V as time goes on.

The form you want is:
$$Vo(t) = V_{start} + \Delta V \left(1 - e^{-\frac{t}{RC}}\right)$$
where $\Delta V = V_{end} - V_{start}$

This will describe a voltage curve that begins with V_start and decays to V_end. (Plug in V_th and V_s for V_start and V_end)

 

[dudforreal]

I got -12.016V instead of the correct answer -11.2066V.

 

[gneill]

I got -12.016V instead of the correct answer -11.2066V.

You'll have to show more of your calculations. Mine show the expected result. Perhaps you should check your time constant value.

 

[dudforreal]

I got -8.125-[(-13+8.125)*(1-e^(-3.6/2.25))] = -12.016

 

[gneill]

I got -8.125-[(-13+8.125)*(1-e^(-3.6/2.25))] = -12.016

Okay, my fault. I didn't pay close enough attention to your time constant calculation above. When the switch is open, only R1 is in the circuit. So the time constant should involve only R1 and the capacitor. Sorry about that. Try again with the new value.

 

[dudforreal]

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

 

[gneill]

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.

 

[dudforreal]

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]

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?

 

[dudforreal]

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

 

[gneill]

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?

 

[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.