The problem statement, all variables and given/known data 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.
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.
You don't provide enough information to see where you went wrong; You'll have to post your calculations.
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.
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.
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?
Okay, that looks good. Now, what happens when the switch is opened? What is the initial value of Vo, and where will it head?
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.
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: [tex] Vo(t) = V_{th} e^{-\frac{t}{RC}}[/tex] 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: [tex] Vo(t) = V_{start} + \Delta V \left(1 - e^{-\frac{t}{RC}}\right) [/tex] where [itex]\Delta V = V_{end} - V_{start}[/itex] 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})