Finding Vo(t) After 125ms in Ideal Circuit w/ Charged Capacitor

In summary, the op amp in the circuit is ideal and operating in the linear region. The capacitor is initially charged to 2V. After 125ms, the output voltage Vo(t) is found. The voltage divisor rule is used to convert to impedance and the transform into phasor domain is used to convert back into time domain. I am weak at circuits and I cannot solve this problem. I would suggest using phasors in the frequency domain or fixing the initial conditions using the capacitor relation i=C dv/dt.
  • #1
Studious_stud
39
0

Homework Statement


If the op amp in the following circuit is ideal and operating in the linear region, and the capacitor is initially charged to 2V, find the output voltage Vo(t) after 125ms
63rfk9.jpg


Homework Equations


Voltage divisor rule, converting to impedance


The Attempt at a Solution



Transform into phasor domain, convert back into time domain?

I really have no idea how to go about this one, the fact that the capacitor is charged and finding the output voltage after 125ms is throwing me off.

I'm quite weak at circuits anyway, but unfortunately I have to take one module in them...

Thanks for any help guys, would be great.
 
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  • #2
Since it's a DC input, it may be easiest to use the relation between current and voltage for a capacitor here.
 
  • #3
Studious_stud said:
Transform into phasor domain, convert back into time domain?

I really have no idea how to go about this one, the fact that the capacitor is charged and finding the output voltage after 125ms is throwing me off.

I'm quite weak at circuits anyway, ...

This is a problem where a strength in math will make up for a weakness in circuits. This circuit is an ideal integrator (with the assumptions given). You could solve this one with phasors or in the frequency domain, but I would suggest staying in the time domain on this one.

You can use the capacitor relation i=C dv/dt, Ohms law i=v/R and the fact that the negative terminal of the opamp is a virtual ground (i.e. it is held at ground potential even though not connected to ground) to derive an output voltage which is an integral of the input voltage. The initial charge on the capacitor will be related to the integration constant to fix the initial conditions.

So here you are just integrating the constant input signal and you know that the integral of a constant is a linear ramp.

Post your attempt at this and we can guide you if you have trouble.
 

FAQ: Finding Vo(t) After 125ms in Ideal Circuit w/ Charged Capacitor

How do you calculate the voltage at a specific time (t) in an ideal circuit with a charged capacitor?

To calculate the voltage at a specific time (t) in an ideal circuit with a charged capacitor, you can use the formula V(t) = Vmax(1-e^(-t/RC)), where Vmax is the maximum voltage of the capacitor, R is the resistance in the circuit, and C is the capacitance of the capacitor.

Can the voltage at a specific time (t) be negative in an ideal circuit with a charged capacitor?

Yes, the voltage at a specific time (t) can be negative in an ideal circuit with a charged capacitor. This can occur if the capacitor is discharging and the voltage drops below zero.

How does the resistance in the circuit affect the voltage at a specific time (t) in an ideal circuit with a charged capacitor?

The resistance in the circuit affects the voltage at a specific time (t) in an ideal circuit with a charged capacitor by determining the rate at which the capacitor discharges. A higher resistance will result in a slower discharge rate, while a lower resistance will result in a faster discharge rate.

Can the voltage at a specific time (t) be greater than the maximum voltage of the capacitor in an ideal circuit with a charged capacitor?

No, the voltage at a specific time (t) cannot be greater than the maximum voltage of the capacitor in an ideal circuit with a charged capacitor. The maximum voltage of the capacitor is the point at which the capacitor is fully charged and cannot hold any more energy.

What happens to the voltage at a specific time (t) as the capacitance of the capacitor increases in an ideal circuit with a charged capacitor?

As the capacitance of the capacitor increases in an ideal circuit with a charged capacitor, the voltage at a specific time (t) will also increase. This is because a larger capacitance allows the capacitor to hold more charge, resulting in a higher voltage.

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