Parallel RC with current source

In summary, the first image shows a circuit with a switch, and the second image shows two circuits with the switch open. The first circuit is Thev transform, and the second circuit uses the Thevenin transformation to find the voltage at t = 0. The results of the calculations show that the voltage is 90 V. The 9 Ω resistor is not part of the Thevenin transformation, and the circuit would still be valid if it was included. The calculations also show that the voltage at t > 0 is 30 V.
  • #1
dwn
165
2

Homework Statement



See images

In the first image, I made a circuit with the switch closed and then two when the switch is opened.
The bottom circuit is the Thev transform.

Homework Equations


The Attempt at a Solution



Let me first explain my understanding of the circuit. Prior to t = 0, the dc source been suppling a current to capacitor for an lengthy amount of time in which case the cap would reach a 'steady state' of voltage. I decided to find this voltage by using a Thevenin transformation ( 90 V and 6Ω series).

vc(0) = 90 V
My concern with this is that when I created the new circuit in iCircuit app, the cap voltage only read 30 V..?

vc(∞) = 0 V (intuitive understanding of the circuit)

In order to find vc(t), I went back to the original circuit and decided to try:
Code:
v[SUB]c[/SUB](t) = 1/C ∫i dt
At this point, I just got lost and wasn't sure how to continue.
 

Attachments

  • Screen Shot 2014-04-28 at 10.27.57 AM.jpg
    Screen Shot 2014-04-28 at 10.27.57 AM.jpg
    14 KB · Views: 1,085
  • Screen Shot 2014-04-28 at 10.29.19 AM.jpg
    Screen Shot 2014-04-28 at 10.29.19 AM.jpg
    9.7 KB · Views: 1,110
Physics news on Phys.org
  • #2
Can you explain your Thevenin equivalent calculations? The result you get doesn't look right.
 
  • #3
Sure thing, I was doing this late in the evening/early morning depending on perspective, so that could be why.

Starting with 18 Ohm and 5 A source and treating the rest as open circuit: V = IR = 18 * 5 = 90 V

1/18 + 1/9 = 1/R R = 6
 
  • #4
dwn said:
Sure thing, I was doing this late in the evening/early morning depending on perspective, so that could be why.

Starting with 18 Ohm and 5 A source and treating the rest as open circuit: V = IR = 18 * 5 = 90 V

Okay, what about the 9 Ω resistor? Won't it form a voltage divider with your new Thevenin model? Did you consider incorporating the 9 Ω into the Thevenin model to make things simpler?
 
  • #5
With a dc voltage source, isn't the capacitor shorted? Which would mean the resistors are in series?

If I did use the voltage divider: V = 9(90/27) = 30V
 
  • #6
dwn said:
With a dc voltage source, isn't the capacitor shorted? Which would mean the resistors are in series?
Nope. If you take only one of the resistors into your Thevenin equivalent the you have a voltage source in series with one resistor which together parallel the remaining resistor and capacitor.

You know that you can change around the depicted order of components in parallel without altering the circuit (the same is true of components in series --- the sum of their resistances and potential drops are constants). If you rearrange and simplify your circuit before applying Thevenin then things will be easier:

attachment.php?attachmentid=69174&stc=1&d=1398698412.gif

If I did use the voltage divider: V = 9(90/27) = 30V

Right. But it's better to simplify the resistor network first and reduce the complexity.
 

Attachments

  • Fig2.gif
    Fig2.gif
    2.3 KB · Views: 1,953
  • #7
Yes, I think I remember that now that you mention it.

This stuff is killing all confidence I had in myself as a human being...may be time for a beer.
 
Last edited:
  • #8
Analyzing the circuit for t > 0

RC equation : v(t) = V0e-t/RC = 30e-t/0.6

Correct?
 
  • #9
dwn said:
Analyzing the circuit for t > 0

RC equation : v(t) = V0e-t/RC = 30e-t/0.6

Correct?

Yes. You got it :approve:
 
  • #10
FINALLY! Are we allowed to make donations to you guys?!? If not, you should talk to someone about that!
 
  • #11
dwn said:
FINALLY! Are we allowed to make donations to you guys?!? If not, you should talk to someone about that!

You can upgrade your Membership to become a paying member (QUICK LINKS ---> Upgrade Membership). Fees contribute to the upkeep of the site infrastructure (all staff are unpaid volunteers; we do it because promoting and encouraging science education is its own reward).
 

What is a parallel RC circuit with a current source?

A parallel RC circuit with a current source is an electrical circuit that consists of a resistor (R) and a capacitor (C) connected in parallel, with a current source (I) providing a constant flow of current. The circuit is used to filter or regulate current in a circuit.

How does a parallel RC circuit with a current source work?

In a parallel RC circuit with a current source, the resistor and capacitor provide different paths for the current to flow. The capacitor stores electrical charge, while the resistor resists the flow of current. The current source ensures a constant flow of current, and the voltage across the resistor and capacitor is determined by the current and the values of R and C.

What is the formula for calculating the impedance of a parallel RC circuit with a current source?

The impedance (Z) of a parallel RC circuit with a current source can be calculated using the formula Z = R + 1/(jωC), where R is the resistance in ohms, C is the capacitance in farads, ω is the angular frequency in radians per second, and j is the imaginary unit.

How does a parallel RC circuit with a current source affect the frequency of a signal?

A parallel RC circuit with a current source acts as a low-pass filter, meaning it allows low-frequency signals to pass through while attenuating high-frequency signals. The cutoff frequency of the circuit is determined by the values of R and C. Signals with frequencies below the cutoff frequency will pass through relatively unchanged, while signals with frequencies above the cutoff frequency will be attenuated.

What are the applications of a parallel RC circuit with a current source?

A parallel RC circuit with a current source is commonly used in electronic circuits to filter out unwanted noise or to regulate current. It is also used in audio equipment to remove unwanted high-frequency signals and in power supplies to smooth out the output voltage. Additionally, it is used in timing circuits, oscillators, and other electronic devices.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
10
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
Back
Top