Calculating the Steady State Value of Vc After Opening a Switch in a RC Circuit

Click For Summary
SUMMARY

The steady-state value of the capacitor voltage (Vc) after opening the switch in an RC circuit is calculated using the formula Vc = IR, where I is the current through the resistor. In this case, with a 1kΩ resistor and a current of 10mA, Vc equals 10V. The capacitor behaves as an open circuit at steady state, and the time taken for Vc to reach within 1 percent of its steady-state value can be determined using the equation Vc = Vie^(-t/RC).

PREREQUISITES
  • Understanding of RC circuits and capacitor behavior
  • Familiarity with Ohm's Law (V = IR)
  • Knowledge of exponential decay in electrical circuits
  • Ability to solve differential equations related to circuit analysis
NEXT STEPS
  • Study the concept of time constants in RC circuits
  • Learn about Norton and Thevenin equivalent circuits
  • Explore the use of Laplace transforms in circuit analysis
  • Investigate transient response in electrical circuits
USEFUL FOR

Electrical engineering students, circuit designers, and anyone studying transient analysis in RC circuits will benefit from this discussion.

compEng
Messages
5
Reaction score
0
Please Help me with this problem or with understanding it. Thank you!

Homework Statement


What is the stead-state value of Vc after the switch opens? Determine how long it takes after the switch opens before Vc is within 1 percent of its stead-state value.

Homework Equations



Vc = Vie-t/RC

The Attempt at a Solution

 

Attachments

  • circ.jpg
    circ.jpg
    23.8 KB · Views: 929
Last edited:
Physics news on Phys.org
It doesn't look like you've attempted the solution. but I'll give you some tips.
when the system is in a steady state the capacitor acts as an open circuit (ie: all the current goes through the 1k resistor.)

what is the voltage across the 1k resistor at steady state: 1x10^3 x 10x10^-3
what do you know about elements connected in parallel.

I've made this part really obvious for you. finish the first part and attempt the second part and I'll give you some more help if you need it.

Good luck
 
okay i think I figured it out.

Because the capitor becomes an open circuit, and because it is in parrelel with the resistor it begins to imitate a Norton Equivilant circuit. and Vc=IR

Vc= 10 * 10-3 * 103 = 10

thank you!
 

Similar threads

How do PI controllers achieve steady state?
  • · Replies 6 ·
Replies
6
Views
2K
Engineering DC Steady State Circuit Analysis: Solving for iL, i, and Vc in a Simple Circuit
  • · Replies 14 ·
Replies
14
Views
4K
Engineering Transient RC Circuit (got answer in s but not in t-domain)
  • · Replies 2 ·
Replies
2
Views
2K
Engineering Initial and final values for second order circuits
  • · Replies 9 ·
Replies
9
Views
4K
What Happens to the Current Through an Inductor When the Switch Opens?
  • · Replies 9 ·
Replies
9
Views
3K
Engineering How to Determine Vc(0) in a DC State RC Circuit?
  • · Replies 1 ·
Replies
1
Views
2K
Voltage through a steady state capacitor.
  • · Replies 10 ·
Replies
10
Views
8K
Engineering Sinusoidal steady state analysis using Laplace transform
  • · Replies 3 ·
Replies
3
Views
2K
Engineering Find di/dt and dr/dt for a DC circuit with indepedent current source
  • · Replies 3 ·
Replies
3
Views
4K
Energy Losses in a Circuit with a Switching Element
  • · Replies 7 ·
Replies
7
Views
1K