Find the charge on the plates a long time after the switch is closed

Click For Summary
SUMMARY

The charge on the plates of a capacitor in a circuit can be determined by analyzing the steady state conditions after the switch is closed. In this case, the current leaving the battery is 0.962 A. By multiplying this current by the resistance of 27 ohms, the voltage across the capacitor can be calculated. In steady state, capacitors behave as open circuits, allowing for simplification of the circuit analysis by eliminating the capacitors from consideration.

PREREQUISITES
  • Understanding of basic circuit theory
  • Knowledge of Ohm's Law
  • Familiarity with capacitors and their behavior in DC circuits
  • Ability to analyze steady state conditions in electrical circuits
NEXT STEPS
  • Study the concept of steady state in RC circuits
  • Learn how to calculate voltage and charge on capacitors
  • Explore complex circuit analysis techniques
  • Review the implications of eliminating capacitors in circuit simplifications
USEFUL FOR

Students studying electrical engineering, circuit designers, and anyone looking to understand capacitor behavior in DC circuits.

Jimmy25
Messages
69
Reaction score
0

Homework Statement



I'm having a problem with the circuit in the attached diagram. I am looking for the charge on the plates of the capacitor a long time after the switch is closed.

Homework Equations





The Attempt at a Solution



I found the current leaving the battery is 0.962 A a long time after the switch is closed. The solution says to multiply this by 27 ohms to find the voltage across the capacitor from which the charge can be calculated. I don't understand how this is possible since the diamond shaped segment is a complex circuit.
 

Attachments

  • Picture 1.png
    Picture 1.png
    7.7 KB · Views: 464
Physics news on Phys.org
A long time after the switch is closed, we have a steady state regime, that is the capacitors are charged up so no current flows through them. You can simply eliminate them from the circuit(mentally) and then calculate the current and the voltage drops between their plates.

Hope this helps.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 62 ·
3
Replies
62
Views
6K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 21 ·
Replies
21
Views
5K