Capacitance homework problem help

Click For Summary

Discussion Overview

The discussion revolves around understanding the behavior of capacitors in electrical circuits, specifically why they act as short circuits when uncharged and as open circuits when fully charged. The scope includes conceptual clarification and mathematical reasoning related to capacitor charging in the context of a DC voltage source.

Discussion Character

  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant questions why a capacitor behaves like a short circuit when uncharged and like an open circuit when fully charged, expressing confusion about the implications of the equation i=Cdv/dt in a DC voltage context.
  • Another participant explains that initially, a capacitor is neutral, allowing current to flow as if it were short-circuited. As charge accumulates, the developing electric field opposes the voltage source, eventually stopping current flow when fully charged, akin to an open circuit.
  • A third participant elaborates on the behavior using Kirchhoff's Voltage Law, indicating that when uncharged, the capacitor can be treated as non-existent (short-circuited), while when fully charged, the current becomes zero (open circuit). They also mention a transient state where current is initially high before stabilizing.
  • A later reply acknowledges the helpfulness of the previous answers, indicating some level of understanding gained from the discussion.

Areas of Agreement / Disagreement

Participants express varying degrees of understanding regarding the behavior of capacitors, but there is no explicit consensus on the explanations provided. The discussion includes multiple perspectives on the underlying principles.

Contextual Notes

Some assumptions about the nature of current flow and the behavior of capacitors under different conditions may not be fully articulated. The discussion also touches on transient states, which may require further exploration to clarify the dynamics involved.

ranju
Messages
221
Reaction score
3
Why capacitance behaves as shortcircuited in uncharged state and why as open circuited when its fully charged..?? I am reallt doubtful..:confused:
i=Cdv/dt...so if we r appyling a dc voltage ..in that case i should be zero.>! thn why short-circuited..??
 
Physics news on Phys.org
A capacitor stores electrical energy by moving charges off of one plate and onto the other. Initially each plate is electrically neutral, just like a normal conductor, so the first few charges require very little work to move and current flows as if the capacitor were short circuited. As more and more charges are moved from one plate to the other, an electric field develops that opposes the voltage source, meaning that each additional charge requires more work to move. When the electric field is equal to the applied voltage, there is no longer any difference in electric potential (voltage) between the voltage source and the plates so you have no more current flow, just like an open.

Does that help?
 
  • Like
Likes   Reactions: 1 person
ranju said:
Why capacitance behaves as shortcircuited in uncharged state and why as open circuited when its fully charged..?? I am reallt doubtful..:confused:
Well for example when we charge a capacitor with a voltage source V via a resistance R, then we have from kirchhoffs Voltage Law
V-I*R-q/C=0 (1).

When the capacitor is uncharged then q=0 the term q/C is also zero, hence what is left from (1) is V-IR=0, it is like the capacitor has been short circuited (or it doesn't exist).

When capacitor is fully charged then q=C*V hence from (1) follows that -I*R=0 thus I=0, it is like the circuit is open and no current can flow.

i=Cdv/dt...so if we r appyling a dc voltage ..in that case i should be zero.>! thn why short-circuited..??
i=0 in the steady state. Before the steady state, there will be a transient state of time dt where the current has very high value (theoretically infinite ) and it charges the capacitor very rapidly (within time dt) to final charge q=C*V. So again the capacitor passes from the "short circuit" stage where the current is very high to the open circuit stage where the current is zero.
 
  • Like
Likes   Reactions: 1 person
Drakkith..and Delta your ans. is quite helpful. thanxx..
 

Similar threads

Undergrad Computing the capacitance of a strange material
  • · Replies 3 ·
Replies
3
Views
1K
Undergrad Charging of capacitors in series
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad Can current flow in an open circuit?
  • · Replies 4 ·
Replies
4
Views
3K
Graduate Theoretical doubt regarding capacitance
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad Voltage & Capacitance: LanguageNerd's Questions
  • · Replies 7 ·
Replies
7
Views
2K
Undergrad Grounding a circuit through a capacitor/parallel RC circuit
  • · Replies 7 ·
Replies
7
Views
3K
Undergrad Help Solve Voltage Homework Questions for Engineering Student
  • · Replies 8 ·
Replies
8
Views
3K
Equivalent capacitance homework problem
  • · Replies 67 ·
3
Replies
67
Views
13K
Undergrad Is this example incorrect in Agawal's book on electric circuits?
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Numerical calculations of a railgun yielded disappointing results
  • · Replies 4 ·
Replies
4
Views
2K