Find v(t) for t > 0. RC Circuit Problem

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    Circuit Rc Rc circuit
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Discussion Overview

The discussion centers around solving for the voltage v(t) in an RC circuit after a switch is opened at t = 0. Participants explore the behavior of the circuit before and after the switch operation, applying relevant equations and concepts related to transient analysis in electrical circuits.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents an initial solution for v(t) using the equation V(t) = Voe-t/RC, calculating specific values for the circuit components.
  • Another participant mentions a similar but more complex problem, indicating the need for further assistance.
  • A participant notes that the capacitor behaves as an open circuit only in steady state (t = infinity) and suggests using superposition to analyze the transient response.
  • There is confusion expressed regarding the application of superposition, particularly about the subtraction of voltage sources in the calculations, with a participant questioning the reasoning behind this approach.
  • A later reply clarifies that the negative sign in the voltage source affects the calculation, indicating that the orientation of the voltage sources must be considered in superposition analysis.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the application of superposition in the context of the problem, with some expressing confusion about the calculations and the reasoning behind them. The discussion remains unresolved regarding the correct interpretation of superposition in this scenario.

Contextual Notes

Some assumptions about the circuit configuration and the behavior of the capacitor may not be fully articulated. The discussion includes references to specific component values and circuit diagrams that are not visible in the text.

bob29
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Homework Statement


The switch in the circuit has been closed for a long time, and it opens at t = 0. Find v(t) for T \geq 0.
[PLAIN]http://img834.imageshack.us/img834/9107/tutorialq2.jpg

Homework Equations


V(t) = Voe-t/RC

\tau = RC

The Attempt at a Solution


@ T<0
Short Circuit due to capacitor
V = V2k
By VDR:
V = 2/(10+2) * 24 = 4V

@ T \geq 0
24V is open therefore
Req = (10//2) = 5/3 ohm
C = 40microFarads

\tau = (5/3)k * 40*10^-6
RC = 1/15

V(t) = 4e-15000t
Or
[PLAIN]http://img713.imageshack.us/img713/5351/28102010077.jpg

Forgot to add in the extra 1000 due to the K
therefore answer should be 1000 less or V(t) = 4e-15t
 
Last edited by a moderator:
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Problem solved.
Having trouble with a similar but more complex problem.

1. Homework Statement .

Question is in the image including diagram.
http://img163.imageshack.us/i/electricalproblem.jpg/
[PLAIN]http://img163.imageshack.us/img163/2886/electricalproblem.jpg

Homework Equations


tau = τ = Req * Capacitor
v(t) = V(∞) + [v(0) - v(∞)]e-t/τ

The Attempt at a Solution



[PLAIN]http://img100.imageshack.us/img100/4793/attempt1t0a.jpg[PLAIN]http://img100.imageshack.us/img100/1591/attempt1t0.jpg
 
Last edited by a moderator:
Capacitor is an open circuit only in steady state (t = infinity).
To solve for the transient you can use superposition.
 
In the working out of the answer.
V(infinity) = (6/2+6)*10 - (2/6+2)*50 = -5
I don't understand how it's subtracting since superposition you have to add.
 
bob29 said:
In the working out of the answer.
V(infinity) = (6/2+6)*10 - (2/6+2)*50 = -5
I don't understand how it's subtracting since superposition you have to add.

Because the 50V source has the - sign at the top and the + sign connected to ground.
 

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