Voltage through a steady state capacitor.

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Discussion Overview

The discussion revolves around the behavior of a capacitor in an AC circuit driven by a sinusoidal source, specifically focusing on the voltage across the capacitor and the implications of steady state conditions. Participants explore the calculations related to reactance, impedance, and voltage division in the context of AC steady state analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asserts that in DC steady state, capacitors are open circuited, leading to a belief that the voltage across the capacitor is the voltage at the open circuit terminals.
  • Another participant clarifies that the circuit is in AC steady state due to the sinusoidal source, indicating that reactances must be considered for the capacitor and inductor.
  • Participants calculate reactances for the inductor and capacitor, with one providing specific expressions for their impedances based on given parameters.
  • There is confusion regarding the variables ε and β, with some participants assuming them to be zero while others use them in their calculations.
  • One participant reports calculated voltages across the capacitor, inductor, and resistor, and questions the form of the voltage expression for the capacitor.
  • Another participant discusses the phase relationships of the voltages, suggesting that the voltages should be expressed in sinusoidal form with phase shifts.
  • Participants check their calculations against the power relationship, noting that the results are approximately consistent with expectations.

Areas of Agreement / Disagreement

There is no clear consensus on the treatment of the variables ε and β, as some participants assume them to be zero while others incorporate them into their calculations. The discussion reflects multiple competing views on the interpretation of steady state conditions and the resulting calculations.

Contextual Notes

Limitations include potential misunderstandings of the term "steady state" in the context of AC circuits, as well as unresolved assumptions regarding the values of ε and β. The discussion also highlights the complexity of calculating voltages in AC circuits, particularly with respect to phase relationships.

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Homework Statement


Homework Equations


The Attempt at a Solution



According to my notes when a circuit is in DC steady state inductors are short circuited and capacitors are open circuited. This leads me to believe that the voltage across the capacitor is basically the voltage looking in through the terminal of the open circuit:

-----------R------o +
|

|
------------------o _

if it's an open circuit then no current flows through R.

(7sin12t - Vc)/R = 0

Vc = 7sin12t

This doesn't seem right to me, can anyone tell me where I'm going wrong?

thanks in advance.
 
Last edited:
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Your circuit isn't in DC steady state. It's being driven by a sinusoidal source, so it's in AC steady state. The inductor and capacitor will have reactances (impedances) at the given driving frequency. Find expressions for them and then do the voltage divider thing (or find the current first, then the voltage across the cap.).
 
This is an AC circuit with a supply voltage given by V = 7Sin12t.
This means that ω = 12.
I don't know what 'steady state' means in this context and would assume it means find the voltage (AC) across the capacitor.
I attempted to find the reactance of L and C and got the following:
Xl = ωL = 12 x (2 +β) = (24 + 12β)Ω
Xc =1/ωC = 1/(12 x 3x 10^-3) = 28Ω
The resistance = (12-ε)Ω
The next logical thing is to get an expression for the impedance, Z,
using Z^2 = [(Xl-Xc)]^2 + R^2
Then use this to get an expression for the current and therefore an expression for Vc
BUT I CAN'T SEE HOW TO DO ALL OF THAT !
Sorry ! Hope you get somewhere with it
 
gneill said:
Your circuit isn't in DC steady state. It's being driven by a sinusoidal source, so it's in AC steady state. The inductor and capacitor will have reactances (impedances) at the given driving frequency. Find expressions for them and then do the voltage divider thing (or find the current first, then the voltage across the cap.).

Thanks!

The phasor voltage of 7sin12t is 7

The impedance of the resistor is the resistance

the impedance of the cap is 1/j(12)(3mF)

impedance of the inductor is j(12)(L)
 
What happened to the ε and β?
 
gneill said:
What happened to the ε and β?

i'm assuming they're 0 for now, they are just variables with given values, sorry i didn't clarify that.
 
technician said:
This is an AC circuit with a supply voltage given by V = 7Sin12t.
This means that ω = 12.
I don't know what 'steady state' means in this context and would assume it means find the voltage (AC) across the capacitor.
I attempted to find the reactance of L and C and got the following:
Xl = ωL = 12 x (2 +β) = (24 + 12β)Ω
Xc =1/ωC = 1/(12 x 3x 10^-3) = 28Ω
The resistance = (12-ε)Ω
The next logical thing is to get an expression for the impedance, Z,
using Z^2 = [(Xl-Xc)]^2 + R^2
Then use this to get an expression for the current and therefore an expression for Vc
BUT I CAN'T SEE HOW TO DO ALL OF THAT !
Sorry ! Hope you get somewhere with it


thanks, sorry i didnt mention that epsilon and beta are variables, my bad.

i have basically the same work as you, i used voltage division to get the voltage through the capacitor and then reverted back from phasor to time domain form.

thanks for all the help
 
It is much easier now.
I got Vc =15.5V, Vl =13.2V and Vr = 6.6V
 
technician said:
It is much easier now.
I got Vc =15.5V, Vl =13.2V and Vr = 6.6V

should the answer not be in the form xsin(wt + y)?

for the voltage across the capacitor i got

vc(t) = 9.2186sin(12t + 0.288)

does this look correct?

edit: i used 6 for epsilon and 2 for beta, but the answer should be roughly there or thereabouts i hope.
 
  • #10
I thought ε and β were being taken to = 0 ?
You are right, my voltages should be given as
Vc = 15.5 Sin(12t)... Strictly should be 15.5Sin(12t - ∏/2) because Vc is ∏2 behind I
Vl = 13.2 Sin(12t)... strictly speaking should be 13.2 Sin(12t + ∏/2) because Vl is ∏/2 ahead of I
Vr = 6.6 Sin(12t)...this is OK as it is because Vr is inphase with I

A good check is that Vs^2 = Vr^2 + [Vl - Vc]^2

ie 7^2 = 6.6^2 + [13.2 - 15.5]^2
49 = 43.6 + 5.3 = 48.9 (near enough)
 
  • #11
technician said:
I thought ε and β were being taken to = 0 ?
You are right, my voltages should be given as
Vc = 15.5 Sin(12t)... Strictly should be 15.5Sin(12t - ∏/2) because Vc is ∏2 behind I
Vl = 13.2 Sin(12t)... strictly speaking should be 13.2 Sin(12t + ∏/2) because Vl is ∏/2 ahead of I
Vr = 6.6 Sin(12t)...this is OK as it is because Vr is inphase with I

A good check is that Vs^2 = Vr^2 + [Vl - Vc]^2

ie 7^2 = 6.6^2 + [13.2 - 15.5]^2
49 = 43.6 + 5.3 = 48.9 (near enough)


thanks.
 

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