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Voltage through a steady state capacitor.

  • #1

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:

Answers and Replies

  • #2
gneill
Mentor
20,792
2,770
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.).
 
  • #3
1,506
17
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
 
  • #4
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)
 
  • #5
gneill
Mentor
20,792
2,770
What happened to the ε and β?
 
  • #6
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.
 
  • #7
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
 
  • #8
1,506
17
It is much easier now.
I got Vc =15.5V, Vl =13.2V and Vr = 6.6V
 
  • #9
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
1,506
17
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
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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