Derive the governing equation for the voltage Vc across the capacitor

In summary: Can you explain explicitly? I still believe there can be current from a to b...There is no potential difference between a and b if the two circuits are not connected to the same ground.
  • #1
annamal
381
33
Homework Statement
Derive the governing equation for the voltage Vc across the capacitor
Relevant Equations
KCL and KVL
I was thinking about doing KVL around the circuit at the right but I noticed when the switch opens, the current through the circuit at the right is not the same throughout
-5 + Ic*2*1-^3 + Ic*10^3 = -Vc

Ic is not the same around the right circuit so I am stuck....

Screenshot 2023-04-23 at 4.41.25 PM.png
 
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  • #2
Hi,

I suppose you did part a and you have ##V_c## and ##I_L## at ##t\uparrow\infty## ?

annamal said:
-5 + Ic*2*1-^3 + Ic*10^3 = -Vc
Care to explain this ? In particular the second term ?
I can't even read it " ##I_c * 2 * 1 - ?^3 ## "?
Do you mean ##I_{b_2} * 2\ k\Omega## ?

I also think you need some more relevant equations ...

##\ ##
 
  • #3
BvU said:
I suppose you did part a and you have Vc and IL at t↑∞ ?
Please note that the problem is oddly stated and the time of interest (steady state) is t <0.
 
  • #4
hutchphd said:
Please note that the problem is oddly stated and the time of interest (steady state) is t <0.
I agree.
 
  • #5
BvU said:
Hi,

I suppose you did part a and you have ##V_c## and ##I_L## at ##t\uparrow\infty## ?Care to explain this ? In particular the second term ?
I can't even read it " ##I_c * 2 * 1 - ?^3 ## "?
Do you mean ##I_{b_2} * 2\ k\Omega## ?

I also think you need some more relevant equations ...

##\ ##
Yeah, I meant KVL:
-5 + Ic*2*10^3 + Ic*10^3 = -Vc

I know what part a is. I am wondering about part b.
 
  • #6
So the 1 and 2 ##k\Omega## resistors have the same current ?

And do you mean ##I_c = I_{b_2} ## ?

##\ ##
 
  • #7
BvU said:
So the 1 and 2 ##k\Omega## resistors have the same current ?

And do you mean ##I_c = I_{b_2} ## ?

##\ ##
I don't know if the resistors have the same current -- that is just an attempt of mine. In fact I think the resistors might have different current.
 
  • #8
Backing up: do you have ##V_C## and ##I_L## for ##t<0## ?
And for ##t\uparrow\infty## ?

##\ ##
 
  • #9
BvU said:
Backing up: do you have ##V_C## and ##I_L## for ##t<0## ?
And for ##t\uparrow\infty## ?

##\ ##
I only know Vc = 3 V; I_L =0.0015 A for t < 0
 
  • #10
Agreed. Now, when the switch is opened, can there still be a current through the 3 ##k\Omega## ?

##\ ##
 
  • #11
BvU said:
Agreed. Now, when the switch is opened, can there still be a current through the 3 ##k\Omega## ?

##\ ##
I think it can though...why can it not?
 
  • #12
Wouldn't the left circuit and the right circuit be isolated from each other ?

1682344794325.png


(thought experiment: imagine the 2 mH wasn't an inductor but a resistor)

##\ ##
 
  • #13
BvU said:
Wouldn't the left circuit and the right circuit be isolated from each other ?

View attachment 325426

(thought experiment: imagine the 2 mH wasn't an inductor but a resistor)

##\ ##
I still don't understand why the current in the middle is zero....if I do KCL at point a or b, it seems like there could be current there....
 
  • #14
annamal said:
I still don't understand why the current in the middle is zero....if I do KCL at point a or b, it seems like there could be current there....
Like here
1682345972722.png
?

##\ ##
 
  • #15
BvU said:
Like here
View attachment 325428 ?

##\ ##
Can you explain explicitly? I still believe there can be current from a to b...
 
  • #16
There is no potential difference between a and b if the two circuits are not connected to the same ground.

[edit] or do KVL for each of the loops ....
 
  • #17
A circuit is called a circuit because there is a return path that completes the circle.
Without the return path, there can be no current in R=3k.
The value of R=3k becomes irrelevant and may as well be a short circuit.
Nodes a and b will then have the same voltage.
 
  • Like
Likes BvU
  • #18
Baluncore said:
A circuit is called a circuit because there is a return path that completes the circle.
Without the return path, there can be no current in R=3k.
The value of R=3k becomes irrelevant and may as well be a short circuit.
Nodes a and b will then have the same voltage.
I don't understand. I drew the circle and currents in blue below:
1682344794325.png
 
  • #19
annamal said:
I don't understand. I drew the circle and currents in blue below:
Would you expect a light globe to work if you only connected one wire to the battery?
 
  • #20
Another tack: what do you find for ##V_C## and ##I_L## for ##t>>0## ?

##\ ##
 
  • #21
First, if the start process is finished the current through L will be d.c. that means ω=2*π*f=0 then the voltage drop through L will be 0.

On the capacitor eventually, after infinite time, the current will be 0 since Zcap=1/ ω/cap and if ω=0 Zcap=∞

Second, if sw1 is open that it is like instead sw1 a resistance of ∞ ohms is inserted here. So, in the loop between a and b the current it is total voltage drop [what ever it is] in the loop divided by ∞ [no current will flow here].
 

Attachments

  • Infinite resistance.jpg
    Infinite resistance.jpg
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  • #22
By-the-way, if between two points in a circuit the voltage drop is 0 then the potential V1=V2. The voltage drop =Z*I. If Z=0 or I=0 the voltage drop is 0.:smile:
 
  • #23
It is not so complicate. You have to find Vc=-(5-Vo)*e^(-γ*t)+5 like formula where Vo it is Vc before sw1 opening.
 

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