Derive the governing equation for the voltage Vc across the capacitor

• Engineering
• annamal
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.
annamal
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....

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 ...

##\ ##

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.

hutchphd said:
Please note that the problem is oddly stated and the time of interest (steady state) is t <0.
I agree.

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.

So the 1 and 2 ##k\Omega## resistors have the same current ?

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

##\ ##

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.

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

##\ ##

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

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

##\ ##

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?

Wouldn't the left circuit and the right circuit be isolated from each other ?

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

##\ ##

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....

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
?

##\ ##

BvU said:
Like here
View attachment 325428 ?

##\ ##
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.

 or do KVL for each of the loops ....

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.

BvU
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:

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?

Another tack: what do you find for ##V_C## and ##I_L## for ##t>>0## ?

##\ ##

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
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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.

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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