RCC circuit, posible mistake with constants

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1f5
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Homework Statement



There's a circuit, with a capacitor C1, Resistor R1, and another capacitor C2.
In t=0 ; C1 is charged with 2 V and C2 with 1 V
The value of resistor is 1 ohm, C1= 1 F C2=0.5 F



Homework Equations



Kirchoff voltage law.
Capacitor voltage at any given time. (integral form)


The Attempt at a Solution



I solve the resulting diferential equation, but don't know constant value. I believed that it was -3

One of the things that the problem request if to find a expresion for v1(t), v2(t). I substituted i(t) (diferential equation solution) in the integral form of capacitor voltage formula, well that was easy, but another incise in the problem is to prove that v1(inf)=v2(inf), with the constants from the last integration it is impossible.
 
1f5 said:

Homework Statement



There's a circuit, with a capacitor C1, Resistor R1, and another capacitor C2.
In t=0 ; C1 is charged with 2 V and C2 with 1 V
The value of resistor is 1 ohm, C1= 1 F C2=0.5 F

Homework Equations



Kirchoff voltage law.
Capacitor voltage at any given time. (integral form)

The Attempt at a Solution



I solve the resulting diferential equation, but don't know constant value. I believed that it was -3

One of the things that the problem request if to find a expresion for v1(t), v2(t). I substituted i(t) (diferential equation solution) in the integral form of capacitor voltage formula, well that was easy, but another incise in the problem is to prove that v1(inf)=v2(inf), with the constants from the last integration it is impossible.

Is there a circuit diagram to go with this problem? There are several ways to connect three components, and various polarities for the charges on the capacitors.
 
Last edited:
RCC picture

Circuit schematic with terminals v1 and v2, a resistor R, switch K, capacitors C1 and C2, and a loop current i drawn around the lower mesh.

upload picture

Later i discovered that I have a mistake with voltage law, then i resolve the DE and had Ce^(-t/r)
After that i make a sim, and see that the current in t=0 is -1A and tends to zero when t goes to infinity. So the function must be -e^(-t).
(r1=equals 1).

I saw voltages too, but i can't figure from which function they come.
 
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The initial current is unlikely to be negative if initially v1 > v2. Also, how do you arrive at a time constant of 1s? By inspection we can see that the current loop has C1 and C2 in series with a resistance R. What's the effective capacitance for C1 and C2 in series? The time constant should then resolve to RCeff when you solve your differential equation.

Perhaps you should show more of your steps used to solve the DE.
 
1f5 said:
Yes, you were right.

Handwritten derivation on lined paper showing a circuit sketch at top with a resistor and capacitor, followed by a series of loop equations and differential equation steps with annotations.

Don't let the initial potentials on the capacitors fool you into making wrong choices regarding potential drops due to current flow through them. Whichever plate of a capacitor the current flows INTO with be the + plate as far as the potential drop for the differential equation is concerned.

Handwritten circuit diagram of a single loop with capacitors C1 and C2 and resistor R, with current i circulating clockwise, and the equation -(1/C1)∫i dt = Ri + (1/C2)∫i dt below it.


If you want to make it more obvious, note that you can model a pre-charged capacitor as an uncharged capacitor in series with a voltage supply:

Circuit schematic with a resistor R across the top between two nodes, each branch containing a capacitor C1 or C2 in series with a battery V1o or V2o, a loop current i drawn as a red arc, and voltages V1 and V2 marked on the outer sides.
 
Attachments
  • Handwritten circuit schematic of a series loop with capacitors C1 and C2 and resistor R, current i circulating clockwise, with polarity signs marked, and below it the equation -(1/C1) integral of i dt = R i + (1/C2) integral of i dt.
    Fig1.gif
    13.1 KB · Views: 1,017
  • Circuit schematic with a resistor R on top between two branches, each containing a capacitor C1 or C2 in series with a battery V1o or V2o; a red clockwise loop current i is drawn in the middle, and voltages V1 and V2 are marked on the left and right.
    Fig2.gif
    1.8 KB · Views: 1,021