RC circuit -- finding the charge on each capacitor

[fedecolo]

Homework Statement

I have to find the charge $q$ on the capacitors (all have capacity $C= 4 \mu F$) and $\mathscr{E} = 12 V$, $R=50 \Omega$

Homework Equations

The Attempt at a Solution

I wrote the nodes current equations
$\begin{cases} i_{1}+i_{2}=i_{3}\\ i_{3}=i_{4}+i_{5}+i_{6}\\ ... \end{cases}$ etc
But when I write the mesh equations I don't know if the current pass or not through the capacitors (because they are all charged). So how can I write the mesh equations?

 

[Merlin3189]

Since you say the capacitors are all charged, I think you are looking for the steady state solution. This means, in a DC circuit all the currents and voltages are constant.

 

[stevendaryl]

Homework Statement

I have to find the charge $q$ on the capacitors (all have capacity $C= 4 \mu F$) and $\mathscr{E} = 12 V$, $R=50 \Omega$

Homework Equations

The Attempt at a Solution

I wrote the nodes current equations
$\begin{cases} i_{1}+i_{2}=i_{3}\\ i_{3}=i_{4}+i_{5}+i_{6}\\ ... \end{cases}$ etc
But when I write the mesh equations I don't know if the current pass or not through the capacitors (because they are all charged). So how can I write the mesh equations?

If the capacitors are fully charged, then the current through the capacitor is zero. So in your drawing, i_4 and i_6 must be zero (after the capacitors are fully charged).

The rules for solving these circuit problems are:

1. The current into a node must be equal to the current coming out of the node.
2. The sum of the voltage changes around any closed loop must be zero.

You have to apply #2 to get the capacitor charges.

 

[fedecolo]

If the capacitors are fully charged, then the current through the capacitor is zero. So in your drawing, i_4 and i_6 must be zero (after the capacitors are fully charged).

The rules for solving these circuit problems are:

1. The current into a node must be equal to the current coming out of the node.
2. The sum of the voltage changes around any closed loop must be zero.

You have to apply #2 to get the capacitor charges.

So, for example the equation of the loop at the bottom right is $3 \mathscr{E} - i_{3}R - \frac{q}{C} - i_{1}R=0$ ?
And $i_{5}$ must not be $0$ as $i_{4},i_{6}$?

 

[stevendaryl]

So, for example the equation of the loop at the bottom right is $3 \mathscr{E} - i_{3}R - \frac{q}{C} - i_{1}R=0$ ?

Yes.

And $i_{5}$ must not be $0$ as $i_{4},i_{6}$?

No. i_5 is zero, also.

 

[gneill]

Hi fedecolo,

A handy trick when analyzing the steady-state of a circuit with L and C components is to suppress those components first. That means removing capacitors (open circuit) and replacing inductors with a wire (short circuit).
With the resulting simplified circuit you should be able to determine the potentials at locations where the capacitors were removed and the currents where the inductors were located.

Removing the capacitors from your circuit yields:

Pick a suitable reference node and determine the potentials at A, B, D, and E.

 

[fedecolo]

Removing the capacitors from your circuit yields:
View attachment 207014

Pick a suitable reference node and determine the potentials at A, B, D, and E.

Thank you so much!
So the equation of the external loop is $3 \mathscr{E}- \mathscr{E} -4i_{1} R=0$.
The potential in D is $18 V$
in A is $12 V$
in B is $6 V$
in E is zero.
To find the charges I only have to substitute the difference of potential in $q= C \Delta V$

 

[gneill]

Thank you so much!
So the equation of the external loop is $3 \mathscr{E}- \mathscr{E} -4i_{1} R=0$.
The potential in D is $18 V$
in A is $12 V$
in B is $6 V$
in E is zero.
To find the charges I only have to substitute the difference of potential in $q= C \Delta V$

That's the idea. Can you show the details of your calculations for the potentials? They don't all look correct to me.

 

[fedecolo]

That's the idea. Can you show the details of your calculations for the potentials? They don't all look correct to me.

Ok.
By the equation of the external loop I find $i_{1}= \frac{\mathscr{E}}{2R}= 0,12 A$
in D: $3 \mathscr{E} - \mathscr{E} + i_{2}R =V_{D}$
but $i_{2}=-i_{1}$
so $V_{D}= 2 \mathscr{E}-i_{1}R= 2 \cdot 12 V - 0,12 A \cdot 50 \Omega = 18 V$

in A: $V_{D}-i_{1}R= 12 V$

in B: $V_{A} - i_{1}R = 6 V$

Are they correct?

 

[gneill]

Ok.
By the equation of the external loop I find $i_{1}= \frac{\mathscr{E}}{2R}= 0,12 A$
in D: $3 \mathscr{E} - \mathscr{E} + i_{2}R =V_{D}$
but $i_{2}=-i_{1}$
so $V_{D}= 2 \mathscr{E}-i_{1}R= 2 \cdot 12 V - 0,12 A \cdot 50 \Omega = 18 V$

in A: $V_{D}-i_{1}R= 12 V$

in B: $V_{A} - i_{1}R = 6 V$

Are they correct?

Yes. You didn't mention that you chose node H as your reference node. That being the case, your calculations are good.

What is the potential at E?

 

[fedecolo]

Yes. You didn't mention that you chose node H as your reference node. That being the case, your calculations are good.

What is the potential at E?

Oh yes I forgot it, sorry

The potential in E is zero, isn't it?

 

[gneill]

The potential in E is zero, isn't it?

How did you arrive at that conclusion? (hint: No, it isn't zero).

 

[fedecolo]

How did you arrive at that conclusion? (hint: No, it isn't zero).

The currents $i_{4},i_{5},i_{6}$ (in my first draw) are zero, and the current that goes in E must goes out equal. So the current that goes in E ($i_{3}$) must be zero, so the potential in E is zero?

 

[gneill]

The currents $i_{4},i_{5},i_{6}$ (in my first draw) are zero, and the current that goes in E must goes out equal. So the current that goes in E ($i_{3}$) must be zero, so the potential in E is zero?

If the current that goes from E to G is zero, what is potential drop across the resistor in that path?

 

[fedecolo]

If the current that goes from E to G is zero, what is potential drop across the resistor in that path?

So it's $4 \mathscr{E}$

 

[gneill]

So it's $4 \mathscr{E}$

How do you arrive at that conclusion?

 

[fedecolo]

How do you arrive at that conclusion?

Because of Kirchhoff's law the sum of currents that goes in E must goes out from E. So $i_{3}=0$. So there is no voltage drop at the resistance.
So E is at the same potential of G? If it's ok, the potential is $3 \mathscr{E}$.
Am I correct now?

 

[gneill]

Because of Kirchhoff's law the sum of currents that goes in E must goes out from E. So $i_{3}=0$. So there is no voltage drop at the resistance.
So E is at the same potential of G? If it's ok, the potential is $3 \mathscr{E}$.
Am I correct now?

Yes; correct.

 

[fedecolo]

Yes; correct.

Thank you!