RC circuit -- finding the charge on each capacitor

In summary, the conversation involves finding the charge on capacitors in a circuit with a given voltage, resistance, and capacitance. The approach involves using the rules for solving circuit problems, such as the current into a node being equal to the current coming out and the sum of voltage changes around a closed loop being zero. The solution involves suppressing the capacitors and simplifying the circuit to find the potentials at different points and then using the equation q=CΔV to find the charges on the capacitors. Through this approach, the potential at points D, A, and B were found to be 18 V, 12 V, and 6 V respectively.
  • #1
fedecolo
61
1

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?
 

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  • #2
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.
 
  • #3
fedecolo said:

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, [itex]i_4[/itex] and [itex]i_6[/itex] 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.
 
  • #4
stevendaryl said:
If the capacitors are fully charged, then the current through the capacitor is zero. So in your drawing, [itex]i_4[/itex] and [itex]i_6[/itex] 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}##?
 
  • #5
fedecolo said:
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. [itex]i_5[/itex] is zero, also.
 
  • #6
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:
upload_2017-7-12_9-32-47.png


Pick a suitable reference node and determine the potentials at A, B, D, and E.
 
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  • #7
gneill said:
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##
 
  • #8
fedecolo said:
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.
 
  • #9
gneill said:
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?
 
  • #10
fedecolo said:
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?
 
  • #11
gneill said:
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?
 
  • #12
fedecolo said:
The potential in E is zero, isn't it?
How did you arrive at that conclusion? (hint: No, it isn't zero).
 
  • #13
gneill said:
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?
 
  • #14
fedecolo said:
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?
 
  • #15
gneill said:
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}##
 
  • #16
fedecolo said:
So it's ## 4 \mathscr{E}##
How do you arrive at that conclusion?
 
  • #17
gneill said:
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?
 
  • #18
fedecolo said:
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.
 
  • #19
gneill said:
Yes; correct.

Thank you!
 

1. What is an RC circuit?

An RC circuit is a type of electrical circuit that contains a resistor (R) and a capacitor (C). These components work together to store and release electrical energy, making an RC circuit useful for various applications such as filtering and time-delay circuits.

2. How do I find the charge on each capacitor in an RC circuit?

To find the charge on each capacitor in an RC circuit, you can use the formula Q = CV, where Q is the charge, C is the capacitance of the capacitor, and V is the voltage across the capacitor. You can also use Kirchhoff's laws and the equations for capacitors in series and parallel to solve for the charge on each capacitor.

3. What is the significance of finding the charge on each capacitor in an RC circuit?

Finding the charge on each capacitor in an RC circuit allows you to understand how the circuit is functioning and how much energy is stored in each capacitor. This information can be useful for troubleshooting and optimizing the circuit for different applications.

4. Can I use the charge on each capacitor to calculate the total energy stored in an RC circuit?

Yes, you can use the charge on each capacitor to calculate the total energy stored in an RC circuit. The formula for calculating energy stored in a capacitor is E = 1/2 CV^2, where E is the energy, C is the capacitance of the capacitor, and V is the voltage across the capacitor.

5. How does the charge on each capacitor change over time in an RC circuit?

The charge on each capacitor in an RC circuit changes over time according to the equation Q = Q0 (1-e^-t/RC), where Q is the charge at any given time, Q0 is the initial charge, t is time, R is the resistance in the circuit, and C is the capacitance of the capacitor. This equation shows an exponential decay in the charge on each capacitor as time increases.

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