Find the functions of V1, V2, V3 in an RC circuit

Click For Summary
The discussion focuses on finding the functions of voltages V1, V2, and V3 in an RC circuit using Kirchhoff's current law (KCL) and matrix representation. The problem involves three capacitors and two resistors, with initial conditions specified for the voltages at the nodes. Participants emphasize the need to express currents in terms of voltage differences across circuit elements to derive the correct differential equations. The conversation highlights that the sum of the potentials V1, V2, and V3 remains constant over time, which relates to the conservation of charge principle. Overall, the thread seeks to clarify the approach to formulating and solving the circuit equations.
Zoja
Messages
2
Reaction score
1
Homework Statement
Consider the following electrical network with three capacitors ##C (1 nF)## and two resistors ##R=1M\Omega##.
It represents a model of three cells of capacitance ##C## connected by a gap junction with resistance##R##.
At time## t=0## (initial condition), the potential at point ##1 (V_1)## is ##100 mV## (with respect to ground) and the potential at point ##2 (V_2)## and at point ##3 (V_3)## is ##0 mV##.
How do the potentials ##V_1## , ##V_2## and ##V_3## evolve with time? Find the functions ##V_1=f_1(t)## , ##V_2=f_2(t)## and ##V_3=f_3(t)##
Prove that## V1 + V2 + V3## does not change with time. To what physical principle does this correspond?

(I apologize if I posted in the wrong section, but it is homework given to me..and I am also new to the forum)
Relevant Equations
##I=\frac{dQ}{dt}##
##V_r=IR##
##V_c=\frac{Q}{C}##

##Q_1+Q_2+Q_3=Q_1(0)##
I tried using Kirchhof's current law, and to pose the problem in matrix form as ##\frac{dv}{dt}=Mv## with## v## the vector of the ##3## potentials at nodes ##1, 2## and ##3##, and ##M## is a ##3x3## matrix.
it would be enough to show me which will be the differential equations, I would proceed by solving them by myself.
 

Attachments

  • 1.png
    1.png
    2.3 KB · Views: 300
Physics news on Phys.org
Zoja said:
Problem Statement: Consider the following electrical network with three capacitors ##C (1 nF)## and two resistors ##R=1M\Omega##.
It represents a model of three cells of capacitance ##C## connected by a gap junction with resistance##R##.
At time## t=0## (initial condition), the potential at point ##1 (V_1)## is ##100 mV## (with respect to ground) and the potential at point ##2 (V_2)## and at point ##3 (V_3)## is ##0 mV##.
How do the potentials ##V_1## , ##V_2## and ##V_3## evolve with time? Find the functions ##V_1=f_1(t)## , ##V_2=f_2(t)## and ##V_3=f_3(t)##
Prove that## V1 + V2 + V3## does not change with time. To what physical principle does this correspond?

(I apologize if I posted in the wrong section, but it is homework given to me..and I am also new to the forum)
Relevant Equations: ##I=\frac{dQ}{dt}##
##V_r=IR##
##V_c=\frac{Q}{C}##

##Q_1+Q_2+Q_3=Q_1(0)##

I tried using Kirchhof's current law
Welcome to the PF. :smile:

Can you show us the equations you got for the KCL nodes?
 
##i_1=-I_1+I_2##

##i_2=I_1-2I_2+I_3##

##i_3=I_2-I_3##
 
Um, no. That's obviously of no help.

KCL equations involve the sum of all currents out of each node, sure, but you need to express those currents in terms of the voltage difference across impedances leading out of each node. If it's all resistors, that's just linear equations. When there are inductors and capacitors, you write the differential equations and solve them.

Have at it! :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
11K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 11 ·
Replies
11
Views
5K