• Support PF! Buy your school textbooks, materials and every day products Here!

Energy dissipated in the resistors in a 2 mesh RC circuit

1. Homework Statement
Hi mates, I have problems solving the third part of this exercise, I've already done all the previous calculations.

Given the following circuit, where the switch S is open, the power supply = 50 volts and:
  • The initial charge in the C capacitor: QC = 0 coulombs
  • The initial charge in the 2C capacitor: Q2C = 20^-6 coulombs
Calculate (with real instruments):
  1. Switch⇒I find the current passing by the amperemeter (IA) and the voltage across the voltmeter VV After a long time t (t>5 time constants). (Done).
  2. Switch⇒II find the voltage across the voltmeter VV passed 10 milliseconds. (Done).
  3. Switch⇒III next get the final reading of the voltmeter VV at a given time t (t >0 ∧ t< 5 time constants) and the energy dissipated in the resistors till that instant.


attachment.jpg




2. Homework Equations
  • Ohm's law: R=ΔV/i
  • Capacitance: C=q/ΔV
  • Kirchoff's laws: Σi=0 ∧ ΣΔV=0
  • Energy gived by the battery untill an instant t
    Image1.gif
  • Energy dissipated in the resistor untill an instant t
    Image2.gif
  • Energy stored in the capacitor untill an instant t
    Image3.gif



3. The Attempt at a Solution
As stated before, I've done the first and the second parts.

For the first:
  • 58_59e4a5e4c7a317067e0e748d4c7303e8.png
    Amps*
  • 55_39ed5ab39f3e1b932d58b8143acf9655.png
    Volts*
For the second:
  • 49_549e15a8fc3c64f775b461c0ce4c83f2.png
    73_a3a7a2e80e63986351b15fb412bb6203.png
    34_76bab7f98db551c5bfbe82b73004a907.png
    Volts*
For the third:
Well, here is where i get lost, because I dont't know how to study and analize the circuit formed at this point, I've only seen discharging RC circuits whis a single current, one resistor and one capacitor due those elements could be joined into their equivalent ones, but here, I get 2 resistors, 2 capacitors and 2 meshes, like this one.



I know that the initial charges at this point are:

In order to get the initial charge in the C capacitor I used the voltage obtained in the voltmeter in the second part at the 10 milliseconds, because are connected parallel. So, multiplying that voltage times the capacitance, we get
49_f93ffa7bf895f1ac07d8dc5df756a1e0.png



The initial charge in the 2C capacitor stills equal, since its an ideal capacitor, so is the same: Q2C = 20^-6 coulombs
 

Attachments

Last edited by a moderator:

Answers and Replies

165
45
It would help things if you first produced a clearer circuit diagram.
 

Related Threads for: Energy dissipated in the resistors in a 2 mesh RC circuit

Replies
2
Views
20K
Replies
4
Views
542
Replies
3
Views
2K
Replies
5
Views
1K
Replies
31
Views
1K
Replies
2
Views
5K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
2
Views
2K
Top