Energy dissipated in the resistors in a 2 mesh RC circuit

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SUMMARY

The discussion focuses on calculating the energy dissipated in resistors within a two-mesh RC circuit, given a power supply of 50 volts and initial charges of 0 coulombs in capacitor C and 20^-6 coulombs in capacitor 2C. The user successfully calculated the current and voltage after a long time and at 10 milliseconds but struggled with the final readings and energy dissipation in the resistors. Key equations utilized include Ohm's law, capacitance formulas, and Kirchhoff's laws, emphasizing the need for a clearer circuit diagram to facilitate analysis.

PREREQUISITES
  • Understanding of Ohm's law (R=ΔV/I)
  • Familiarity with capacitance calculations (C=q/ΔV)
  • Knowledge of Kirchhoff's laws (Σi=0 and ΣΔV=0)
  • Experience with RC circuit analysis and energy calculations
NEXT STEPS
  • Study energy dissipation in RC circuits with multiple resistors and capacitors
  • Learn to create and analyze circuit diagrams for complex circuits
  • Explore the concept of time constants in RC circuits
  • Investigate simulation tools for circuit analysis, such as LTspice or Multisim
USEFUL FOR

Students and educators in electrical engineering, circuit designers, and anyone involved in analyzing RC circuits and energy dissipation in resistive networks.

Eduardo Leon
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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

Homework Equations


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

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.

14v5xa2k62r2x


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
 

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Last edited by a moderator:
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It would help things if you first produced a clearer circuit diagram.
 

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