RLC Circuit DC Homework Solution

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Homework Help Overview

The discussion revolves around an RLC circuit in a DC context, focusing on the behavior of the circuit at the moment the switch is closed and the initial conditions for current and charge. Participants explore the implications of applying Kirchhoff's loop and junction rules to derive equations governing the circuit's behavior.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the initial conditions of the circuit, particularly the current through the inductor and capacitor at time t=0. Questions arise regarding the behavior of the circuit elements just before and after the switch is closed, and whether certain terms in their equations can be simplified or eliminated.

Discussion Status

The discussion is active, with participants questioning assumptions about initial currents and the necessity of certain equations. Some guidance has been offered regarding the behavior of inductors and capacitors at t=0, but there is no explicit consensus on the interpretation of the results or the need for further calculations.

Contextual Notes

Participants note the absence of a power source before the switch is closed, leading to discussions about the initial values of current and charge. There is also mention of the need to consider the effects of resistors in the circuit when analyzing the behavior of the capacitor and inductor over time.

  • #31
Arman777 said:
Well ##i_2=0## since capacitor is full.There can't be current.I couldn't find ##i_1##...I think its 1A but I am ot sure...I thought ##ε=i_1(R_1+R_3)+L(di_1/dt)##
After very long time, the circuit is in stationary state. Nothing changes, the time derivative of the currents. i 1=1 A and i2=0 are correct.
Arman777 said:
Part (e) I guess Q=εC
Part (f) If its one ampere and I find charge correctly then ##U_L=1/2Li^2## and ##U_C=Q^2/2C##
The capacitor is not connected directly to the battery, there are two resistors in between. Is some potential drop across those resistors?
 
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  • #32
ehild said:
The capacitor is not connected directly to the battery, there are two resistors in between. Is some potential drop across those resistors?
Oh you are right yes, then ##Q/C=ε-i_3R_3-i_2R_2## ##i_2=0## and ##i_3=1A## ?
 
  • #33
Arman777 said:
Oh you are right yes, then ##Q/C=ε-i_3R_3-i_2R_2## ##i_2=0## and ##i_3=1A## ?
Yes.
 
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  • #34
ehild said:
Yes.
Thanks a Lot.
 
  • #35
Well done!
 
  • #36
Second well done

upload_2017-5-22_17-59-25.png


was what I was fishing for in #21
 
  • #37

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