RLC Circuit DC Homework Solution

AI Thread Summary
The discussion focuses on solving a homework problem related to an RLC circuit and the behavior of currents at the moment the switch is closed. Participants clarify that at time t=0, both the current through the inductor and the charge on the capacitor are zero, which simplifies the equations needed to find the initial values of the currents. They emphasize that while the equations are important for later calculations, understanding the initial conditions is crucial for solving the problem. The conversation also touches on the behavior of the circuit as it reaches a steady state over time, confirming that the current through the capacitor will be zero once fully charged. Overall, the thread highlights the importance of applying fundamental circuit principles to analyze transient and steady-state conditions in RLC circuits.
Arman777
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Homework Statement


upload_2017-5-21_15-17-45.png


Homework Equations




The Attempt at a Solution


I used the loop rule where ##ΣΔV=0## and junction rule.
From here I get 4 equations
##i_3=i_1+i_2##
##ε-i_3R_3-i_1R_1-L(di\dt)=0##
##ε-i_3R_3-i_2R_2-Q\C=0##
##i_1R_1-L(di\dt)+Q\C+i_2R_2=0##

1-I said Q\C will disappear since Q=0 at t=0. But what about (di\dt) ? Is (di\dt)=0 ? I think it should be, but If its zero I didnt understand why it asked again since we need to claim its zero to solve the question.If its not zero then I couldn't solve the upper 4 equations.İts like something is missing.
 
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You do not need any equations here.
Just apply what you know about the behaviour of L and C at the moment of energization i.e. t=0.

What is the current through the inductor "just before" the switch is closed?
 
cnh1995 said:
You do not need any equations here.
Just apply what you know about the behaviour of L and C at the moment of energization i.e. t=0.

What is the current through the inductor "just before" switch is closed?
Isnt it a time depented value ? Function like I(t)=Awcos(wt)+Bwsin(wt) ? I don't know...
 
Arman777 said:
Isnt it a time depented value ?
Yes.
But you only need to know what happens at t=0.
cnh1995 said:
What is the current through the inductor "just before" the switch is closed?
 
cnh1995 said:
Yes.
But you only need to know what happens at t=0.
Can you help a bit more..I couldn't see the solution
 
There was no source in the circuit before the switch is closed.
So what can you say about the inductor current "before" the switch is closed?
 
Zero ?
 
Arman777 said:
Zero ?
Yes. So what is the inductor current just after closing the switch?
 
cnh1995 said:
Yes. So what is the inductor current just after closing the switch?
zero ?
 
  • #10
Arman777 said:
zero ?
Yes.
 
  • #11
Arman777 said:
1-I said Q\C will disappear since Q=0 at t=0. But what about (di\dt) ? Is (di\dt)=0 ? I think it should be, but If its zero I didnt understand

You could have just say yes.. ? Are my equations wrong ? Cause I need to find currents.
 
  • #12
Arman777 said:
You could have just say yes.. ?
I just answered your question. You need to be more specific about your problems.
Arman777 said:
Cause I need to find currents.
Only at t=0. As I said, you do not need any of the equations you've written.

What can you say about the current through the capacitor at t=0 i.e. just after the switch is closed?
 
  • #13
Ok. I think you do need to solve the equations for solving b and c because of presence of R1. The current through R1 is governed by both L and C values. My apologies.
But you do not need them to know i1 and i2 at t=0.
 
  • #14
cnh1995 said:
you do not need them to know i1 and i2 at t=0.
Question asks that,I need to know
 
  • #15
Arman777 said:
Question asks that,I need to know
Yes. So what is the current through the capacitor at t=0?
How does a capacitor behave when energized at t=0?
 
  • #16
Arman777 said:

Homework Statement


View attachment 203932

Homework Equations




The Attempt at a Solution


I used the loop rule where ##ΣΔV=0## and junction rule.
From here I get 4 equations
##i_3=i_1+i_2##
##ε-i_3R_3-i_1R_1-L(di\dt)=0##
##ε-i_3R_3-i_2R_2-Q\C=0##
##i_1R_1-L(di\dt)+Q\C+i_2R_2=0##

The fourth equation is not needed. And di/dt should be di1/dt.
So you have the system of equations
##i_3=i_1+i_2##
##ε-i_3R_3-i_1R_1-L(di_1/dt)=0##
##ε-i_3R_3-i_2R_2-Q/C=0##
Substitute ##i_3=i_1+i_2## for i3. You might need the relation between Q and i2. Then you have a system of equations for two currents and their derivatives.
The problem asks the initial values. You found that i1(0)=0. What about i2(0)? Answer @cnh1995's question.
 
Last edited:
  • #17
cnh1995 said:
Yes. So what is the current through the capacitor at t=0?
How does a capacitor behave when energized at t=0?
Zero ? At the beginning there should be no current ? Seems wrong but also makes sense.
ehild said:
The fourth equation is not needed. And di/dt should be di1/dt.
So you have the system of equations
##i_3=i_1+i_2##
##ε-i_3R_3-i_1R_1-L(di_1/dt)=0##
##ε-i_3R_3-i_2R_2-Q/C=0##
Substitute ##i_3=i_1+i_2## for i3. You might need the relation between Q and i2. Then you have a system of equations for two currents and their derivatives.
The problem asks the initial values. You found that i1(0)=0. What about i2(0)? Answer @cnh1995's question.
I see well I find ##-i_2R_2=Q\C## but in any case at t=0,Q=0
 
  • #18
Yes but for t>0 you should get something else
 
  • #19
BvU said:
Yes but for t>0 you should get something else

upload_2017-5-22_10-48-55.png

Rest of the question.Well at ##t=∞## there would be no current on the capacitor brench.

In the brench where inductor exist, there would be current.I can write an equation like, ##ε-iR_3-iR_1-L(di\dt)=0##.Do I need to solve this ? I couldn't directly see the answer.
For (e) probably its ##εC##.
For (f), I need to find that current passes through inductor,I know the equations and ıf my answer to e is true then I can find it easily.
 
  • #20
Arman777 said:
Zero ? At the beginning there should be no current ? Seems wrong but also makes sense.

I see well I find ##-i_2R_2=Q\C## but in any case at t=0,Q=0
No, it is not true . You ignored i3, and it is not zero at t=0.
 
  • #21
Arman777 said:
can write an equation like, ##\varepsilon-iR_3-iR_1-L(di/dt)=0##. Do I need to solve this ? I couldn't directly see the answer
Yes, no and Oh? :smile: What do you expect of the ##dI/dt## in the last term ?
(e) No. draw the circuit a little differently: ##\varepsilon## on the left and all the R etc on the right
(f) What you say is true...
 
Last edited:
  • #22
ehild said:
No, it is not true . You ignored i3, and it is not zero at t=0.
Shame on me o:). Wake up first and then check PF. Over to you, Ersbeth
 
  • #23
I didnt understand...I ll ask my proffesor I guess...
 
  • #24
Arman777 said:
I didnt understand...I ll ask my proffesor I guess...
Substitute I3=I1+I2 into your second and third equations. What do you get?
Use that i1=0 and Q=0 at t=0 in the second equation. What is i2 at t=0?
Knowing i2(0) and i1(0) you get d(i1)/dt from the first equation.
 
  • #25
ehild said:
Substitute I3=I1+I2 into your second and third equations. What do you get?
Use that i1=0 and Q=0 at t=0 in the second equation. What is i2 at t=0?
Knowing i2(0) and i1(0) you get d(i1)/dt from the first equation.
I find ##i_2=2A## and ##di_1\dt=6A\s##
 
  • #26
Correct, but use / in fractions, instead of \.
an oblique stroke (/) in print or writing, used between alternatives (e.g., and/or ), in fractions (e.g., 3/4 ), in ratios (e.g., miles/day ), or between separate elements of a text.
So ##di_1/dt=6A/s##.
Now take the derivative of the second equation, and use that i2=dQ/dt. Calculate di2/dt.
 
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  • #27
ehild said:
Calculate di2/dt.
Why do we need this ? .I ll do that in a moment
 
  • #28
In Part c it asks ##q_{upper}/dt## which that's equal to ##i_2=2A## as you said.
 
  • #29
Arman777 said:
Why do we need this ? .I ll do that in a moment
Oh, well, it was not asked.
Then go to the questions about infinite time. What are i1 and i2?
 
  • #30
ehild said:
Oh, well, it was not asked.
Then go to the questions about infinite time. What are i1 and i2?
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)##
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##
 
  • #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?
 
  • #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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