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Question about free source RLC circuit

  • Engineering
  • Thread starter e0ne199
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  • #1
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Homework Statement


below is my question :
Scrapbook_1492781963940.png


Homework Equations


all of the equatios used to solve this problems are equations for free source RLC circuit.

The Attempt at a Solution


so far i have solved problem (a), whose answer is -1400V/s, the problem is when i want to solve (b) and (c)
INKredible_20170430_170617.png

do you know how to solve the remaining questions?? any help is really appreciated, thx before
 

Answers and Replies

  • #2
gneill
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do you know how to solve the remaining questions?? any help is really appreciated, thx before
Looks like you'll need to analyze the circuit. What methods have you covered for solving second order circuits?
 
  • #3
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I calculated it to get answers (b) and (c) using overdamped response equations but i don't know why the answers i get is not the same as the answers displayed above...did you know how to solve it?
Btw the question i posted is about free source RLC circuit when it enters transient conditions
 
  • #4
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Looks like you'll need to analyze the circuit. What methods have you covered for solving second order circuits?
Can you help me?
 
  • #5
gneill
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I calculated it to get answers (b) and (c) using overdamped response equations but i don't know why the answers i get is not the same as the answers displayed above...did you know how to solve it?
Btw the question i posted is about free source RLC circuit when it enters transient conditions
My intuition tells me that the circuit will be underdamped, not overdamped. A hint that this might be the case is that in part (c) they ask for the first value of t where the capacitor voltage is zero. That implies that there's at least one more, so there are oscillations.

You'll have to show how you used the standard response equations to solve the problem. It's not obvious to me how you would incorporate and account for the initial conditions (both capacitor voltage and inductor current).

Looking at the circuit, my own approach would likely be to use Laplace Transforms and nodal analysis. The Laplace transform method allows you to easily incorporate the initial conditions. If you haven't covered Laplace yet, you'll have to slog through writing and solving the differential equation.
 
  • #6
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My intuition tells me that the circuit will be underdamped, not overdamped. A hint that this might be the case is that in part (c) they ask for the first value of t where the capacitor voltage is zero. That implies that there's at least one more, so there are oscillations.

You'll have to show how you used the standard response equations to solve the problem. It's not obvious to me how you would incorporate and account for the initial conditions (both capacitor voltage and inductor current).

Looking at the circuit, my own approach would likely be to use Laplace Transforms and nodal analysis. The Laplace transform method allows you to easily incorporate the initial conditions. If you haven't covered Laplace yet, you'll have to slog through writing and solving the differential equation.
ah sorry i was wrong...you are right, the circuit is underdamped,sorry i didn't realize i typed a wrong circuit response...btw i have done calculations involving simple differential equations based on the book but i am unable to get solutions that fit the answers for (b) and (c)...i have done the calculations several times and check if there is anything wrong within my calculations but i am still unable to find where my mistake is....could you please help me find the way to get answer (b) and (c)?
 
  • #7
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i am really out of clue this time and i am so desperate to find the answer for this question...
 
  • #8
gneill
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You'll have to show your work so that helpers can evaluate your approach and what you've tried. Helpers cannot do the work for you, they can only offer help with work that you show.
 
  • #9
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IMG_20170523_220900.png


sorry for the late post...here is my result for part (b)....the answer is not the same as the book so do tou know what actually is the problem for my answer? thx before
 
  • #10
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hello? anyone?
 
  • #11
44
1
You'll have to show your work so that helpers can evaluate your approach and what you've tried. Helpers cannot do the work for you, they can only offer help with work that you show.
i have just posted my solution above, could you please check where the wrong is on my answer?
 
  • #12
gneill
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It's very hard to follow your handwriting (at least for me).

From what I can tell your result looks almost okay: The exponential damping term should use the damping coefficient α = 100, not an ω. Otherwise it looks good.
 
  • #13
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1
It's very hard to follow your handwriting (at least for me).

From what I can tell your result looks almost okay: The exponential damping term should use the damping coefficient α = 100, not an ω. Otherwise it looks good.
but the answer is not the same as the question i posted for the first time...do you know the real solution for this? i think my answer is still wrong...hope you (or anyone here) can help solve this with correct solution...
 
  • #14
44
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View attachment 204109

sorry for the late post...here is my result for part (b)....the answer is not the same as the book so do tou know what actually is the problem for my answer? thx before
about this...i tried to input t=1 ms and i can't get the same answer (ans: 0.695) as the question i posted before...still need a helping hand to solve this problem..
 
  • #15
gneill
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Can you show details for your calculation of the voltage at time t = 1 ms? I arrive at the correct answer when I use your expression for v(t).
 
  • #16
44
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Can you show details for your calculation of the voltage at time t = 1 ms? I arrive at the correct answer when I use your expression for v(t).
i am sorry again for the late post....
here it is :
IMG_20170604_111809.png


can you show me your answer?
 
  • #17
44
1
Can you show details for your calculation of the voltage at time t = 1 ms? I arrive at the correct answer when I use your expression for v(t).
hello? are you still there? [emoji4][emoji4]
 
  • #18
gneill
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Your handwriting is really very difficult to read. Please type out your equations or write them larger and more carefully.

From what I can tell, you must have to have made a calculator error.
 
  • #19
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perhaps this is clearer?
could you point out which one is wrong?
ep21mgWds.png
 
Last edited:
  • #20
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Your handwriting is really very difficult to read. Please type out your equations or write them larger and more carefully.

From what I can tell, you must have to have made a calculator error.
i have just posted my revision, could you please check it out? :)
i really want to know what mistake i have done here because i have struggled so hard to get the right answer for this :(
 
  • #21
gneill
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Just your result is incorrect. That means you have a calculator error. Check to see if your calculator is set to degrees or radians mode for the trig functions.
 
  • #22
44
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Just your result is incorrect. That means you have a calculator error. Check to see if your calculator is set to degrees or radians mode for the trig functions.
sorry for the late response again...i have checked my calculator and everything including the configuration is alright...how did you get the correct result? could you show it to me?
or could you show me the correct expression of V (t) before you put t=1ms into the expression?
 
  • #23
gneill
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sorry for the late response again...i have checked my calculator and everything including the configuration is alright...how did you get the correct result? could you show it to me?
or could you show me the correct expression of V (t) before you put t=1ms into the expression?
Your expression looks fine. From what I can make out you have written:

##v(t) = e^{-100 t}\left(2 cos(200 t) - 6 sin(200 t) \right)##

which is correct. What you wrote for t = 1 ms also looks good. So that means you have a calculator issue.

What value do you calculate for ##2 cos(0.2)## ?
 
  • #24
44
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Your expression looks fine. From what I can make out you have written:

##v(t) = e^{-100 t}\left(2 cos(200 t) - 6 sin(200 t) \right)##

which is correct. What you wrote for t = 1 ms also looks good. So that means you have a calculator issue.

What value do you calculate for ##2 cos(0.2)## ?
the result of 2*cos (0.2) is 1.99 using my calculator (TI89 emulator)...is that wrong?
 
  • #25
gneill
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the result of 2*cos (0.2) is 1.99 using my calculator (TI89 emulator)...is that wrong?
Yes. Looks like it's assuming an argument of degrees. Switch to radians.
 

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