Simple RLC Circuit problem


by Raihan
Tags: circuit, simple
Raihan
Raihan is offline
#1
Oct16-06, 11:08 PM
P: 19
Please help me to solve this RLC circuit problem. I am completely confused.If you give me the direct answer it would be much appreciated.
For the series RLC circuit in Figure, find the input/output
difference equation for

1.[tex] y(t)=v_{R} [/tex]
2.[tex] Y(t)=i(t) [/tex]
3.[tex] y(t)=v_{L} [/tex]
4.[tex] y(t)=v_{C} [/tex]

I have attached the Circuit diagram in a .jpg file.
Attached Thumbnails
RLC circuit.JPG  
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berkeman
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#2
Oct17-06, 09:35 AM
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You must show your own work in order for us to help you (PF homework forum rules). Would KCL or KVL be the best way to start?
Raihan
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#3
Oct17-06, 08:07 PM
P: 19
Hey first I tried taking the KVL around the loop
something like
[tex] -x(t) + v_c(t) + v_L (t) + v_R (t) = 0 [/tex]----(1)
replaced v_L(t) with first order Ldi_L(t)/dt and make an equation for
[tex] v_C(t) [/tex]
and then as its in series I tried to write a function for
[tex] i_L(t) = \frac {v_R (t)} R [/tex]------(2)
and for [ tex ] v_R(t)/R=C \frac {dv_c(t)} {dt} [/tex]----(3)
Then tried sub (3) in (1)
and got
[tex] v_C(t) = x(t) - \frac {L} {R} dv_R(t)/dt - v + R(t)[/tex]----(4)
and then tried sub it i eqn 3. and didnt come up with a satisfactory result.
Please help.
Thanks

SGT
#4
Oct18-06, 07:20 AM
P: n/a

Simple RLC Circuit problem


Quote Quote by Raihan
Hey first I tried taking the KVL around the loop
something like
[tex] -x(t)+v_c(t)+v_L(t)+v_R(t)=0 [/tex]----(1)
replaced v_L(t) with first order Ldi_L(t)/dt and make an equation for
[tex] v_C(t) [/tex]
and then as its in series I tried to write a function for
[tex] i_L(t)=v_R(t)/R [/tex]------(2)
and for [tex] v_R(t)/R=Cdv_c(t)/dt [/tex]----(3)
Then tried sub (3) in (1)
and got
[tex] v_C(t)=x(t)-\frac {L} {R}dv_R(t)/dt-v+R(t)[/tex]----(4)
and then tried sub it i eqn 3. and didnt come up with a satisfactory result.
Please help.
Thanks
In series circuits you should always use [tex]v_C[/tex] as the independent variable (and [tex]i_L[/tex] in parallel circuits).
Since the current is the same for all elements, write [tex]v_L[/tex] and [tex]v_R[/tex] as functions of the current. Finally write the current as a function of [tex]v_C[/tex].
Raihan
Raihan is offline
#5
Oct18-06, 11:11 AM
P: 19
Thank you very much for your info SGT, would you please help little bit more.
SGT
#6
Oct18-06, 03:03 PM
P: n/a
Quote Quote by Raihan
Thank you very much for your info SGT, would you please help little bit more.
Make the substitutions I suggested in your equation 1. More help will only be provided after you show some work.
Raihan
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#7
Nov23-06, 03:05 PM
P: 19
I tried And I am not going anywhere. please help
SGT
#8
Nov23-06, 06:20 PM
P: n/a
Post what you have done and I will give you more hints.
Raihan
Raihan is offline
#9
Nov27-06, 11:48 PM
P: 19
heres what I got so far.. please help after this point..
thanks
Attached Thumbnails
solution.gif  
SGT
#10
Nov28-06, 02:41 AM
P: n/a
In the second equation don't use the integral term. Keep it as [tex]V_C(t)[/tex].
In the two other terms replace i by [tex]C\frac{dV_C}{dt}[/tex]. You get a second order equation in [tex]V_C[/tex]
Raihan
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#11
Nov28-06, 12:56 PM
P: 19
Would you please not mind to show me please. I have tried this so far. please help after this.
thanks
SGT
#12
Nov28-06, 02:00 PM
P: n/a
The rules of the forum are that you must do your work. We only give hints. Rewrite the second equation with the suggestions I made and post it here.
serienumerica
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#13
Dec6-06, 07:48 AM
P: 2
The easiest way to solve any RCL circuit with an input vs(t) is by a difference equation.

Let curr= (q1-q0)/dt


q2=2.*q1-q0 + dt**2*( -q1/(L*c) -(R/L)*curr +vs(t-dt) ).

Then everything else follows ,

Vc(t) = q2/C , VL = L * ( q2-2*q1+q0)/dt^2 , VR = R*(q2-q1)/dt
SEE http://www.geocities.com/serienumerica/RCLfree.doc


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