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Maximum current amplitude in RLC circuit

  • Thread starter John 123
  • Start date
  • #1
36
0
Using the following o.d.e
[tex]
L\frac{d^2i}{dt^2}+R\frac{di}{dt}+\frac{1}{C}i=\frac{d}{dt}E(t)
[/tex]
The following problem has several parts all of which I have solved except for the one below.
L=1/20
R=5
[tex]
C=4.10^{-4}
[/tex]
[tex]
\frac{dE}{dt}=200\cos100t
[/tex]
Where L is an inductance in henries, R is a resistance in ohms, C is a capacitance in farads and E is the emf in volts.
The part I cannot agree with the book is as follows.
Firstly:
What should the frequency of the input E(t) be in order that it be in resonance with the system? [This I have solved correctly as :
[tex]
100\sqrt5
[/tex]
radians/sec
But this part leads to the next which I can't agree.
What is the maximum value of the current amplitude for this resonant frequency?
Book Answer= 2/5 amp.
John
 
Last edited:

Answers and Replies

  • #2
121
1
Have you solved the differential equation to obtain i (t) ? No doubt that frequency enters somewhere in the equation, then substitute 100sqrt(5), differentiate and set equal to 0 to find the maximum or minimum current, then differentiate a second time and check to see if it's a negative value, so that you're sure you're at a maximum, not a minimum.
 
  • #3
rl.bhat
Homework Helper
4,433
5
If you integrate dE= 200cos100t*dt. you will get Emax.
At resonance inductive reactance XL cancels capacitive reactance XC leaving only resistance in the circuit. Now find the maximum current.
 
  • #4
36
0
Reply to Lennox Lewis
Yes the steady state current is:
[tex]
i_s=\frac{2}{85}(\sin100t+4\cos100t)
[/tex]
When you substitute the frequency
[tex]
100\sqrt5
[/tex]
Differentiate and set to zero you get
[tex]
t=\frac{\tan^{-1}0.25}{100\sqrt5}
[/tex]
But this leads to a max current of 0.097 Amp[The book answer is 2/5 Amp]?
Regards
John
 
  • #5
36
0
Reply to rl.bhat
You have used the frequency 100 rad/sec whereas the frequency for resonance is
[tex]
100\sqrt5
[/tex]
?
Regards
John
 
  • #6
rl.bhat
Homework Helper
4,433
5
If E(t) = 2sin100t, what is dE(t)/dt ?
In the given problem Emax = 2V.
The maximum current at resonance is Emax/R.
 
  • #7
36
0
Hi again
Am I misunderstanding these two parts of the question?
Part 1.
What should the frequency of the input E(t) be in order that it be in resonance with the system?
ANSWER
[tex]
100\sqrt5 rad.sec^{-1}
[/tex]
Part 2.
What is the maximum value of the amplitude for THIS RESONANT FREQUENCY?[My caps bold]
ANSWER
Well if
[tex]
\frac{dE(t)}{dt}=200\cos{(100\sqrt5)t}
[/tex]
then
[tex]
E(t)=\frac{2}{\sqrt5}\sin{(100\sqrt5)t}
[/tex]
so
[tex]
i_max=\frac{2}{5\sqrt5}amps?
[/tex]
Regards
John
 
  • #8
36
0
My apologies there is an error in the last posting.
It should be:
[tex]
\frac{dE(t)}{dt}=200\cos(100\sqrt5)t
[/tex]
and
[tex]
E(t)=\frac{2}{\sqrt5}\sin(100\sqrt5)t
[/tex]
This leads to
i(max)=2/5xsqrt5

John
 
  • #9
36
0
Here is another question with the same problem.
a.Find the steady state current if L=1/20,R=20,C=1/10000,E=100COS200t.
b.What is the frequency of the input E(t) in order that it be in resonance with the system.
c. What is the maximum value of the amplitude for this resonant frequency?
I have answered parts a. and b. correctly as:
a. [tex]
i=\cos200t-2\sin200t
[/tex]
b.
[tex]
\omega=200\sqrt5
[/tex]
But once again I cannot see the book answer for part c, which is 5 amp?
Regards
John
 
  • #10
rl.bhat
Homework Helper
4,433
5
In both the problem E is given.You can change E keeping Eo constant by changing either L or C. At a particular setting of L, C and R, E will be maximum. At that instant, they have asked, what is the frequency of E. You can change E by either changing Eo keeping frequency constant or by changing frequency keeping Eo constant. In the problem they have adopted the second method. At resonance impedance will be purely resistive. Hence Imax = Eo/R = 100/20
 
  • #11
36
0
Many thanks.
Yes the amplitude E remains the same. I think the wording of the question is confusing by asking for the maximum amplitude FOR THIS FREQUENCY. However, as you say, whatever the frequency the amplitude hasn't been changed.
Regards
John
 

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