Parallel RL circuit

  • Thread starter ng
  • Start date
  • #26
SGT
ng said:
V=Acos(wt+phi)=Acoswt+Bsinwt


L d^2i/dt +Rdi/dt +i/C=dV/dt

i=A1exp(s1t)+A2exp(s2t)

di/dt=s1A1exp(s1t)+s2A2exp(s2t)

d^2i/dt^2= s1^2A1exp(s1t)+s2^2A2exp(s2t)

i(0)=V/R
if its dc dV/dt=0
ac then dV/dt=-Awsinwt+wBcoswt

is this going wrong?
It's wrong!
i(0) is the current in the inductor prior to the application of the excitation.
[tex]\frac{di}{dt}(0)[/tex] is the derivative of the current in the inductor prior to the application of the excitation.
Since [tex]v_L(t) = L\frac{di}{dt}[/tex], we can say that
[tex]\frac{di}{dt}(0) = \frac{v_L(0)}{L}[/tex]
 
  • #27
ng
31
0
I am sorry but I donno how to go abt with the equations.
 
  • #28
SGT
ng said:
I am sorry but I donno how to go abt with the equations.
What part of the equations you don't understand?
 
  • #29
ng
31
0
how u get the steady state and coomplementary solution.
i understood quite a few things from ur replies but not how to reach the
final equations.i am going wrong whenever i try it.
so cud u please help out?thanku again
 
  • #30
SGT
The steady state solution is of the form:
[tex]i_P = K cos(\omega t + \phi) = A cos \omega t + B sin \omega t[/tex]
You differentiate it two times and replace [tex]i_P[/tex] and its two derivatives in the differential equation:
[tex]L\ddot i_P + R\dot i_P + \frac{i_P}{C} = \frac{dV}{dt}[/tex]
You can calculate the two unknowns: [tex]K[/tex] and [tex]\phi[/tex] or [tex]A[/tex] and [tex]B[/tex].
You have the solution of the homogeneous equation, for instance in the overdamped case:
[tex]i_H = A_1 e^{s_1 t} + A_2 e^{s_2 t}[/tex].
The general solution is:
[tex]i = i_H + i_P = A_1 e^{s_1 t} + A_2 e^{s_2 t} + K cos(\omega t + \phi)[/tex]
You differentiate it to get [tex]\dot i[/tex]
In order to calculate [tex]A_1[/tex] and [tex]A_2[/tex] you must have the initial values of the current and its derivative: [tex]i(0)[/tex] and [tex]\dot i(0)[/tex]
You make t = 0 in the equations of the current and its derivative and set them to the initial values. You have two equations in two unknowns.
 
  • #31
ng
31
0
SGT,Thankyou very much indeed.
I think i got the stuff in my head atlast.
Thankyou. :smile:
 
  • #32
ng
31
0
but then sgt,i also read that ip shud be assumed as
acos(wt+phi) + bsin(wt+phi)
not Kcos(wt+phi) as the supply is Vcos(wt+phi).they need not be in phase...
i am confused now.
 
  • #33
SGT
ng said:
but then sgt,i also read that ip shud be assumed as
acos(wt+phi) + bsin(wt+phi)
not Kcos(wt+phi) as the supply is Vcos(wt+phi).they need not be in phase...
i am confused now.
No, you make
[tex]i_P = K cos(\omega t + \phi_1)[/tex]
where [tex]\phi_1 \neq \phi[/tex]
if you make
[tex]i_P = A cos \omega t + B sin \omega t[/tex]
the phase angle [tex]\phi_1[/tex] will be automatically calculated:
[tex]A = K cos \phi_1[/tex] and [tex]B = K sin \phi_1[/tex]
 
  • #34
ng
31
0
okay sgt,now i get it.
thanx a loooooooooooooot!
 

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