Parallel RL circuit

  • Thread starter ng
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  • #1
ng
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Hi there,

Can someone help me with parallel RL circuit across ac supply?wats the differential equation .......
Thanx for all help.

ng
 

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  • #2
Astronuc
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Try this - http://www.play-hookey.com/ac_theory/ac_rl_parallel.html [Broken] to start.
 
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  • #3
ng
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Thanx for the response.I have already seen that and din help me much.
The problem is with getting the transient solution.
RC parallel -If the input is Vcoswt then current throu R will be V/Rcoswt.throu C is
-CwV sinwtThis is te steady state i understand.But the transient and getting to it throu te differential equations is my problem for both RL and Rc parallel across ac supply.
 
  • #4
Astronuc
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In the doc file, you have a differential equation.

V = VR(t) = VL(t) = L [itex]\frac{di(t)}{dt} [/itex] = ( I(t) - i(t) )/R,

but I(t) = V(t)/R + i(t), and let i(t) = VL/ZL where Z is the impedance.

Also consider i(t) = io cos ([itex]\omega[/itex] t + [itex]\theta[/itex]), where [itex]\theta[/itex] is the phase angle.

The voltages are in phase, the currents are out of phase.

Evaluate at time, t = 0.

See if that helps.
 
  • #5
SGT
In your attached file you have already arrived to the differential equation, you just didn't write it in a form to solve it. Write your equation as:
[tex]L\frac{di}{dt} + Ri = RIcos\omega t[/tex].
This is a non homogeneous differential equation. Find the solution of the homogeneous equation iH(t) and a particular solution that will be of the form iP(t) = Acos(ωt + φ).
The general solution is i(t) = iH(t) + iP(t)
Replacing the initial conditions you have the solution. It contains two terms: the exponential term is the transient and the cosine term is the steady state.
 
  • #6
ng
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hey sgt,thankyou for that reply.yeah,how stupid of me.....u r right.
thanx again.
 
  • #7
321
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For your last circuit. I started it out like this,

Apply KCL at the top supernode and you have

[tex]I_{in} = i_R + i_L + i_c[/tex]

Note, that all cicuit components have the same potential and recall that

[itex]L \frac {di_L}{dt} = v_L[/itex] and [itex]C \frac {dv_c}{dt} = i_c[/itex]
 
  • #8
ng
31
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thanx for the reply, corneo
hmm... okay.but i was wondering why shud i assume the solution
Acos(wt+Phi)?

Is this what is assumed for first order circuits subject to ac supply?i read its Acoswt+Bsinwt for second order circuits.am i right?
:confused:
 
  • #9
321
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You should assume the solution is [tex]A \cos (\omega t + \phi)[/tex] because I think your going to use methods of undetermined coefficients. I think cosines and sines appear in your solution to 2nd order ODE because the circuit is underdamped ([itex]\zeta < 1[/itex]). I just took a class in circuits so a lot of this is still new to me.
 
  • #10
ng
31
0
Thankyou corneo,
Okay
so i get

Ldi/dt+Ri=RIcoswt

to get the complementary solution assume
Ldi/dt+Ri=0

int di/i=- intR/L dt

ln i= -R/Lt+K

i=exp(-t/tau+k)

at t=0 initial condition i is V/R

then its..?is it right?

for particular solution i assume i=Acoswt
putting in first eqn

-Lwsinwt+RAcoswt=RIcoswt...

where am i going wrong?
please help.
 
  • #11
SGT
ng said:
thanx for the reply, corneo
hmm... okay.but i was wondering why shud i assume the solution
Acos(wt+Phi)?

Is this what is assumed for first order circuits subject to ac supply?i read its Acoswt+Bsinwt for second order circuits.am i right?
:confused:
Both solutions can be used with first and second order circuits (or higher orders). They are equivalent.
[tex]Acos(\omega t + \phi) = Acos\omega t cos\phi - Asin\omega t sin\phi = A_1 cos\omega t + A_2 sin\omega t[/tex]
where
[tex]A_1 = Acos\phi[/tex]
[tex]A_2 = -Asin\phi[/tex]
 
  • #12
ng
31
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Okay i get it.thanx SGT.
but ...heheh :biggrin: how do u find the value of A?
 
  • #13
SGT
ng said:
Okay i get it.thanx SGT.
but ...heheh :biggrin: how do u find the value of A?
You replace the solution and its derivative in the differential equation. When expanding the sine and the cosine of [tex]\omega t + \phi[/tex], you will have in both members of the equation terms in sine and in cosine. Equalling the coefficients of like terms you get the two unknowns A and [tex]\phi[/tex].
 
  • #14
ng
31
0
SGT,Thanx again ALOT!!!
So,
for particular solution i assume i=Acoswt
putting in first eqn

Ldi/dt+Ri=RIcoswt

-Lwsinwt+RAcoswt=RIcoswt
equating the coefficients on LHS and RHS
looks like i am going wrong...is it?
please help.
 
  • #15
SGT
ng said:
SGT,Thanx again ALOT!!!
So,
for particular solution i assume i=Acoswt
putting in first eqn

Ldi/dt+Ri=RIcoswt

-Lwsinwt+RAcoswt=RIcoswt
equating the coefficients on LHS and RHS
looks like i am going wrong...is it?
please help.
No, you put [tex]i = Acos(\omega t + \phi)[/tex]
or [tex]i = Acos(\omega t) + Bsin(\omega t)[/tex]
calculate [tex]\frac{di}{dt}[/tex]
and replace i and [tex]\frac{di}{dt}[/tex] in the equation.
 
  • #16
ng
31
0
:smile:
SGT...thanx a million.

i hope i am right this time....

-L(Awsinwt+Bwcoswt)+R(Acoswt+Bsinwt)=RIcoswt

(-LBw+RA)coswt=RIcoswt

-LBw+RA=RI
-LAw+RB=0

solve this to get A and B
w=2 pi f

this approach is used for second order also right?

thankyou once again!!! :approve:
 
  • #17
ng
31
0
:biggrin: just one more question...
why cant it be Acoswt?why the phi?
why shud it be coswt?why not sinwt?

(am i being really stupid? :rolleyes: )
 
  • #18
SGT
ng said:
:biggrin: just one more question...
why cant it be Acoswt?why the phi?
why shud it be coswt?why not sinwt?

(am i being really stupid? :rolleyes: )
When you differentiate cosωt you get -sinωt. If you replace both in the equation you get a term in sinωt in one side that must be equalled to zero on the other side, this is inconsistent. Just try it in your equation and you will see.
Of course you can use the sine instead of the cosine, they are the same function, only with different phases.
cosωt = sin(π/2 - ωt)
 
  • #19
ng
31
0
okay i understand.
Is this used in second order equations as well?
hmm.... an underdamped case for instance
there r two coefficients in it right?
how is that found?

i hope i am notgetting onto ur nerves!
thankyou for all the help.
 
  • #20
SGT
ng said:
okay i understand.
Is this used in second order equations as well?
hmm.... an underdamped case for instance
there r two coefficients in it right?
how is that found?

i hope i am notgetting onto ur nerves!
thankyou for all the help.
Both in first and second order equations the steady state response to a sinusoidal input have two parameters to be evaluated: A and φ or A1 and A2.
You have two equations: one with the coeficients of the terms in sine and one with the coeficients of the terms in cosine. Two equations, two unknowns ...
 
  • #21
ng
31
0
thanx for the quick response.
For example,
The overdamped case has the solution
v=A1exp(s1t)+A2exp(s2t)

and the equation is of the form
series rlc
L d^2i/dt +Rdi/dt +i/C=dV/dt

cud u please help me?
 
  • #22
SGT
ng said:
thanx for the quick response.
For example,
The overdamped case has the solution
v=A1exp(s1t)+A2exp(s2t)

and the equation is of the form
series rlc
L d^2i/dt +Rdi/dt +i/C=dV/dt

cud u please help me?
The solution you gave is of the homogeneous equation. The total solution is the sum of iH and iP.
[tex]i = A_1 e^{s_1 t} + A_2 e^{s_2 t} + K cos(\omega t + \phi)[/tex]
 
  • #23
ng
31
0
yes,but how to find the values of A1 and A2?
 
  • #24
SGT
ng said:
yes,but how to find the values of A1 and A2?
You obtain the derivative [tex]\frac{di_L}{dt}[/tex] and then replace the initial values [tex]i_L (0)[/tex] and [tex]\frac{di_L}{dt}(0)[/tex] in the equations.
 
  • #25
ng
31
0
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?
 

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