Parallel RL Circuit: Get Help with AC Supply & Differential Equation

[ng]

Hi there,

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

ng

 

[Astronuc]

Try this - http://www.play-hookey.com/ac_theory/ac_rl_parallel.html to start.

 

[ng]

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.

 

[Astronuc]

In the doc file, you have a differential equation.

V = V_R(t) = V_L(t) = L \frac{di(t)}{dt} = ( I(t) - i(t) )/R,

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

Also consider i(t) = i_o cos (\omega t + \theta), where \theta is the phase angle.

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

Evaluate at time, t = 0.

See if that helps.

 

[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:
L\frac{di}{dt} + Ri = RIcos\omega t.
This is a non homogeneous differential equation. Find the solution of the homogeneous equation i_H(t) and a particular solution that will be of the form i_P(t) = Acos(ωt + φ).
The general solution is i(t) = i_H(t) + i_P(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.

 

[ng]

hey sgt,thankyou for that reply.yeah,how stupid of me...u r right.
thanx again.

 

[Corneo]

For your last circuit. I started it out like this,

Apply KCL at the top supernode and you have

I_{in} = i_R + i_L + i_c

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

L \frac {di_L}{dt} = v_L and C \frac {dv_c}{dt} = i_c

 

[ng]

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?

 

[Corneo]

You should assume the solution is A \cos (\omega t + \phi) 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 (\zeta < 1). I just took a class in circuits so a lot of this is still new to me.

 

[ng]

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.

 

[SGT]

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?

Both solutions can be used with first and second order circuits (or higher orders). They are equivalent.
Acos(\omega t + \phi) = Acos\omega t cos\phi - Asin\omega t sin\phi = A_1 cos\omega t + A_2 sin\omega t
where
A_1 = Acos\phi
A_2 = -Asin\phi

 

[ng]

Okay i get it.thanx SGT.
but ...heheh how do u find the value of A?

 

[SGT]

Okay i get it.thanx SGT.
but ...heheh 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 \omega t + \phi, 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 \phi.

 

[ng]

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.

 

[SGT]

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 i = Acos(\omega t + \phi)
or i = Acos(\omega t) + Bsin(\omega t)
calculate \frac{di}{dt}
and replace i and \frac{di}{dt} in the equation.

 

[ng]

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!

 

[ng]

just one more question...
why can't it be Acoswt?why the phi?
why shud it be coswt?why not sinwt?

(am i being really stupid? )

 

[SGT]

just one more question...
why can't it be Acoswt?why the phi?
why shud it be coswt?why not sinwt?

(am i being really stupid? )

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)

 

[ng]

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.

 

[SGT]

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 ...

 

[ng]

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?

 

[SGT]

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 i_H and i_P.
i = A_1 e^{s_1 t} + A_2 e^{s_2 t} + K cos(\omega t + \phi)

 

[ng]

yes,but how to find the values of A1 and A2?

 

[SGT]

yes,but how to find the values of A1 and A2?

You obtain the derivative \frac{di_L}{dt} and then replace the initial values i_L (0) and \frac{di_L}{dt}(0) in the equations.

 

[ng]

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?

 

[SGT]

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.
\frac{di}{dt}(0) is the derivative of the current in the inductor prior to the application of the excitation.
Since v_L(t) = L\frac{di}{dt}, we can say that
\frac{di}{dt}(0) = \frac{v_L(0)}{L}

 

[ng]

I am sorry but I donno how to go abt with the equations.

 

[SGT]

I am sorry but I donno how to go abt with the equations.

What part of the equations you don't understand?

 

[ng]

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

 

[SGT]

The steady state solution is of the form:
i_P = K cos(\omega t + \phi) = A cos \omega t + B sin \omega t
You differentiate it two times and replace i_P and its two derivatives in the differential equation:
L\ddot i_P + R\dot i_P + \frac{i_P}{C} = \frac{dV}{dt}
You can calculate the two unknowns: K and \phi or A and B.
You have the solution of the homogeneous equation, for instance in the overdamped case:
i_H = A_1 e^{s_1 t} + A_2 e^{s_2 t}.
The general solution is:
i = i_H + i_P = A_1 e^{s_1 t} + A_2 e^{s_2 t} + K cos(\omega t + \phi)
You differentiate it to get \dot i
In order to calculate A_1 and A_2 you must have the initial values of the current and its derivative: i(0) and \dot i(0)
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.

 

[ng]

SGT,Thankyou very much indeed.
I think i got the stuff in my head atlast.
Thankyou.

 

[ng]

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.

 

[SGT]

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
i_P = K cos(\omega t + \phi_1)
where \phi_1 \neq \phi
if you make
i_P = A cos \omega t + B sin \omega t
the phase angle \phi_1 will be automatically calculated:
A = K cos \phi_1 and B = K sin \phi_1

 

[ng]

okay sgt,now i get it.
thanx a loooooooooooooot!