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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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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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V = V

but I(t) = V(t)/R + i(t), and let i(t) = V

Also consider i(t) = i

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

Evaluate at time, t = 0.

See if that helps.

- #5

SGT

[tex]L\frac{di}{dt} + Ri = RIcos\omega t[/tex].

This is a non homogeneous differential equation. Find the solution of the homogeneous equation i

The general solution is i(t) = i

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.

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

thanx again.

thanx again.

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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

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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?

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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

Both solutions can be used with first and second order circuits (or higher orders). They are equivalent.ng said:

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?

[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

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Okay i get it.thanx SGT.

but ...heheh how do u find the value of A?

but ...heheh how do u find the value of A?

- #13

SGT

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].ng said:Okay i get it.thanx SGT.

but ...heheh how do u find the value of A?

- #14

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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

No, you put [tex]i = Acos(\omega t + \phi)[/tex]ng said:

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.

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

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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!!!

- #17

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why cant it be Acoswt?why the phi?

why shud it be coswt?why not sinwt?

(am i being really stupid? )

- #18

SGT

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.ng said:

why cant it be Acoswt?why the phi?

why shud it be coswt?why not sinwt?

(am i being really stupid? )

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

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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

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 Ang said:

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.

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

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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

The solution you gave is of the homogeneous equation. The total solution is the sum of ing said:

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?

[tex]i = A_1 e^{s_1 t} + A_2 e^{s_2 t} + K cos(\omega t + \phi)[/tex]

- #23

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

- #24

SGT

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.ng said:yes,but how to find the values of A1 and A2?

- #25

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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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