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niga

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Hi everyone
i am back with some more dumber qs. :(

I think its really silly but i dono why i am getting stuck.
I hope someone out there will HELP me.

Thanx to all.
 

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Is the supply across RLC circuit DC?
Then I think you will burn either power supply or inductor if your power supply can supply very large current

Normally for an inductor conductance is given as

XL = 2*pi*f*L
in DC case F = 0
so your XL = 0
which means a short across the DC supply :)
 
thanx for the response.
Yes i understand that....so there is absolutely no transient state for RLC series or parallel across dc?
 
there is a transient state when the DC is switched from 0 to some value (say 5V) but when it is constant there wont be any

why dont you try simulating this circuit in one of the spice softwares ?
you may understand it better
 
thanx instruite but i was trying to get the equations written down.
i dono what am missing.hmm..
 
could somone post the document in question ..
i still cannot display graphics in a .doc file.
 
oh sorry abt that.but its just an R and L in parallel and an RC parallel to sinusoidal source.
 
Here is the steady state solution for the RC circuit.

Do you need the transient solution also?

I"ll do the RL one tomorrow.

Len
 

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Thankyou sooooooooo much Len,
I do know the steady state solution for RL and RC.. but i am stuck with the differential equations like the one we figured out before.

parallel circuits are @$#@@ for me.
I feel i am bothering u a lot.Do tell me what i shud do.
 
ok i think i have something cleared.

Ldi/dt=R(I-i)

Ldi/dt=R(Icosωt-i)

Ldi/dt+Ri=RIcoswt

now,
do i assume particular soluion as Acoswt..?this is assumed for first order equations?
then for the complementary solution

Ldi/dt+Ri=0

hmm...is the solution k exp(-t/tau)?

am i gettn somewhere? :D
 
niga said:
ok i think i have something cleared.

Ldi/dt=R(I-i)

Ldi/dt=R(Icosωt-i) Divide both sides by L

di/dt+R/Li=RIcoswt/L

now,
do i assume particular soluion as Acoswt..?this is assumed for first order equations?

Yes, A = I/tau e^(-t/tau) and integral e^(t/tau) coswt dt

then for the complementary solution

di/dt+Ri/L=0 Yes

hmm...is the solution k exp(-t/tau)? yes; tau = L/R

am i gettn somewhere? Yes :D
 
Thanx again,Len

"Yes, A = I/tau e^(-t/tau) and integral e^(t/tau) coswt dt "
hmm..?why and?
 
is this right?

1.integral[di/i}=integral(-R/L) dt

ln i= -R/Lt+K

i=exp(-t/tau+k)

i=exp(-t/tau).exp(k)
at t=0 depends om the initial conditions.


current throu inductor is zero....am i right?

...?


2.how did u get that value for A?
let I the input be Icoswt.
for getting the particular solution from
di/dt+R/Li=RIcoswt/L

assume
i=Acoswt

substituting in above eqn

-Lwsinwt+RAcoswt=RIcoswt

equating the 2 sides

RAcoswt=RIcoswt
which gives A

this looks absurd....don u think?
 
niga said:
is this right?

1.integral[di/i}=integral(-R/L) dt

ln i - ln k= -R/Lt

i= k exp(-t/tau)

i= k exp(-t/tau)
at t=0 depends om the initial conditions.


current throu inductor is zero....am i right? Yes, but don't evaluate k at this point. Wait until you find the particular function and then i = ic + ip where c = complementry, p = particular

Then make i = 0 and t = 0 to determine k


...?


2.how did u get that value for A? I think you may be misunderstanding what I meant by "A"

The particular function is ip = I/tau exp (-t/tau) integral exp(t/tau) cos wt dt where tau = L/R



let I the input be Icoswt.
for getting the particular solution from
di/dt+R/Li=RIcoswt/L

assume
i=Acoswt

substituting in above eqn

-Lwsinwt+RAcoswt=RIcoswt

equating the 2 sides

RAcoswt=RIcoswt
which gives A

this looks absurd....don u think?
 
is that the particular solution for all first order equations?
then why do we assume A=I coswt?

looks like am a very difficult student! :wink:
 
i read that Acoswt is the solution assumed for first and second order equations.
and its found by putting into the first equation and equating the coefficients on RHS and LHS....is it correct?

but then the value of A; ip = I/tau exp (-t/tau) integral exp(t/tau) cos wt dt where tau = L/R

when i equated the two sides i din get this...sheesh i think i am a dumbo. :lol:
 
if i may just ask

i tot this is wat we wud get

ln i=-R/Lt+k

how does it become ln k?


ln i - ln k= -R/Lt

i= k exp(-t/tau)

i= k exp(-t/tau
)
 
Taking your last question first, if k is a constant, then ln k is also a constant.

It is better expressed as: ln i = -Rt/L + C, let C = ln k
Thus ln i - ln k = -Rt/L
so ln (i/k) = -Rt/L
and hence i = k exp (-Rt/L) = k exp (-t/tau)

I have done the Particular Function part of the problem in the attachment.

This is the general solution for finding the particular function for cases where the right hand side is cos and/or sin.

If the the RHS was expotential, then the PF would be expotential.

If the the RHS was ax^2 + b x + c then the PF would be of this form.

Re your second question, The method I used is a short cut.

Note that in the attachment, it is R cos (wt - phi); the "-" looks like a "+".

Len
 

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