Both solutions can be used with first and second order circuits (or higher orders). They are equivalent.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?
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?
No, you put [tex]i = Acos(\omega t + \phi)[/tex]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.
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: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? )
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 A_{1} and A_{2}.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.
The solution you gave is of the homogeneous equation. The total solution is the sum of i_{H} and i_{P}.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?
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?