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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 can't 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 A1 and A2.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 iH and iP.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?
ng said: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?
What part of the equations you don't understand?ng said:I am sorry but I donno how to go abt with the equations.
No, you makeng said: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.
A parallel RL (resistor-inductor) circuit is a type of electrical circuit where the components are connected in parallel, meaning they share the same voltage but have different current paths. It consists of a resistor and an inductor connected in parallel, with an AC power supply providing the voltage.
An AC (alternating current) supply provides a constantly changing voltage to the circuit, which causes the current to fluctuate as well. In a parallel RL circuit, the inductor causes a phase shift between the voltage and current, resulting in a different behavior compared to a purely resistive circuit.
The differential equation for a parallel RL circuit is dI/dt + (R/L)I = (1/L)V, where I is the current, R is the resistance, L is the inductance, and V is the voltage. This equation describes the relationship between the current and voltage in the circuit.
To solve a parallel RL circuit using differential equations, you can use the above equation to find the current as a function of time. You can then use this current function to find other important quantities, such as voltage and power, in the circuit.
Parallel RL circuits are commonly used in electronics, such as in power supplies and audio amplifiers. They are also used in AC motors and generators, as well as in electrical power distribution systems. In addition, parallel RL circuits are used in various scientific and engineering fields for modeling and analyzing dynamic systems.