# Parallel RL circuit

SGT
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?

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
$$\frac{di}{dt}(0) = \frac{v_L(0)}{L}$$

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

SGT
ng said:
I am sorry but I donno how to go abt with the equations.
What part of the equations you don't understand?

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.

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.

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

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
ng 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.
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 $$B = K sin \phi_1$$

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