Parallel RL Circuit: Get Help with AC Supply & Differential Equation

[ng]

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

 

[ng]

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]

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 $$B = K sin \phi_1$$

 

[ng]

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