Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
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
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?