Due to Kirchhoff’s law:
Ldtdi+Ri=e
5dtdi+12i=120sin200t
The solutions to a nonhomogeneous equation are of the form
i(t)=ic(t)+ip(t)
where ic(t) is the general solution to the associated homogeneous equation and ip(t) is a particular solution.
The associated homogeneous equation:
5dtdi+12i=0
ic(t)=Ce−512t
ip(t)=Asin200t+Bcos200t
ip’(t)=A⋅200cos200t−B⋅200sin200t
Now put these into the original differential equation to get:
5(A⋅200cos200t−B⋅200sin200t)+ 12(Asin200t+Bcos200t)=120sin200t
A=6250990 ; B=−625097500.
Answer: i(t)=Ce−512t+6250990sin200t−625097500cos200t
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