Solution;
Given;
L=2H
R=16Ω
C=0.02F
E=100sin33t
At t=0,Q=0 and I=0
Also from E;
V0=100V
w=33
From Kirchoff's Loop Law;
LdtdI+IR+CQ=V0sinwt
Since the capacitor is intially uncharged,I=dtdQ , Substituting;
Ldt2d2Q+RdtdQ+CQ=V0sinwt
Which is a second order differential equation.
One possible solution of the above differential equation is;
Q(t)=Q0cos(wt−ϕ)
Q0=wR2+(XL−XC)2V0
tanϕ=RXL−XC
XL=wL
XC=wC1
By direct substitution of values;
XL=33×2=66
XC=33×0.021=1.515
tanϕ=1666−1.515=4.03
Q0=33162+(66−1.515)2100=0.0456 ϕ=tan−14.03=76.07
Hence , charge at any time t;
Q(t)=0.0456cos(33t−76.07)
But ;
I=+dtdQ
Hence;
I(t)=1.505sin(33t−76.07)
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