Solution;
Given;
L=2H
R=16 ohms
C=0.02farads
E(t)=100sin33t
From which;
V0=100
w=33t
From Kirchoff's Law;
LdtdI+IR+CQ=V0sinwt
But since the capacitor is initially uncharged,
I=dtdQ
By substitution;
Ldt2d2Q+RdtdQ+CQ=V0sinwt
Which is a second order differential equation whose solution is;
Q(t)=Q0cos(wt−ϕ)
Q0=wZV0
Z=R2+(XL−XC)2
Z=16+(33×2−33×0.021)2
Z=66.44
Q0=33×66.44100=0.0456
tanϕ=RXL−XC=1666−1.515=4.03
ϕ=tan−1=76.07
Therefore;
Q(t)=0.0456sin(33t−76.07)
But ;
I=+dtdQ
I(t)=1.505sin(33t−76.07)A
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