Solution;
Given;
E(t)=100sin33t
From which;
V0=100V
w=33
Then current at time t is;
I(t)=Imsin(wt−ϕ)
Where,
Im=ZV0
Z=R2+(XL−XC)2
R=16Ω
XL=wL=33×2=66
Xc=wC1=33×0.021=1.515
By substitution;
Z=162+(66−1.515)2=66.44
Therefore;
Im=66.44100=1.505A
Also;
tanϕ=RXL−XC=1666−1.515=4.03
ϕ=tan−1(4.03)=76.1
And;
I(t)=1.505sin(33t−76.1)
Since the capacitor is intially uncharged;
I(t)=+dtdQ
Therefore;
Q=∫I(t)dt=∫1.505sin(33t−76.1)dt=1.505[∫sin(33t−76.1)
Q=1.505[−33cos(33t−76.1]
Q=0.0456cos(33t−76.1)
Comments