Solution;
Given;
E ( t ) = 100 s i n 33 t E(t)=100sin33t E ( t ) = 100 s in 33 t
From which;
V 0 = 100 V V_0=100V V 0 = 100 V
w = 33 w=33 w = 33
Then current at time t is;
I ( t ) = I m s i n ( w t − ϕ ) I(t)=I_msin(wt-\phi) I ( t ) = I m s in ( wt − ϕ )
Where,
I m = V 0 Z I_m=\frac{V_0}{Z} I m = Z V 0
Z = R 2 + ( X L − X C ) 2 Z=\sqrt{R^2+(X_L-X_C)^2} Z = R 2 + ( X L − X C ) 2
R = 16 Ω R=16\Omega R = 16Ω
X L = w L = 33 × 2 = 66 X_L=wL=33×2=66 X L = w L = 33 × 2 = 66
X c = 1 w C = 1 33 × 0.02 = 1.515 X_c=\frac{1}{wC}=\frac{1}{33×0.02}=1.515 X c = wC 1 = 33 × 0.02 1 = 1.515
By substitution;
Z = 1 6 2 + ( 66 − 1.515 ) 2 = 66.44 Z=\sqrt{16^2+(66-1.515)^2}=66.44 Z = 1 6 2 + ( 66 − 1.515 ) 2 = 66.44
Therefore;
I m = 100 66.44 = 1.505 A I_m=\frac{100}{66.44}=1.505A I m = 66.44 100 = 1.505 A
Also;
t a n ϕ = X L − X C R = 66 − 1.515 16 = 4.03 tan\phi=\frac{X_L-X_C}{R}=\frac{66-1.515}{16}=4.03 t an ϕ = R X L − X C = 16 66 − 1.515 = 4.03
ϕ = t a n − 1 ( 4.03 ) = 76.1 \phi=tan^{-1}(4.03)=76.1 ϕ = t a n − 1 ( 4.03 ) = 76.1
And;
I ( t ) = 1.505 s i n ( 33 t − 76.1 ) I(t)=1.505sin(33t-76.1) I ( t ) = 1.505 s in ( 33 t − 76.1 )
Since the capacitor is intially uncharged;
I ( t ) = + d Q d t I(t)=+\frac{dQ}{dt} I ( t ) = + d t d Q
Therefore;
Q = ∫ I ( t ) d t = ∫ 1.505 s i n ( 33 t − 76.1 ) d t = 1.505 [ ∫ s i n ( 33 t − 76.1 ) Q=\int I(t)dt=\int1.505sin(33t-76.1)dt=1.505[\int sin(33t-76.1) Q = ∫ I ( t ) d t = ∫ 1.505 s in ( 33 t − 76.1 ) d t = 1.505 [ ∫ s in ( 33 t − 76.1 )
Q = 1.505 [ − c o s ( 33 t − 76.1 33 ] Q=1.505[-\frac{cos(33t-76.1}{33}] Q = 1.505 [ − 33 cos ( 33 t − 76.1 ]
Q = 0.0456 c o s ( 33 t − 76.1 ) Q=0.0456cos(33t-76.1) Q = 0.0456 cos ( 33 t − 76.1 )
Comments