Solution;
Given;
L=2H
R=16 ohms
C=0.02farads
E(t)=100sin33t
From which;
V 0 = 100 V_0=100 V 0 = 100
w = 33 t w=33t w = 33 t
From Kirchoff's Law;
L d I d t + I R + Q C = V 0 s i n w t L\frac{dI}{dt}+IR+\frac{Q}{C}=V_0sinwt L d t d I + I R + C Q = V 0 s in wt
But since the capacitor is initially uncharged,
I = d Q d t I=\frac{dQ}{dt} I = d t d Q
By substitution;
L d 2 Q d t 2 + R d Q d t + Q C = V 0 s i n w t L\frac{d^2Q}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=V_0sinwt L d t 2 d 2 Q + R d t d Q + C Q = V 0 s in wt
Which is a second order differential equation whose solution is;
Q ( t ) = Q 0 c o s ( w t − ϕ ) Q(t)=Q_0cos(wt-\phi) Q ( t ) = Q 0 cos ( wt − ϕ )
Q 0 = V 0 w Z Q_0=\frac{V_0}{wZ} Q 0 = wZ 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
Z = 16 + ( 33 × 2 − 1 33 × 0.02 ) 2 Z=\sqrt{16+(33×2-\frac{1}{33×0.02})^2} Z = 16 + ( 33 × 2 − 33 × 0.02 1 ) 2
Z = 66.44 Z=66.44 Z = 66.44
Q 0 = 100 33 × 66.44 = 0.0456 Q_0=\frac{100}{33×66.44}=0.0456 Q 0 = 33 × 66.44 100 = 0.0456
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 = 76.07 \phi=tan^{-1}=76.07 ϕ = t a n − 1 = 76.07
Therefore;
Q ( t ) = 0.0456 s i n ( 33 t − 76.07 ) Q(t)=0.0456sin(33t-76.07) Q ( t ) = 0.0456 s in ( 33 t − 76.07 )
But ;
I = + d Q d t I=+\frac{dQ}{dt} I = + d t d Q
I ( t ) = 1.505 s i n ( 33 t − 76.07 ) A I(t)=1.505sin(33t-76.07)A I ( t ) = 1.505 s in ( 33 t − 76.07 ) A
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