Solution;
Given;
L = 2 H L=2H L = 2 H
R = 16 Ω R=16\Omega R = 16Ω
C = 0.02 F C=0.02F C = 0.02 F
E = 100 s i n 33 t E=100sin33t E = 100 s in 33 t
At t=0,Q=0 and I=0
Also from E;
V 0 = 100 V V_0=100V V 0 = 100 V
w = 33 w=33 w = 33
From Kirchoff's Loop 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
Since the capacitor is intially uncharged,I = d Q d t I=\frac{dQ}{dt} I = d t d Q , Substituting;
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.
One possible solution of the above differential equation 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 R 2 + ( X L − X C ) 2 Q_0=\frac{V_0}{w\sqrt{R^2+(X_L-X_C)^2}} Q 0 = w R 2 + ( X L − X C ) 2 V 0
t a n ϕ = X L − X C R tan\phi=\frac{X_L-X_C}{R} t an ϕ = R X L − X C
X L = w L X_L=wL X L = w L
X C = 1 w C X_C=\frac{1}{wC} X C = wC 1
By direct substitution of values;
X L = 33 × 2 = 66 X_L=33×2=66 X L = 33 × 2 = 66
X C = 1 33 × 0.02 = 1.515 X_C=\frac{1}{33×0.02}=1.515 X C = 33 × 0.02 1 = 1.515
t a n ϕ = 66 − 1.515 16 = 4.03 tan\phi=\frac{66-1.515}{16}=4.03 t an ϕ = 16 66 − 1.515 = 4.03
Q 0 = 100 33 1 6 2 + ( 66 − 1.515 ) 2 = 0.0456 Q_0=\frac{100}{33\sqrt{16^2+(66-1.515)^2}}=0.0456 Q 0 = 33 1 6 2 + ( 66 − 1.515 ) 2 100 = 0.0456 ϕ = t a n − 1 4.03 = 76.07 \phi=tan^{-1}4.03=76.07 ϕ = t a n − 1 4.03 = 76.07
Hence , charge at any time t;
Q ( t ) = 0.0456 c o s ( 33 t − 76.07 ) Q(t)=0.0456cos(33t-76.07) Q ( t ) = 0.0456 cos ( 33 t − 76.07 )
But ;
I = + d Q d t I=+\frac{dQ}{dt} I = + d t d Q
Hence;
I ( t ) = 1.505 s i n ( 33 t − 76.07 ) I(t)=1.505sin(33t-76.07) I ( t ) = 1.505 s in ( 33 t − 76.07 )
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