Due to Kirchhoff's law:
U L + U R + U C = E U_L+U_R+U_C=E U L + U R + U C = E
where
U L = L d i d t U_L=L\frac{di}{dt} U L = L d t d i U R = i R U_R=iR U R = i R U C = 1 C ∫ i d t U_C=\frac1C\int idt U C = C 1 ∫ i d t
Let's differentiate (1):
L d 2 i d t 2 + R d i d t + 1 C i = 0 L\frac{d^2i}{dt^2}+R\frac{di}{dt}+\frac 1C i=0 L d t 2 d 2 i + R d t d i + C 1 i = 0
d 2 i d t 2 + R L d i d t + 1 L C i = 0 \frac{d^2i}{dt^2}+\frac RL\frac{di}{dt}+\frac {1}{LC} i=0 d t 2 d 2 i + L R d t d i + L C 1 i = 0
Characteristic equation:
λ 2 + R L λ + 1 L C = 0 \lambda^2+\frac RL\lambda+\frac {1}{LC}=0 λ 2 + L R λ + L C 1 = 0
R L = 20 0.05 = 400 \frac RL=\frac {20}{0.05}=400 L R = 0.05 20 = 400 1 L C = 1 0.05 ⋅ 100 ⋅ 1 0 − 6 = 2 ⋅ 1 0 5 \frac {1}{LC}=\frac {1}{0.05\cdot 100\cdot 10^{-6}}=2\cdot10^5 L C 1 = 0.05 ⋅ 100 ⋅ 1 0 − 6 1 = 2 ⋅ 1 0 5
λ 2 + 400 λ + 2 ⋅ 1 0 5 = 0 \lambda^2+400\lambda+2\cdot 10^5=0 λ 2 + 400 λ + 2 ⋅ 1 0 5 = 0
λ = − 200 ± 400 j \lambda=-200\pm400j λ = − 200 ± 400 j j = − 1 j=\sqrt{-1} j = − 1
i ( t ) = e − 200 t ( C 1 cos 400 t + C 2 sin 400 t ) i(t)=e^{-200t}(C_1 \cos {400t}+C_2 \sin {400t}) i ( t ) = e − 200 t ( C 1 cos 400 t + C 2 sin 400 t )
i ∣ t = 0 = C 1 = 0 ; i|_{t=0}=C_1=0; i ∣ t = 0 = C 1 = 0 ;
L d i d t ∣ t = 0 + R i ∣ t = 0 + U C ∣ t = 0 = E L\frac{di}{dt}|_{t=0}+Ri|_{t=0}+U_C|_{t=0}=E L d t d i ∣ t = 0 + R i ∣ t = 0 + U C ∣ t = 0 = E
U C ∣ t = 0 = 1 C Q ∣ t = 0 = 0 U_C|_{t=0}=\frac 1C Q|_{t=0}=0 U C ∣ t = 0 = C 1 Q ∣ t = 0 = 0
R i ∣ t = 0 = 0 Ri|_{t=0}=0 R i ∣ t = 0 = 0
d i d t = C 2 e − 200 t ( − 200 ⋅ sin 400 t + 400 ⋅ cos 400 t ) \frac{di}{dt}=C_2e^{-200t}(-200\cdot\sin{400t}+400\cdot\cos{400t}) d t d i = C 2 e − 200 t ( − 200 ⋅ sin 400 t + 400 ⋅ cos 400 t )
d i d t ∣ t = 0 = 400 ⋅ C 2 \frac{di}{dt}|_{t=0}=400\cdot C_2 d t d i ∣ t = 0 = 400 ⋅ C 2
Substitute (3), (4) and (5) into (2):
L ⋅ 400 ⋅ C 2 = E L\cdot400\cdot C_2=E L ⋅ 400 ⋅ C 2 = E
C 2 = E 400 ⋅ L = 100 400 ⋅ 0.05 = 5 C_2=\frac{E}{400\cdot L}=\frac{100}{400\cdot 0.05}=5 C 2 = 400 ⋅ L E = 400 ⋅ 0.05 100 = 5
i ( t ) = 5 ⋅ e − 200 t ⋅ sin 400 t . i(t)=5\cdot e^{-200t}\cdot \sin{400t}. i ( t ) = 5 ⋅ e − 200 t ⋅ sin 400 t .
Q ( t ) = ∫ 5 ⋅ e − 200 t ⋅ sin 400 t d t = Q(t)=\int5\cdot e^{-200t}\cdot \sin{400t} dt= Q ( t ) = ∫ 5 ⋅ e − 200 t ⋅ sin 400 t d t =
− 1 200 e − 200 t ( sin 400 t + 2 cos 400 t ) + C 3 -\frac{1}{200}e^{-200t}(\sin{400t}+2\cos{400t})+C_3 − 200 1 e − 200 t ( sin 400 t + 2 cos 400 t ) + C 3
Q ∣ t = 0 = − 1 200 ⋅ 2 + C 3 = 0 Q|_{t=0}=-\frac{1}{200}\cdot2+C_3=0 Q ∣ t = 0 = − 200 1 ⋅ 2 + C 3 = 0
C 3 = 1 100 C_3=\frac{1}{100} C 3 = 100 1
Q ( t ) = 1 200 [ 2 − e − 200 t ( sin 400 t + 2 cos 400 t ) ] Q(t)=\frac{1}{200}[2-e^{-200t}(\sin{400t}+2\cos{400t})] Q ( t ) = 200 1 [ 2 − e − 200 t ( sin 400 t + 2 cos 400 t )]
Answer:
i ( t ) = 5 ⋅ e − 200 t ⋅ sin 400 t i(t)=5\cdot e^{-200t}\cdot \sin{400t} i ( t ) = 5 ⋅ e − 200 t ⋅ sin 400 t
Q ( t ) = 1 200 [ 2 − e − 200 t ( sin 400 t + 2 cos 400 t ) ] Q(t)=\frac{1}{200}[2-e^{-200t}(\sin{400t}+2\cos{400t})] Q ( t ) = 200 1 [ 2 − e − 200 t ( sin 400 t + 2 cos 400 t )]
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