Question

电路答题，求I(t)

求高手解答 不需要画相量图
我的思路用相量法吧电压源时域化成相量然后求把求总阻抗，再求电流 U/Z =I ，，，，，，i-》转化成时序模式同理求IL

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Answer 1

iC(t)=Cdu(t)/dt,
iL(t)=∫u(t)dt/L,
i(t)=iC(t)+iL(t)=Cdu(t)/dt+∫u(t)dt/L

u(t)=us(t)-i(t)
du(t)/dt=-20√2sin(2t)-i'(t)
i'(t)=C[-20√2sin(2t)-i'(t)]'+[us(t)-i(t)]/L=C[-40√2cos(2t)-i''(t)]+[us(t)-i(t)]/L
Ci''(t)+i'(t)+i(t)/L=us(t)/L-40√2Ccos(2t)=(1/L-4C)10√2cos(2t)
i''(t)+i'(t)/C+i(t)/(LC)=[1/(LC)-4]10√2cos(2t)
求出上述微分方程的解j就好了
可以设i(t)=Acos(2t)+Bsin(2t)
则i''(t)=-4Acos(2t)-4Bsin(2t),i'(t)=2Bcos(2t)-2Asin(2t)
则B/(LC)-2A/C-4B=0,A=(1-4LC)B/(2L)
而A/(LC)+2B/C-4A=1/(LC)-4]10√2=(1-4LC)A/(LC)+2B/C=[(1-4LC)^2/(2L^2C)+2/C]B
=(1-4LC)10√2/(LC)=[(1-4LC)^2+4L^2]B/(2L^2C)
B=20√2L(1-4LC)/[(1-4LC)^2+4L^2]=-5√2
A=10√2(1-4LC)^2/[(1-4LC)^2+4L^2]=5√2
i(t)=5√2[cos(2t)-sin(2t)]

追问

不满意，请用相量法

Answer 2

ω=2,-->XL=jωL=j,Xc=1/jωC=-0.5j
Z=R+XL//Xc=1-j=(2)^(1/2)/-π/4
I=Us/Z=10A/π/4
Us=UR+UL=IR+UL--->UL=10V/-π/4
IL=UL/j=10A/-3π/4（正好和总电流I反相）
Ic=UL/-0.5j=20A/π/4（正好和总电流I同相）