如右圖所示之電路,vs(t)= 10cos10t V,求電流i(t)為何?

A−4-0.707cos(10t-45°) A
B−4-0.707cos(10t+45°) A
C−4+0.707cos(10t-45°) A正確答案
D−4+0.707cos(10t+45°) A
答案與詳解

DC部分6A電流源:10Ω與串聯支路分流,i_DC=−4A;AC部分ω=10,Z_L=j×10×1.5=j15Ω,Z_C=1/(j×10×0.01)=−j10Ω,並聯Z_C∥Z_R=(−j10×10)/(10−j10)=5−j5Ω,總阻抗=5+j15+5−j5=10+j10Ω,I=10∠0°/(10+j10)=10/(10√2∠45°)=0.707∠−45°A,合計−4+0.707cos(10t−45°)
