A、图D-1电路中以一电感线圈作滤波器,已知负载电阻R<sub>f2sub>=22Ω,滤波电感L=0.5H(线圈电阻忽略),AB端接f=50Hz、有效值U<sub>1sub>=220V的交流电流,求负载端电压U<sub>2sub>。
A、图D-1电路中以一电感线圈作滤波器,已知负载电阻R<sub>f2sub>=22Ω,滤波电感L=0.5H(线圈电阻忽略),AB端接f=50Hz、有效值U<sub>1sub>=220V的交流电流,求负载端电压U<sub>2sub>。
A、如图所示电路中,u<sub>Ssub>=10sin314tV,R<sub>1sub>=2Ω,R<sub>2sub>=1Ω,L=637mH,C=637μF,求电流i<sub>1sub>,i<sub>2sub>和电压u<sub>csub>。
A、e<sub>1sub>c<sub>1sub>F<sub>1sub>=e<sub>2sub>c<sub>2sub>F<sub>2sub> B、c<sub>1sub>F<sub>1sub>=c<sub>2sub>F<sub>2sub> C、c<sub>1sub>/c<sub>2sub>=F<sub>2sub>/F<sub>1sub> D、c<sub>1sub>/c<sub>2sub>=F<sub>1sub>/F<sub>2sub>
A、必有F<sub>Rsub>=F<sub>1sub>+F<sub>2sub> B、不可能有F<sub>Rsub>=F<sub>1sub>+F<sub>2sub> C、必有F<sub>Rsub>>F<sub>1sub>,F<sub>Rsub>>F<sub>2sub> D、可能有F<sub>Rsub><F<sub>1sub>,F<sub>Rsub><F<sub>2sub>
A、图示阶梯形圆截面杆,承受轴向载荷F<sub>1sub>=50kN与F<sub>2sub>作用,AB与BC段的直径分别为d1=20mm和d2=30mm,如欲使AB与BC段横截面上的正应力相同,试求载荷F<sub>2sub>之值。
A、F<sub>1sub>=F<sub>2sub> B、F<sub>1sub>-F<sub>2sub>=F<sub>esub> C、F<sub>1sub>/F<sub>2sub>=efα D、F<sub>1sub>+F<sub>2sub>=F<sub>0sub>
A、F<sub>1sub>=6N,F<sub>2sub>=0 B、F<sub>1sub>=0,F<sub>2sub>=6N C、F<sub>1sub>=F<sub>2sub>=8N D、F<sub>1sub>=6N,F<sub>2sub>=8N
A、F<sub>isub>V<sub>isub>=F<sub>isub>+1V<sub>isub>+1 B、F<sub>isub>V<sub>isub><f<sub>i+1V<sub>isub>+1 C、F<sub>isub>V<sub>isub>>F<sub>isub>+1V<sub>isub>+1f<sub>
A、F<sub>0.05,10,20sub> = 1/F<sub>0.95,10,20sub> B、F<sub>0.05,10,20sub> = 1/F<sub>0.05,20,10sub> C、F<sub>0.95,10,20sub> = 1/F<sub>0.95,20,10sub> D、F<sub>0.95,10,20sub> = 1/F<sub>0.05,20,10sub>