+i2g0.125F 解 m-(2+ j0.125o0.5g2 10∠0° 2+j4 4.165H 2∠-60° =(0.5+j8.33)2
N0 i + - u (b) 2 0.125F 解： = + − + − = = − + (0.5 j8.33) 2 j4 2 60 10 0 ) j0.125 1 (2 m m 0 I U Z 4.165H 0.5
OL =(00712s418H8142
(0.007 j0.12) S 1 j 1 0 0 = − = = − L G Z Y 4.18H 143
2、如下图,当(t)=10√2cos100时, 各电流表读数相同,且电路消耗功率为 866W,试求R、L、C的值 Ai A R CL E b 解:由题知O=1000rad/s U=10∠0°Vn1=/2+3画相量图
2、如下图，当 时， 各电流表读数相同，且电路消耗功率为 86.6W,试求R、L、C的值 u(t) =10 2 cos1000t V R A3 A2 A1 C L a b + - u 解： U =100 V 由题知 =1000 rad/s I 1 I 2 I 3 . . . = + 画相量图
a A3 + 由于I=l2=1=I A2 R qn1=30 C L 912=90 b 30 P=Uh1cos1=10/cos(-30°)=866 解之得=10A 即I,=10∠90°A U 10∠-30°A
10 30 A 10 90 A 3 2 = − = I I 即 R A3 A2 A1 C L a b + - u 由于I1 =I2 =I3 =I = − = = 30 90 30 3 2 1 i i i 解之得 I=10 A P =UI1 cos1 =10I cos(−30) = 86.6 1 I U 2 I 3 I
a P=1 2R=866 A R →R=0.866 cL 3 a 6 10∠0° 2=10∠90° jac/j1000C U 10∠0° 10∠-30 R+jaL0.866+j1000L 解上式得C=1mFL=0.5mH
R L L U I j C C U I 0.866 j1000 10 0 j 10 30 j1000 1 10 0 1 10 90 32 + = + = − = = = = 解上式得 C = 1mF L = 0.5 mH P I R R= = =32 86 6 0866 . . R A 3 A 2 A 1 C L ab +-u