R, joM R2解:(a)去耦等效后, * L I r +列网孔方程 sol oL,凵U ∫(8+j48)i1-1412=1 j2411+(12+j24)I2=0 89 20 ∠90° 28.1∠81.87° 0(b24 j12 U=102=0.35∠8.13 109U j24
解：(a)去耦等效后， 列网孔方程： − + + = + − = j24 (12 j24) 0 (8 j48) j24 1 1 2 1 2 I I I I US U j24 − j12 j24 8 2 10 + － + － I1 I2 = = = 10 0.35 8.13 28.1 81.87 90 2 2 U I I US U 1 jL 2 jL R1 jM R2 * R + * － + －
IJOM R R 解:(b)戴维南定理 R U joLE joL Uoc=j0M.1=124.~1 0.494∠946 8+148 O 24 =2+j12+ 8+j48 3.94+j0.32g R OC 0.35∠8.13°V Z+r
解：(b)戴维南定理： 0.35 8.13 V 3.94 j0.32 8 j48 24 2 j12 ( ) 0.494 9.46 8 j48 1 j j24 O C O 2 1 1 2 O 2 2 O C 1 0 = + = = + + = + = + + = + = = U Z R R U Z M Z Z U M I US U 1 jL 2 jL R1 jM R2 * R + * － + －
反映阻抗法: =R+JOL,+ (0M R2+R+joL 24 8+j48+ =32+1242 2+10+j12 32+124 JoM1=1247 0.6∠53.1° 32+1244+3 U=0.6∠53.1° R 10 0.6∠53.1° =0.35∠8.13°V R,+R+10L 12+j12
0.35 8.13 V 12 j12 10 0.6 53.1 j 0.6 53.1 0.6 53.1 4 j3 j3 32 j24 1 j j24 32 j24 1 32 j24 2 10 j12 24 8 j48 j ( ) j 2 1 1 2 2 2 2 1 1 = + = + + = = + = + = + = = + + + = + + + + = + + R R L R U MI I R R L M Zi R L 反映阻抗法：