... jnωoL + Cn 1 jRR)Cn 1 j(ooω−ω− = jn100π( 31, 8 .10 -3) + 66 10 . 318 .10 0n 1 j10 10 ) 10 . 318 .10 0n 1 j(−−π−π−= j10n + n 10 j10n 10 0j−− → Z 1 = j10 + 10 j10 10 0j−− = 5 ... jCn 1 oω + LjnRR)Ljn(ooω+ω= - j )10 .5 (10 n 1 63 − + 1, 0 .10 jn100 10 0 )1, 0 .10 jn(33+ = - jn200 + jn1n100j+→ Z 1 = - j200 + j1 10 0j+ = 50 – j150 = 15 8 ,11 4∠- 71, 57o (Ω)→ 1 I= ... j )10 .5,22( 1 6−ω = - j)5,22 (10 0 10 6π = - j1 41, 47 = 14 1,47∠- 90o (Ω)→ 1C11CZIU= = (1, 54∠85,5o) (14 1,47∠- 90o) = 218 ∠- 4,5o (V) hay uC1(t) = 218 sin(ωt – 4,5o) (V) 17 Mạch điện Ngô Ngọc...