... jn103.0 ,1) 100 200 j100n + =-j + + jn n103 (5 .10 − ) 10 0 + jn103.0 ,1 n → Z1 = - j200 + j100 = 50 – j150 = 15 8 ,11 4∠- 71, 57o (Ω) 1+ j 2∠30o 1 = E1 = I → = 12 ,65 10 1,57o (mA) o Z1 15 8 ,11 4∠ − 71, 57 ... Điện áp R1: U = U ' + U ' ' = R1. 'R1 + R1. ' 'R1 I I I I I I I I = R1 ( 'R1o + 'R 11 + 'R12 ) + R1 ( ' 'R1o + ' 'R 11 + ' 'R12 ) = 10 3(0 + 7,0 71. 10-3∠45o + 1, 789 .10 -3∠- 63,43o) + 10 3(- ... nωo C n100π. 318 .10 − -3 Zn = jnωoL + = jn100π( 31, 8 .10 ) + 1 R−j 10 − j nωo C n100π. 318 .10 − 10 0 −j n → Z1 = j10 + − j100 = + j5 = ∠45o (Ω) = j10n + 10 10 − j10 10 − j n 200 E1 I → 1 = = = 9∠-...