Impedance may be combined in series and parallel by the same rules we have already established for resistance Example: 4925mH500uF 1=j5 O=101/s ∴Zm=4+15-12=4+j3=5∠369°2 smH z1=j5 j2 500uF a=1031/s j5(-j2)_10 ZI+Zc j5 乃-3.332Impedance may be combined in series and parallel by the same rules we have already established for resistance. ZR = 4 ZL = j5 ZC = − j2  = + − = + =    Zeq 4 j5 j2 4 j3 5 36.9 ZC = − j2 = = −  − − = + = 3.33 3 10 5 2 5( 2) j j j j j j Z Z Z Z Z L C L C eq 10 1/ s 3  = 4 5mH 500F 5mH 500F 10 1/ s 3  = ZL = j5 Example: