§7-7 Impedance We define the ratio of the phasor voltage to the phasor current as impedance R: V=RI or R→Z=R L:V= jaLI or V→2=jo=t (XL=OL) C: y i or →2 jaC I JaC jac =Xc C (X OC
§7-7 Impedance We define the ratio of the phasor voltage to the phasor current as impedance. • • = I V Z R Z R I V R V = R I or = → = • • • • : L j L Z j L jX I V L V = j L I or = → = = • • • • : C jX C j j C Z j C I V I or j C C V = = → = = − = • • • • 1 1 1 1 : ) C ( XC 1 = − ( X L) L =
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-3.332
Impedance 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 500F 5mH 500F 10 1/ s 3 = ZL = j5 Example:
Z=R+ⅸ→Z=Z∠9 Rectangular form Polar form X=X+X R--resistance X--reactance R Z=VR2+X--impedance amplitude >o inductive 9= tan impedance angle 93<0 capacitive R 0 resistive
Z = R+ jX Z = Z Rectangular form Polar form Z R C L X XL XC 1 = − = + R--resistance X--reactance Z = R + X − −impedance amplitude 2 2 impedance angle R X = − − −1 tan = resistive capacitive inductive 0 0 0
Example: Find i. Solution 5 1 zea=1.5k jlk/1k- j2k) jIk +1k-j2k 40√2sin3000rJ H F 2k+1.5k=25k∠369° 1.5k 1 40∠-90 2500∠36.9 2 40∠-90°V )jik 16×10-3∠-126.9° i(t)=16×103√2cos(300-126.9°)A ori(t)=16×103√2sin(30004-369°)A
Example: Find i. Solution: 2 1 5 2 5 36 9 1 1 2 1 1 2 1 5 k j . k . k . j k k j k j k( k j k ) Z . k eq = + = + − − = + 16 10 126.9 2500 36.9 40 90 3 = − − = = − • • eq s Z V I or i(t) 16 10 2 sin(3000t 36.9 )A 3 = − − i(t) 16 10 2 cos(3000t 126.9 )A 3 = − −
