Lectures 16 & 17 Ph Ci it AC P Phasor Circuits, AC Power, Thevenin

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Lectures 16 & 17
PPhhasor CCiircuiitts, AACC PPower,
TThheevveenniinn
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Kirchhoff’s Laws in Phasor Form
We can apply KVL directly to phasors.
TThhe sum off tthhe phhasor vollttages equalls
zero for any closed path.
V + V − V = 0
1 2 3
The sum of the phasor currents entering a
nnooddee mmuusstt eeqquuaall tthhee ssuumm ooff tthhee pphhaassoorr
currents leaving.
II = II
in out
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuit Analysis Using Phasors
and Impedances
1. Replace the time descriptions of the voltage
andd currentt sources wiitthh tthhe corresponddiing
phasors. (All of the sources must have the
same ffrequency.))
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
22. RRepllace iindducttances bby tthheiir compllex
impedances Z = jωL. Replace capacitances by
L
tthheiir compllex iimpeddances ZZ = 11//((jjωCC)).
C
Resistances have impedances equal to their
resiisttances.
3. Analyze the circuit using any of the techniques
studied earlier in Chapter 2, performing the
calculations with complex arithmetic.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.3
Find i(t):
V =100cos(500t + 30o) → V =100∠30o
S S
Z = jωL = j(500)(0.3) = j150Ω
L
1 1
Z = − j = − j = − j50Ω
C
ωC ((500))((40))((10−6))
Z = R + Z + Z =100 + j150 − j50 =100 + j100
eq L C
=141.4∠45o
V 100∠30o
S
I = = = 0.707∠ −15o → i(t) = 0.707cos(500t −15o)
Z 141.4∠45o
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.3
Find the phasor
voltage across each
eelleemmeenntt::
VV ==110000∠∠3300o
SS
I = 0.707∠ −15o
S
V = RI = (100)(0.707∠ −15o)
R
V = ( jwL)I = (ωL∠90o)I = (ωL∠90o)(0.707∠ −15o) =106.1∠75o
L
⎛ 1 ⎞ ⎛ 1 ⎞
V = ⎜− j ⎟I = ⎜ ∠ −90o ⎟I = (50∠ −90o)(0.707∠ −15o) = 35.4∠ −105o
CC
⎝⎝ ωCC ⎠⎠ ⎝⎝ωCC ⎠⎠
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
Find the voltage v (t) in steady state:
c
v (t) = 10sin(1000t) = 10cos(1000t − 90) → V = 10∠ − 90
s s
Z = jwL = j(1000)(0.1) = j100Ω
L
11 11
Z = − j = − j = − j100Ω
C ωC (1000)(10x10−6)
1 1 1
ZZ == == ==
RC 1 1 1 1 0.01+ j0.01
+ +
R Z 100 − j100
C
1∠00
= = 7700.7711∠∠ − 4455 = 5500 − jj5500
0.01414∠45
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
⎛ Z ⎞ ⎛ 70.71∠ − 45o ⎞ ⎛ 70.71∠ − 45o ⎞
V = ⎜ RC ⎟V = ⎜ ⎟(10∠ −90o) = ⎜ ⎟(10∠ −90o)
C ⎜ ⎟ S ⎜ ⎟ ⎜ ⎟
Z + Z (50 − j50) + j100 50 + j50
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
RC L
⎛⎛ 70.71∠ − 45o ⎞⎞
= ⎜ ⎟(10∠ −90o) = (1∠ −90o)(10∠ −90o) =10∠ −180o
⎜ ⎟
70.71∠45o
⎝ ⎠
v (t) =10cos(1000t −180o) = −10cos(1000t)
C
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
Find the phasor current through each element:
V 10∠ − 90o 10∠ − 90o 10∠ − 90o
I = S = = = = 0.1414∠ −135
Z + Z j100 + (50 − j50) 50 + j50 70.71∠45o
L RC
VV 1100∠∠ −118800o
I = CC = = 00.11∠∠ −118800oo
R
R 100
V 10∠ −180o 10∠ −180o
I = C = = = 0.1∠ − 90o
C
ZZ − jj110000 110000∠∠ − 9900
C
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.5
Use the node voltage technique to find v (t):
1
KCL at node 1:
VV VV − VV
1 + 1 2 = 2∠ − 90o → (0.1+ j0.2)V − j0.2V = − j2
1 2
10 − j5
KKCCLL aatt nnooddee 22 ::
V V − V
2 + 2 1 = 1.5∠0o → − j0.2V + j0.1V = 1.5
1 2
jj1100 − jj55
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.