Ac circuit analysis on electrical circuits

ac circuit analysis

1. AC Circuit Analysis
(also called sinusoidal steady-state analysis or phasor analysis)
Before reviewing AC Circuit Analysis, a review of complex numbers is useful.
Complex Numbers
A complex number can be expressed in two forms:
1)Rectangular form:
X A jB   where j -1 
Re
Im (j)
A
B X
X
2)Polar form:
jA
Note that A X cos( )
and B X sin( ) so
X A jB X cos( ) j X sin( )
X X (cos( ) j sin( ))
[recall Euler's Identity: e cosA jsinA]


 
 



   
 
 
so X X (true polar form)
j
e


or X X (shorthand polar form)  
Converting between rectangular form and polar form:
Polar to Rectangular:
A = X cos( )
B = X sin( )


2 2
-1
Rectangular to Polar:
X A B
B
tan
A

 
 

 
 

2. t
v
sec/3162
sin
max
radf
tVv




2
V
VV
max
rms

AC Phasor Representation
Definition: A phasor is a complex number in polar form that represents magnitude and
phase angle of a sinusoidal voltage or current.

3. )(sin
sin
max22
max11




tVv
tVv

V
1
V
2



22
11
0
VV
VV
t
v1
v2

Reference

4. Sinusoids can be visualized as the real-axis projection of
vectors rotating in the complex plane. The phasor for a
sinusoid is a snapshot of the corresponding rotating vector
at t = 0.

5. )(cos
cos
max22
max11




tVv
tVv

V
1
V
2



22
11
0
VV
VV
t
v1
v2

Reference

6.   
111 cos :function Time θtωVtv 
111
:Phasor θVV
  

45cos20
1
 ttv 
  

60sin10
2  ttv 

4520
1
V

3010
2 V


7.3997.29
14.1906.23
5660.814.1414.14
30104520
21s





j
jj
VVV
  

7.39cos97.29  ttv
s


7. To determine phase relationships from a phasor diagram,
consider the phasors to rotate counterclockwise. Then when
standing at a fixed point, if V
1 arrives first followed by V
2
after a rotation of θ , we say that V
1
leads V
2
by θ .
Alternatively, we could say that V
2
lags V
1
by θ . (Usually, we
take θ as the smaller angle between the two phasors.)

8. V-I Relations in a Resistor
RR
RIV

9. V-I Relations in an Inductor
LL
LjIV 
LL LjIV 
LLLZIV

90 LLjZ
L


11. V-I Relations in a Capacitor
CCCZIV

90
111

CCjC
jZ
C


13. Circuit Element Phasor Relations
ElementV/I RelationPhasor RelationPhase
CapacitorI = C dV/dtI = j ω C V
= ωCV

90°
I leads V
by 90º
InductorV = L dI/dtV = j ω L I
= ωLI

90°
V leads I
by 90º
ResistorV = I RV = R I
= R I

0°
In-phase

14. Resistors:
R
Z R 0 R   
Capacitors:
Inductors:
C
1 1 -j 1
Z 90
jwC j2 fC wC wC
     

L Z jwL j2 fL wL 90    

Complex Impedances: Components are represented in AC circuits as follows:
Z = impedance or complex impedance (in )
V phasor voltage
Z
phasor currentI
 
Z V
+
_
I
KVL and KCL in AC Circuits:
KVL and KCL are satisfied in AC circuits using phasor voltages and currents. They are
not satisfied using the magnitudes of the voltages and the currents. Consider how KVL
applies to the RL circuit below in both the time-domain and in the phasor domain.
v(t) = VPcos(wt)
+
_
R
L
+ vL (t) -
+

vR (t)

-
+
_
R
jwL
+ VL -
+

VR

-
V
Time-Domain Circuit Phasor Circuit
L R
L R
Phasor circuit:
V V V (KVL is satisfied using phasors)
V V V (KVL is NOT satisfied using magnitudes only)
 
 
L R
Time-domain circuit:
v(t) = v (t) + v (t) (KVL is satisfied)

15. 








0k2090-k65.2
90-k65.2
0 10V
C
V
Z
R
= 20k= 20k 0
Z
C = 1/j (377·1) = 2.65k -90
20k 0
+
–
2.65k -90
10V  0
V
C
+
–
 4.82- 1.31V
C
V
Example:  = 377rad/sec, find V
C