Download Alternating Current

Alternating Current
Voltage Source : v(t) = V cos t
Current Source : i(t) = I cos t
Phasors: a graphical method for (combinations) of
trigonometric functions

I
t
i(t)=I cos t
p212c32: 1
Full Wave Rectifier
G
Irav = (2/ I
i
t
p212c32: 2
RMS values
i2(t)
Root-Mean-Square:
I rms  ((i (t )) 2 ) average
i (t )  I cos t
i(t)
i 2 (t )  I 2 cos 2 t
1
 I 2 ( (1  cos2t ))
2
2
1
I
average(i 2 (t ))  I 2 ( (1  0)) 
2
2
I
I rms 
2
V
Vrms 
2
Household Power
120V = Vrms
V  2Vrms  2120V
 170V
p212c32: 3
i  I cos  t
v  iR

 IR cos  t
 V cos  t
V  IR
v(t)=
Vcos(t)
I

R

Current is in phase with Voltage
V=I
t
R
i(t)=I cos t
v(t)=V cos t
p212c32: 4
v  EMF  0  v  ( L
di
)
dt
I cos  t
di
v  L  LI sin  t
dt
 LI cos( t  90)
 V cos( t  90)
V  LI  IX L

i(t)=
Icos(t)
L
X L  Inductive Reactance
V=IXL


I
t
Current lags Voltage
Voltage leads Current
v(t)=V cos (t+90°)
i(t)=I cos t
p212c32: 5
vq C
i  I cos 
dq
i
dt
I
q  sin 

t

i(t)=
Icos(t)

q
q
C
I
t
I
v
sin  t
C
I

cos( t  90)
C
 V cos( t  90)
I
V
 IX C
C
X C  Capacitive Reactance

t
V=IXL
i(t)=I cos t
v(t)=V cos t
Current leads Voltage
Voltage lags Current
p212c32: 6
Element
Resistor
Inductor
Amplitude
V = IR
V  IX L
Capacitor
V  IX C
XC
Reactance Phase
R
in phase
X L  L v leads i
1
XC 
i leads v
C
XL
R

p212c32: 7
L-R-C Circuit
i = I cos(t)
i(t)=Icos(t)
v(t)=Vcos(t+)
= IXLcos(t+)
+ IRcos(t)
+ IXCcos(t)
L
vL(t)=IXLcos(t+)
R
vR(t)=IRcos(t)
C
vC(t)=IXCcos(t)
p212c32: 8
v(t)=Vcos(t+)
= IXLcos(t+)
+ IRcos(t)
+ IXCcos(t)
Z2 = R2+ X2
= R2+ (XLXC)2
tan() = X/R
VL=IXL VC=IXC
VL- VC =IX
I
V=I

Z
VR=IR
VL=IXL
V=I
Z 
VC=IXC
VL- VC =IX
VR=IR
I
p212c32: 9
V  IZ
Z  R2  X 2
 R 2  ( X L  X C )2
 R 2  (L  1 C ) 2
X
tan  
R
X L  X C L  1 C


R
R
v  V cos(t   )
voltage leads (lags) current if
X L  XC ( X L  XC )
p212c32: 10
LRC series circuit example
R  300 C .5F
V  50V
L  60mH f  1591 Hz

XL 
XC 
X 

Z
I
VR 
VC 
VC 
p212c32: 11
Power
Instantane ous Power :
p  iv  I cos(t )V cos(t   )
1

1
 IV  cos( )  cos( 2t   ) 
2

2
Average Power :
1
P  IV cos( )  I rmsVrms cos( )
2
cos( ) " Power factor"
=R Z
p212c32: 12
Series Resonance
Z
XL
XC
X
R
log()
Z
phi
I
I = V/Z
log()
p212c32: 13
I
o
1
o 
LC
Width of resonance :    I  I o

2
o
Quality factor Q 

o L

R
HW: add Q,  calculations to all rlc series HW problems
p212c32: 14
LRC series circuit example (more)
R  300
C  .5F
L  60mH
f  1591 Hz   cos( 53)
Z  500
I  .1A
V  50V
P
o 
fo =
Q
 
p212c32: 15
LRC series circuit example (and more)
R  160 C  .1F
L  .6 H
o 
fo =
Q
 
p212c32: 16
Parallel L-R-C Circuit
iL(t)=
ILcos(t-90° )
L
iR(t)=
IRcos(t)
R
iC(t)=
ICcos(t )
C
v(t)=
Vcos(t)
i = I cos(t+)
= IL cos(t-90° )
+ IR cos(t)
+ IC cos(t)
= V/XLcos(t-90° )
+ V/R cos(t)
+ V/XC cos(t)
p212c32: 17
IL
i = I cos(t+)
= IL cos(t-90° )
+ IR cos(t)
+ IC cos(t)
I2 = IR2+ (ICIL)2
tan() = (ICIL)/IR
IC
I C- I L
V
I=V/Z

IR
IC
IL
I=V/Z

IR
I C- I L
V
p212c32: 18
I 1

V Z
1
2
2
 (1 R)  (1 X C  1 X L )
Z
 (1 R)  (C  1 L)
2
2
IC  I L
tan  
 R(C  1 L)
IR
i  I cos(t   )
current leads (lags) voltage if
X L  XC ( X L  XC )
p212c32: 19
LRC parallel circuit example
R  300 C .5F
V  50V
L  60mH f  1591 Hz

XL 
XC 

Z
I
IR 
IL 
IC 
P
p212c32: 20
Parallel “Resonance”
Z
XL
XC
R
log()
Z
I
log()
phi
p212c32: 21
V
I
Z
V
  VC
R

o 
1
LC
p212c32: 22
Transformers:
Ferromagnetic Materials Strengthen Flux
 B1   B 2
all field lines go through both areas
d B1
V1   N1
dt
d B 2
V2   N 2
dt
V1 N1

V2 N 2
N1
V1, I1
Primary
B
N2
V2, I2
Secondary
p212c32: 23
Transformers
V1 N1

V2 N 2
P1  P2
conservati on of energy
I1V1  I 2V2
B
N1
V1, I1
Primary
N2
V2, I2
Secondary
(rms values)
V1 N1 I 2


V2 N 2 I1
N2 >N1 => V2 > V1 Step-Up Transformer
N2 <N1 => V2 < V1 Step-Down Transformer
p212c32: 24
A coffee maker from Europe is designed to operate on a 240-V line (rms) to obtain
960W of power.
(a) Determine what characteristics are needed by a transformer so that the proper
delivery voltage be obtained from the US standard voltage of 120 V (rms)?
(b) What current is drawn at the secondary?
(c) What is the resistance of the coffee maker?
(d) What current is drawn from the 120 V outlet by the primary?
(e) What is the power delivered by the 120 V source?
p212c32: 25