Download Alternating Current

Rotating Generators and
Faraday’s Law
 B   B dA  B A  B A cos 

 B  B A cos t
d B
d

  B A cos t
dt
dt
 B
0
  o  t
A  sin t
   N B
A  sin t

For N loops of wire
Alternating Current
V  t   Vo
I  t   Io
V  t   Vo sin t
V  t  Vo
I(t) 

sin t  Io sin t
R
R
AC Generator and a Resistor
  peak cos  t 
VR  VR peak cos  t 
VR VR peak
IR 

cos  t   I peak cos  t 
R
R
AC Power


P  I2 R  I2peak cos2  t  R
2
V
1 peak
1 2
P  Pav  I peak R 
2
2 R
T/2
1
1
2
cos  t dt  ?

T T / 2
2
1 1
cos    cos 2
2 2
2
Root Mean Square (rms)
V  t   Vpeak cos t
V2 
Vrms  V 
2
I(t)  I peak cos t
2
Vpeak
I2 
2
2
Vpeak
2

Vpeak
2
Irms 
1 2
2
P  I peak R  I rms
R
2
2
V
1 peak
2
Vrms
P

2 R
R
I 
2
I 2peak
2
I2peak
2

Ipeak
2
Inductive Circuits
VL  VL peak cos  t   L
dI
dt
  peak cos  t 
I
VL peak
L
sin  t  
I peak
VL

XL
VL peak
XL
dI
VL  L
dt


sin  t   Ipeak cos  t  
2

XL  L
Inductive Reactance
Capacitive Circuits
Q  CVC
  peak cos  t 
I  CVC peak sin  t   
I peak
VC

XC
VC peak
XC


sin  t   I peak cos  t  
2

1
XC 
C
Capacitive Reactance
Voltage transformers
 solenoid
 NP 
  o
IP  A


d B
Vs  N s
dt
d B
VP  N P
dt
NS  NP  step up transformer
NP  NS  step down transformer
d B VP VS


dt
N P NS
Current in transformers
PPr imary  PSecondary
VP IP  VSIS
IP
NS
VS 
VP
NP
IS
NP IP  NSIS
Actually currents are 180 degrees out of phase
Example: transformers
Vp  110V
IP
N p  916
Ns  100
IS
What is Vs ?
LC Circuits
Kirchhoff Loop Equation:
Q
dI
L 0
C
dt
2
dQ Q

0
2
dt
LC
Solution:
Q  Qmax cos  t   
1

LC
I  t  0  0
Q(t  0)  Qmax
Energy in an LC circuit
Q  Qmax cos  t 
dQ
I
 Q max  sin  t 
dt
1

LC
Imax  Qmax 
1 Q2 Q2max
UE 

cos 2  t 
2 C
2C
2
Q
1 2 L2Q2max
U B  LI 
sin 2  t   max sin 2  t 
2
2
2C
2
2
Q2max
Q
Q
UE  UB 
cos 2  t   max sin 2  t   max
2C
2C
2C
Active Figure 32.17
(SLIDESHOW MODE ONLY)
LRC Circuits
Kirchhoff Loop Equation:
Q
dI
 RI  L  0
C
dt
d 2Q
dQ Q
L 2 R
 0
dt
dt C
Solution:
Q  Qmax ebt cos   ' t  
R
b 
2L
1
R2
' 
 2
LC 4L
Damped RLC Circuit
• The maximum value
of Q decreases after
each oscillation
– R < RC
• This is analogous to
the amplitude of a
damped spring-mass
system
Active Figure 32.21
(SLIDESHOW MODE ONLY)
LRC Circuits
Q  Qo e

R
t
2L
cos   ' t   
1
R2
' 
 2
LC 4L
• Underdamped
• Critically Damped
• Overdamped
1
R2
 2
LC 4L
1
R2
 2
LC 4L
4L
 R2
C
4L
 R2
C
1
R2
 2
LC 4L
4L
 R2
C
Driven RLC Circuit
dI
Q
Vapp peak cos t  L  IR   0
dt
C
d 2Q
dQ 1
L 2 R
 Q  Vapp peak cos t
dt
dt C
Phasor Diagrams
Z  R   X L  XC 
2
 XL  XC 
  tan 

R


1
2
Resonance
X L  XC
1

LC
I
Vapp peak
R   X L  XC 
2
2
cos  t   
Power:
2
1 2
1 Vapp peak
1 Vapp peak Vapp peak R 1
Pav  I peak R 
R
 I peak Vapp peak cos 
2
2
2 Z
2
Z
Z 2
Pav  I rms Vapp rms cos 
Power Factor
What is Power factor at Resonance?
More Resonance
Pav 

2
2
Vapp
R

rms
L  
2
2
2
o

2
 2 R 2
o o L
Q


R
41.
An emf of 96.0 mV is induced in the windings of a coil
when the current in a nearby coil is increasing at the rate of 1.20
A/s. What is the mutual inductance of the two coils?
49.
A fixed inductance L = 1.05 μH is used in series with a
variable capacitor in the tuning section of a radiotelephone on a
ship. What capacitance tunes the circuit to the signal from a
transmitter broadcasting at 6.30 MHz?
55.
Consider an LC circuit in which L = 500 mH and C =
0.100 μF. (a) What is the resonance frequency ω0? (b) If a
resistance of 1.00 kΩ is introduced into this circuit, what is the
frequency of the (damped) oscillations? (c) What is the percent
difference between the two frequencies?
LC Demo
R = 10 W
C = 2.5 F
L = 850 mH
1. Calculate period
2. What if we change
C = 10 F
3. Underdamped?
4. How can we
change damping?