Download Chapter 33 - A.C. Circuits

Electromagnetic Oscillations
and
Alternating Current
Chapter 33
Oscillations in an LC Circuit
We will discover that charge sloshes back and forth.
As this happens the current goes one way then the other.
Analogy: a block moving on a spring. Here total energy
(kinetic + potential) is constant.
For the LC circuit total energy (electric + magnetic) is
constant.
i
i
+
+
+
time
i
i
LC Circuit
i
i
+
+
+
Analyzing an LC Circuit
Total energy in the circuit:
Differentiate :

1 2 1 q2
U  U B  U E  LI 
2C
2
dU d 1 2 1 q 2
)  0 No change
 ( LI 
in energy
2C
dt dt 2
dq d 2q q dq
dI q dq
 0  L( ) 2 
 LI 
C dt
dt dt
dt C dt
2
dq 1
L 2  q  0
C
dt
Analyzing an LC Circuit
Total energy in the circuit:
Differentiate :

1 2 1 q2
U  U B  U E  LI 
2C
2
dU d 1 2 1 q 2
)  0 No change
 ( LI 
in energy
2C
dt dt 2
dq d 2q q dq
dI q dq
 0  L( ) 2 
 LI 
C dt
dt dt
dt C dt
2
dq 1
L 2  q  0
C
dt
Analyzing an LC Circuit
Total energy in the circuit:
Differentiate :

1 2 1 q2
U  U B  U E  LI 
2C
2
dU d 1 2 1 q 2
)  0 No change
 ( LI 
in energy
2C
dt dt 2
dq d 2q q dq
dI q dq
 0  L( ) 2 
 LI 
C dt
dt dt
dt C dt
2
dq 1
L 2  q  0
C
dt
Analyzing an LC Circuit
Total energy in the circuit:
Differentiate :
d 2q
2

w
q0
2
dt
w 2  1 LC
q  q p coswt
1 2 1 q2
U  U B  U E  LI 
2C
2
dU d 1 2 1 q 2
)  0 No change
 ( LI 
in energy
2C
dt dt 2
dq d 2q q dq
dI q dq
 0  L( ) 2 
 LI 
C dt
dt dt
dt C dt
2
dq 1
L 2  q  0
C
dt
The charge sloshes back and
forth with frequency w= (LC)-1/2
Analyzing an LC Circuit
q  q p coswt with
w  LC
1
2
dq
Current I 
 qpw sin wt
dt
 Current is maximum when charge is zero, and vice versa.
Energy:
2
q
Q  p
UC 

cos2 wt
2C 2C
2
q


1 2 1 2 2 2
1 2 1
U B  LI  Lq pw sin wt  Lq p  sin 2 wt  p sin 2 wt
LC 
2
2
2
2C
2
q 2p
U B  UC 
 const .
2C
RLC Circuit: Damped Oscillations
R
C
L
The change here is that energy is dissipated in the resistor:
d
2
U

U

I
R
 B E
dt
A similar analysis gives current and charge that continue
to oscillate but with amplitudes that decay exponentially:

q  q p eRt / 2L coswt 
w w 2  (R /2L) 2
Alternating Current Circuits
An “AC” circuit is one in which the driving voltage and
hence the current are sinusoidal in time.
V(t)
Vp
fv
p
2p
wt
V = VP sin (wt - fv )
I = IP sin (wt - fI )
-Vp
w is the angular frequency (angular speed) [radians per second].
Sometimes instead of w we use the frequency f [cycles per second]
Frequency  f [cycles per second, or Hertz (Hz)]
w  2p f
Alternating Current Circuits
V = VP sin (wt - fv )
I = IP sin (wt - fI )
I(t)
V(t)
Ip
Vp
Irms
Vrms
fv
-Vp
p
2p
wt
fI/w
t
-Ip
Vp and Ip are the peak current and voltage. We also use the
“root-mean-square” values: Vrms = Vp / 2 and Irms=Ip / 2
fv and fI are called phase differences (these determine when
V and I are zero). Usually we’re free to set fv=0 (but not fI).
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz.
Write this in sinusoidal form, assuming V(t)=0 at t=0.
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz.
Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V.
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz.
Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V.
This 60 Hz is the frequency f: so w=2pf=377 s -1.
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz.
Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V.
This 60 Hz is the frequency f: so w=2pf=377 s -1.
So V(t) = 170 sin(377t + fv).
Choose fv=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).
Resistors in AC Circuits
R
V
~
EMF (and also voltage across resistor):
V = VP sin (wt)
Hence by Ohm’s law, I=V/R:
I = (VP /R) sin(wt) = IP sin(wt)
(with IP=VP/R)
V
I
p
2p
wt
V and I
“In-phase”
Capacitors in AC Circuits
C
Start from:
q = C V [V=Vpsin(wt)]
Take derivative: dq/dt = C dV/dt
So
I = C dV/dt = C VP w cos (wt)
V
~
I = C wVP sin (wt + p/2)
V
I
p
2p wt
This looks like IP=VP/R for a resistor
(except for the phase change).
So we call
Xc = 1/(wC)
the Capacitive Reactance
The reactance is sort of like resistance in
that IP=VP/Xc. Also, the current leads
the voltage by 90o (phase difference).
V and I “out of phase” by 90º. I leads V by 90º.
Capacitor Example
A 100 nF capacitor is
connected to an AC supply
of peak voltage 170V and
frequency 60 Hz.
C
V
What is the peak current?
What is the phase of the current?
What is the dissipated power?
~
Inductors in AC Circuits
~
L
V = VP sin (wt)
Loop law: V +VL= 0 where VL = -L dI/dt
Hence:
dI/dt = (VP/L) sin(wt).
Integrate: I = - (VP / Lw cos (wt)
or
V
Again this looks like IP=VP/R for a
resistor (except for the phase change).
I
p
I = [VP /(wL)] sin (wt - p/2)
2p
wt So we call
the
XL = wL
Inductive Reactance
Here the current lags the voltage by 90o.
V and I “out of phase” by 90º. I lags V by 90º.
Inductor Example
A 10 mH inductor is
connected to an AC supply
of peak voltage 10V and
frequency 50 kHz.
What is the peak current?
What is the phase of the current?
What is the dissipated power?
V
~
L
Circuit
element
Resistor
Capacitor
Inductor
Resistance
or
Amplitude
Reactance
R
Xc=1/wC
XL=wL
Phase
VR= IP R
I, V in
phase
VC=IP Xc
I leads V
by 90°
VL=IP Xc
I lags V by
90°
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Vp
Ip
wt
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Capacitor
Vp
Ip
Ip
wt
wt
Vp
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Capacitor
Inductor
Vp
Ip
Vp
Ip
wt
Ip
wt
wt
Vp