Download Lecture 31

Phys 272
Alternating Current
A.C. Versus D.C
(Natural) Frequency, Period, etc…
I(t) = I p sin(w t + f )
w : Angular frequency (rad /s º s-1)
T : Period ( s)
f : Frequency (Hz)
2p
w = 2pf =
T
d
sin w t = w cos w t
dt
1
ò sin(w t )dt = - cos w t
w
Alternating Current
Vd = Vp sin w d t
Vd is the driving emf of the power source
Vp is the peak value
w d is the driving angular frequency
In general, this will cause a current:
I = I p sin(w d t + f )
where I p is the peak current
and f is the constant phase
Voltage and Current
US: 120V rms, 60Hz
Europe: 230V rms, 50Hz
Question
Given f = p , amplitude = 20A, frequency = 30Hz,
write the expression of I(t).
w = 2p f = 60p
I p = 20A
I(t) = I p sin(w t + f )
Þ I(t) = (20A)sin(60p t + p ) = -(20A)sin(60p t)
Question
Given I(t) = 10sin(30t + p / 2), all in SI units,
find the amplitude, frequency, period, angular frequency and phase.
Amplitude = 10A
w = 30rad / s
2p
T=
= 0.209s
w
1
f = = 4.77Hz
T
f=+
p
2
Resistive AC circuit
Vd = Vp sin w d t
VR
I=
by Ohm's Law
R
but VR = Vd
Vp sin w d t Vp
ÞI =
= sin w d t
R
R
Compare with I = I p sin(w d t + f ) :
Vp
I p = , f = 0 (in phase)
R
Capacitive AC circuit
Vd = Vp sin w d t
q
C=
Þ q = CVC = CVd
VC
dq
dVd
ÞI=
=C
dt
dt
Þ I = w d CVp cos w d t
p
Þ I = w d CVp sin(w d t + )
2
Compare with I = I p sin(w d t + f ) :
I p = w d CVp , f =
p
2
(I leads V by
p
2
)
Capacitive Reactance
Recall that resistance was defined by:
V
R=
I
We now define "reactance" X by:
Vp
1
XC =
=
I p w dC
Unit :
W
Inductive AC circuit
Vd = Vp sin w d t
dI
Vd - L = 0
dt
1
Þ I = ò Vd dt
L
Vp
ÞI=cos w d t
wdL
Vp
p
ÞI=
sin(w d t - )
wdL
2
Compare with I = I p sin(w d t + f ) :
Vp
p
p
Ip =
, f = - (I lags V by )
wd L
2
2
Inductive Reactance
Vp
Ip =
wdL
Vp
Þ XL =
= wdL
Ip
Unit :
W
RLC
q
dI
- IR - L = 0
C
dt
q
dI
Þ + IR + L = Vd
C
dt
d 2q
dq 1
Þ L 2 + R + q = Vp sin w d t
dt
dt C
Vd -
Solution :
I = I p sin(w d t + f )
where
Vp
Ip =
R2 + (
=
1
- w d L)2
w dC
Vp
R 2 + (XC - X L )2
1
- wdL
-1 w d C
f = tan
R
X - XL
= tan -1 C
R
RLC Circuit and Impedance
Impedance: Z = R + (XC - X L )
2
Þ Ip =
Vp
Z
Phase:
XC - X L
tan f =
R
2
Summary:
Reactance and Impedance
XR = R
XL = w d L
1
XC =
w dC
Z = R 2 + (XC - X L )2
All measured in W. Used just like Ohm's Law:
Vp = I p X
Z is essentially Xtotal .
RLC Circuit and Impedance
Vp
Impedance: Z = R + (XC - X L ) Þ I p =
Z
VRp = I p R
2
VLp = I p X L
VCp = I p XC
2
High Pass and Low Pass Filter
Impedance: Z = R 2 + (w d L VLp = I p X L Þ VLp = Vp
1 2
)
w dC
wdL
1 2
R + (w d L )
w dC
2
VLp ® Vp when w d ® ¥ [High Pass]
VLp ® 0 when w d ® 0
VCp = I p XC
Þ VCp = Vp
1 / w dC
R 2 + (w d L -
1 2
)
w dC
VCp ® 0 when w d ® ¥
VCp ® Vp when w d ® 0 [Low Pass]
Resonance
Vp
Z
To maximize I p we need to minimize Z, which is the smallest (Z = R) when:
Z = R 2 + (XC - X L )2 Þ I p =
1
X L - XC = 0 Û w d L =0
w dC
Þ wdL =
1
w dC
1
LC
1
Þ wd =
º w 0 [Driving frequency = Natural frequency]
LC
Þ w d2 =
XC - X L
Þf = 0
R
R
Circuit becomes purely resistive (as if the capacitor and inductor is absent).
At resonance: I p =
Vp
and tan f =
Resonance
Happens when driving frequency equals natural frequency
1
wd =
º w0
LC
Vp
At resonance: I p =
R
Find XL and XR at resonance
wd =
1
LC
XL = w d L =
L
C
1
L
XC =
=
w dC
C
XC - X L
Þf = 0
R
R
Circuit becomes purely resistive (as if the capacitor and inductor is absent).
At resonance: I p =
Vp
and tan f =
Root-mean-square
I rms º
I2
where I 2 is the mean (average) of I 2 .
Suppose I = I p sin(w d t + f ), then:
I 2 = I p 2 sin 2 (w d t + f )
Þ I 2 = I p 2 sin 2 (w d t + f )
1
but sin (w d t + f ) =
2
1 2
2
Þ I = Ip
2
2
Therefore I rms º
I
2
1
=
Ip
2
Power and rms
P = I 2R
Þ P = I R =R I
2
P = I rms
2
1 2
R = Ip R
2
2
= I rms R
2
More rms
1
I rms =
Ip
2
Similarly:
1
Vrms =
Vp
2
Pavg = Vrms I rms cos f
cos f : power factor
Transformer
V1 N1
=
V2 N 2
I1V1 = I 2V2
Þ I 1 N1 = I 2 N 2
Reducing current during
transmission
Saving energy, why?
»10 V
5
Solution
R : Appliances at home
r : Resistance of the power lines
Suppose we need power P at home.
Without transformers (I1 rms current):
P = I 12 R Þ I 1 2 = P / R
Power loss: Ploss
r
= I1 r =
P
R
2
With transformers:
P = I12 R Þ I12 = P / R
N home
I 2 N lines = I1 N home Þ I 2 =
I1
N lines
N home 2
N home 2 r
2
Power loss: Ploss = I 2 r = (
I1 ) r = (
)
P
N lines
N lines R
Complex reactance:
A more general “Ohm’s Law”
V = IR
Now generalized to:
V = IX
where X is the reactance.
In V = IX, everything is written in complex number.
Complex reactance
XR = R
1
XC =
jwC
X L = jwL
Complex Numbers Review
1
j = -1 Þ j = -1, = - j, e jq = cosq + j sin q
j
2
p
-j
p
1
e = j, e = - j =
j
(usually denoted "i",
j
2
2
but we use "j" to avoid confusion with the current)
Cos and Sin in Complex Number
e jq = cosq + j sin q
- jq
e = cos(-q ) + j sin(-q ) = cosq - j sin q
Adding and subtracting gives:
e jq + e - jq
cosq =
2
e jq - e - jq
sin q =
2j
Complex Plane
Im
z = a + jb = R e jq
R
q
b
a
Re
z = a + jb = R e jq
Re(z) = a = R cosq : the real part of z
Im(z) = b = Rsin q : the imaginary part of z
z = R = a 2 + b 2 : the modulus (or magnitude) of z
Complex Plane (Special Cases)
Im
j=e
j
e jq = cosq + j sin q
p
2
Þ Im(e jq ) = sin q
Re
-j = e
-j
Putting q = ±
p
2
e
e
j
p
=j
2
-j
p
2
1
= -j =
j
p
2
:
Finding Imaginary Part
What is the imaginary part of z = jR e jq ?
Method 1 (expand):
z = jR e jq = Rj(cosq + j sin q )
Þ z = -Rsin q + jR cosq
Þ Im(z) = R cosq
j
p
Method 2 (use j = e 2 , recommended):
jq
j
p
jq
z = jR e = Re 2 e = Re
p
p
j (q + )
2
p
Þ z = R cos(q + ) + jRsin(q + )
2
2
p
Þ Im(z) = Rsin(q + )
2
Example: Rewrite in Polar Form
Im
z = 3 + j = 1.73 + j
R
q
a= 3
z = R = 1.732 +12 = 2
b =1
1
p
q = tan ( ) =
6
3
-1
Re
Þ z = 2e
j
p
6
Finding Imaginary Part
R e jq
What is the imaginary part of z =
?
3+ j
Trick: write the denominator in polar form!
3 + j = 2e
j
p
6
(see earlier example)
R e jq
R e jq R j (q - p6 )
Þz=
=
e
p =
j
2
3+ j
2e 6
R
p
R
p
Þ z = cos(q - ) + j sin(q - )
2
6
2
6
R
p
Þ Im(z) = sin(q - )
2
6
Complex Conjugate
The complex conjugate z* of a number z is found by flipping
all the j into - j.
Examples:
z = 3 + 4 j Þ z* = 3 - 4 j
-2 j
e2 j
e
z=
Þ z* =
3j
3j
1+
17-9j
7+9j
Conjugate to Modulus
z = z* z Þ z = z* z
2
Example 1:
z = 3+ 4j Þ z = 3- 4j
*
Þ z = (3 - 4 j)(3 + 4 j) = 32 + 4 2 = 25 Þ z = 5
2
Example 2:
2
2
*
z=
Þz =
3+ 4 j
3- 4j
2
2
4
4
2
2
Þ z =(
)(
)= 2
=
Þz=
2
3- 4 j 3+ 4j
3 +4
25
5
Translating between Real and
Complex Worlds
Real:
Complex:
Vp sin(w t)
«
I p sin(w t + f )
«
Vp e jw t
I p e j (w t+f )
I p = Magnitude of complex current = I complex
The Process
Real:
Complex:
Vp sin w t
Vp e jw t
Real:
Differential Eqn
Ohm's Law:
XR = R
Calculus
1
XC =
jw C
X L = jw L
Real:
Complex:
I p sin(w t + f )
I p e j (w t+f )
The Rule
1. Replace the voltage by complex version.
2. Use Ohm’s law to find the complex current.
Pretend R, L, C are all resistors with
complex reactance.
3. The imaginary part of the complex current
gives you the real current I(t).
4. The magnitude of the complex current gives
you the amplitude of the real current.
Capacitive Circuit
Convert to complex: Vd = Vp sin w d t ® Vd = Vp e jw d t
Pretend the capacitor is a resistor with resistance XC =
1
jw d C
Vd
Find complex current: I =
= jw d CVp e jw d t = w d CVp je jw d t
XC
j
p
Write in polar form using: j = e 2 , get:
j
p
I = w d CVp e 2 e
jw d t
= w d CVp e
p
j (w d t+ )
2
p
Convert back to real: I = w d CVp sin(w d t + )
2
Inductive Circuit
Convert to complex: Vd = Vp sin w d t ® Vd = Vpe jw d t
Pretend the capacitor is a resistor with resistance X L = jw d L
jw t
Vp 1 jw d t
Vd Vp e d
Find complex current: I =
=
=
e
XL
jw d L w d L j
p
-j
1
Write in polar form using = - j = e 2 , get:
j
Vp - j p2 jw d t
Vp j (w d t- p2 )
I=
e e =
e
wdL
wdL
Convert back to real: I =
Vp
p
sin(w d t - )
wdL
2
RLC
Circuit
Convert to complex: V = V sin w t ® V = V e
d
p
d
d
p
jw d t
Pretend everything are resistors with resistance X.
Total "resistance"
(actually called "impedance" in this context):
1
Xtotal = X R + X L + XC = R + jw d L +
jw d C
1
Because = - j :
j
Xtotal = R + j(w d L -
1
)
w dC
Vp e jw d t
Vd
=
Find complex current: I =
Xtotal R + j(w L - 1 )
d
w dC
[Continue on next page]
RLC Circuit V(amplitude)
e
jw t
d
V
p
We found: I =
=
Xtotal R + j(w L - 1 )
d
w dC
If we only need the amplitude of the current, we could
take just the magnitude of the complex current:
Ip = I =
Vp e jw t
R + j(w d L -
1
)
w dC
Vp
=
R 2 + (w d L -
1 2
)
w dC
Now that we are back in the real world, using Vp = I p Xreal ,
we could find the peak voltage across L and C:
wdL
VLp = I p (w d L) =
Vp
1 2
2
R + (w d L )
w dC
VLp = I p (
1
)=
w dC
1 / w dC
R 2 + (w d L -
1 2
)
w dC
Vp
RLC Circuit (full treatment)
Vp e jw d t
V
We found: I =
=
Xtotal R + j(w L - 1 )
d
w dC
To convert to real current, we need to write Xtotal in polar form:
Xtotal = R + j(w d L -
1
) = Xtotal e jq , where:
w dC
Xtotal = R 2 + (w d L -
1 2
) and q = tan -1 (
w dC
wdL -
1
w dC
R
)
Vp e jw d t
Vp e jw d t
Vp j (w d t-q )
Put back into I : I =
=
=
e
Xtotal
Xtotal e jq
Xtotal
Vp
Vp
Convert into real: I =
sin(w t - q ) º
sin(w t + f )
Xtotal
Xtotal
The phase: f = -q = -tan -1 (
The amplitude: I p =
wdL -
Vp
=
Xtotal
R
1
w dC
1
- wdL
w
C
) = tan -1 ( d
)
R
Vp
R 2 + (w d L -
1 2
)
w dC