Download Ch 33 - A.C. Circuits

Alternating Current Circuits
Chapter 33
(continued)
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 current or voltage.
Resistor
Capacitor
Inductor
VRp
Ip
VLp
Ip
wt
Ip
wt
wt
VCp
Impedance in AC Circuits
R
V
~
C
L
The impedance Z of a circuit or circuit element
relates peak current to peak voltage:
Vp
Ip 
Z
(Units: Ohms)
(This is the AC equivalent of Ohm’s law.)
Phasor Diagrams
Circuit element
Impedance
Amplitude
Phase
Resistor
R
VR= IP R
I, V in phase
Capacitor
Inductor
Xc=1/wC
XL=wL
VC=IP Xc
VL=IP Xc
I leads V by 90°
I lags V by 90°
Resistor
Capacitor
Inductor
VRp
Ip
VLp
Ip
wt
Ip
wt
wt
VCp
RLC Circuit
R
V
~
C
L
Use the loop method:
V - V R - VC - VL = 0
I is same through all components.
BUT: Voltages have different PHASES
 they add as PHASORS.
RLC Circuit
Ip
VLp
VRp
f
(VCp- VLp)
VP
VCp
RLC Circuit
Ip
VLp
VRp
f
(VCp- VLp)
VP
VCp
By Pythagoras’s theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
= Ip2 R2 + (Ip XC - Ip XL) 2
RLC Circuit
R
Solve for the current:
Ip 
Vp
Vp

Z
R 2  (X c  X L ) 2
V
~
C
L
RLC Circuit
R
Solve for the current:
Ip 
V
Vp

Z
R 2  (X c  X L ) 2
Impedance:
Vp
~
C
L
 1

Z  R 
 wL
 wC

2
2
RLC Circuit
Ip 
Vp
The current’s magnitude depends on
the driving frequency. When Z is a
minimum, the current is a maximum.
This happens at a resonance frequency:
Z
2


1
2
Z  R 
 wL 
 wC

The circuit hits resonance when 1/wC-wL=0: w r=1/ LC
When this happens the capacitor and inductor cancel each other
and the circuit behaves purely resistively: IP=VP/R.
IP
R =10W
L=1mH
C=10mF
R = 1 0 0 W
0
1 0
wr
2
1 0
3
1 0
4
1 0
5
w
The current dies away
at both low and high
frequencies.
Phase in an RLC Circuit
Ip
VLp
VRp
f
VP
We can also find the phase:
tan f = (VCp - VLp)/ VRp
= (XC-XL)/R
(VCp- VLp)
VCp
= (1/wC - wL) / R
Phase in an RLC Circuit
Ip
VLp
We can also find the phase:
VRp
f
VP
tan f = (VCp - VLp)/ VRp
= (XC-XL)/R
(VCp- VLp)
VCp
= (1/wC - wL) / R
More generally, in terms of impedance:
cos f  R/Z
At resonance the phase goes to zero (when the circuit becomes
purely resistive, the current and voltage are in phase).
Power in an AC Circuit
The power dissipated in an AC circuit is P=IV. Since both
I and V vary in time, so does the power: P is a function of
time.
Use V = VP sin (wt) and I = IP sin (w t+f ) :
P(t) = IpVpsin(wt) sin (w t+f )
This wiggles in time, usually very fast. What we usually
care about is the time average of this:
1
P 
T

T
0
P(t)dt
(T=1/f )
Power in an AC Circuit
Now:

sin( wt  f)  sin( wt)cos f  cos(wt)sin f
Power in an AC Circuit
Now:
sin( wt  f)  sin( wt)cos f  cos(wt)sin f
P(t)  IPVP sin( w t)sin( w t  f )


 IPVP sin 2 (w t)cos f  sin( w t)cos(w t)sin f
Power in an AC Circuit
Now:
sin( wt  f)  sin( wt)cos f  cos(wt)sin f
P(t)  IPVP sin( w t)sin( w t  f )
 IPVP sin 2 (w t)cos f  sin( w t)cos(w t)sin f

Use:
and:
So

sin (w t) 
2
1
2
sin( w t)cos(w t)  0
1
P  IPVP cos f
2
Power in an AC Circuit
Now:
sin( wt  f)  sin( wt)cos f  cos(wt)sin f
P(t)  IPVP sin( w t)sin( w t  f )
 IPVP sin 2 (w t)cos f  sin( w t)cos(w t)sin f

Use:
and:
So

sin (w t) 
2
1
2
sin( w t)cos(w t)  0
1
P  IPVP cos f
2
which we usually write as
P  IrmsVrms cos f
Power in an AC Circuit
P  IrmsVrms cos f
(f goes from -900 to 900, so the average power is positive)
cos(f)is called the power factor.
For a purely resistive circuit the power factor is 1.
When R=0, cos(f)=0 (energy is traded but not dissipated).
Usually the power factor depends on frequency, and
usually 0<cos(f)<1.
Power in a purely resistive circuit
V
f= 0
p
I
2p
wt
V(t) = VP sin (wt)
I(t) = IP sin (wt)
(This is for a purely
resistive circuit.)
P
P(t) = IV = IP VP sin 2(wt)
Note this oscillates
twice as fast.
p
2p
wt
Power in a purely reactive circuit
P  IrmsVrms cos f
The opposite limit is a purely reactive circuit, with R=0.
I
V
P
This happens with an
LC circuit.
0
Then
f
90
wt
The time average of
P is zero.
Transformers
Transformers use mutual inductance to change voltages:
Iron Core
Np turns
Ns turns
Vs
Vp
Primary
(applied voltage)
Secondary
(produced voltage)
Faraday’s law on the left:
If the flux per turn is f then
Vp=Np(df/dt).
Faraday’s law on the right:
The flux per turn is also f, so
Vs=Ns(df/dt).
Ns
Vp
 Vs 
Np
Transformers
Transformers use mutual inductance to change voltages:
Np turns
Iron Core
Ns turns
Vs
Vp
Primary
(applied voltage)
Secondary
(produced voltage)
Ns
Vs 
Vp
Np
In the ideal case, no
power is dissipated in
 itself.
the transformer
Then IpVp=IsVs

Np
Is 
Ip
Ns
Transformers & Power Transmission
Transformers can be used to “step up” and “step
down” voltages for power transmission.
110 turns
Power
=I1 V1
V1=110V
20,000 turns
V2=20kV Power
=I2 V2
We use high voltage (e.g. 365 kV) to transmit electrical
power over long distances.
Why do we want to do this?
Transformers & Power Transmission
Transformers can be used to “step up” and “step down”
voltages, for power transmission and other applications.
110 turns
Power
=I1 V1
V1=110V
20,000 turns
V2=20kV Power
=I2 V2
We use high voltage (e.g. 365 kV) to transmit electrical
power over long distances.
Why do we want to do this?
P = I2R
(P = power dissipation in the line - I is smaller at high voltages)