Download Chapter 32 - Alternating Current (AC)

Chapter 31 - Alternating Current (AC)
I.
Look at the behavior of resistors, inductors, and capacitors in alternating current (AC) circuits. We
will find that the relationship between current and voltage for the elements looks very much like
Ohm's Law.
II.
Alternating current or AC means that there is a periodic variation in the current and voltage - any
type of periodic variation whether it be sinusoidal or square or triangular or….
A.
We will look at sinusoidal types of signals since any periodic signal can be analyzed as a
combination of sinusoidal and/or cosinusoidal signals. For example,
1  2n 
sin
t
 n 1,3,5,... n  T 
4

1.
square wave with period of T and amplitude of 1: v(t ) 
2.
triangular wave with a period of T and amplitude of 1: vt  


2
n 1
B.
 1n1 sin 2n t 
n
Mathematical representation of a sinusoidal voltage.
v = Vm sin t = Vm sin 2ft
where
Vm = amplitude, maximum value, peak value
f = frequency of the AC signal (hertz, cycles/sec, osc/sec, vib/sec)
 = 2f = angular frequency (radians/sec)
T = 1/f = period of the oscillation
The voltage can also be written in a more general way as
v = Vm sin (t + ) , where  = initial phase angle
31-1

 T


C.
Graphical representations of a sinusoidal voltage.
There are different ways to graphically represent a sinusoidal voltage. One way is to graph the
voltage versus time on rectangular axes. A second way is to represent the oscillations with a
rotating phasor (vector or rotor). The phasor is an arrow whose length equals the amplitude of the
voltage and is rotating around the origin with an angular velocity . The projection of this phasor
onto the y-axis gives the instantaneous values of voltage.
Graph of voltage vs. time (v vs. t)
1.
Corresponding phasor or rotor diagram
 = 0
v
t





2.


2
v
t
31-2
3.
 =-

2
v
t
4.
=
3
4
v
t
D.
Another way to express the value of the voltage (or any sinusoidal signal) is give the root-meansquare (rms) value of the sinusoidal signal: Vrms 
v2
If v = Vm sin t , then v2 = Vm2 sin2 t . The average value of a sine squared function is ½ the
V2
value of that function, therefore, v 2  m , and
2
Vm
Vrms 
 0.707Vm .
2
31-3
III. Single Element circuits - We are interested in determining how the current and voltage are related
to each other. In each of these cases we will be given a sinusoidal voltage source v = Vm sin t , and
the resulting current in the circuit is to be found. Specifically, we want to find 1) the amplitude of
the current relative to the amplitude of the voltage, and 2) the phase angle between the current and
the voltage.
A.
Resistor, R
v = Vm sin t
B.
v
t
i
t
Capacitor, C
v = Vm sin t
v
t
i
t
31-4
C.
Inductor, L
v = Vm sin t
v
t
i
t
C’. Inductor, L using a different approach
i = Im sin t
D.
v
t
i
t
Mnemonic device to remember the relative phase between currents (i) and voltages (v) for the
inductor (L) and capacitor (C): ELI the ICEman
31-5
IV. Combination of the three: the RLC series circuit for AC
R
L
C
i = Im sin t
31-6
Summary:
XL  L  2fL and
Finally:
R2  XL  XC 2
Vm = ImZ , where Z =
XC 
1
1

C 2 fC
If i = Im sin t , then v = Vm sin (t + ) .
If v = Vm sin t , then i = Im sin (t - ) .
In both cases:
  tan 1
XL  XC
R
Example: Suppose we have an RLC series circuit where the resistor is 900 ohms, the inductor is 2
henries, and the capacitor is 1 microfarad. A signal generator supplies a sinusoidal voltage v = 100
sin 1000t where v is in volts and t is in seconds. Find the instantaneous value of current in the
circuit. That is find Im ,  , and  . Find also the amplitudes of the voltages across the resistor,
inductor, and capacitor.
31-7
V.
More on the RLC circuit.
A.
Look at the variation of impedance Z, current amplitude Im , and phase angle  as a function of the
frequency of the AC signal. The voltage amplitude Vm is fixed (constant).
Im , Z, 
f, 
O
B.
fo

Power delivered to the RLC circuit.
Let i = Im sin t and v = Vm sin (t + ) , find the instantaneous power delivered to the circuit
and the average power delivered to the circuit.
31-8
C.
Graph of the average power delivered to an RLC series circuit as a function of frequency.
2
Pavg  I rmsVrms cos   Vrms
R

Z2
2
Vrms
R
1 

R 2   L 

C 

2
Note that maximum power is delivered at the resonant frequency.
Pavg
f, 
fo
o
The sharpness of the peak is defined as the quality factor Q =
o oL
.


R
What is the meaning of the sharpness of the peak? That is, what is the difference between a high Q
and low Q circuit?
AM
FM
31-9
KFI
740 kHz
KNX
1070
KCSM
88.5 MHz
KPCC
89.3
KCRW
89.9
KPFK
90.7
KZLA
93.9
VI. Filter Circuits
A.
High Pass Filter
C
R
Vin
Vout
Vout
Vin

B.
Low Pass Filter
R
Vin
C
Vout
Vout
Vin

31-10
VII. The parallel RLC circuit
R
C
L
VIII. The Transformer
transformer
V1
primary
31-11
N1
N2
V2
secondary