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Chapter 33
Alternating Current (AC)
R, L, C in AC circuits
AC, the description

A DC power source, like the one from a
battery, provides a potential difference (a
voltage) that does not change its polarity with
respect to a reference point (often the ground)

An AC power source is sinusoidal voltage
source which is described as
V
V
t
v  Vmax sin t 
Here
v
Vmax

is the instantaneous voltage with respect to a reference (often not the ground).
is the maximum voltage or amplitude.
is the angular frequency, related to frequency f and period T as
v
or
Symbol in a circuit diagram:
v
The US AC system is 110V/60Hz.
Many European and Asian countries use
220V/50Hz.
  2 f 
2
T
Resistors in an AC Circuit, Ohm’s Law
The voltage over the resistor:
vR  v  Vmax sin t 
Apply Ohm’s Law, the current through the
resistor:
iR 
vR
R

Vmax
R
sin t   I max sin t 
The current is also a sinusoidal function of
time t. The current through and the voltage
over the resistor are in phase: both reach
their maximum and minimum values at the
same time.
The power consumed by the resistor is
pR  vR  iR 
vR2
R
i R
2
R
2
Vmax
R
sin 2 t 
We will come back to the power discussion later.
PLAY
ACTIVE FIGURE
Phasor Diagram, a useful tool.

y
The projection of a circular motion with
a constant angular velocity on the yaxis is a sinusoidal function.
R
t
O


To simplify the analysis of AC circuits,
a graphical constructor called a phasor
diagram is used. A phasor is a vector
whose length is proportional to the
maximum value of the variable it
represents
The phasor diagram of a resistor in AC
is shown here. The vectors
representing current and voltage
overlap each other, because they are
in phase.
The projection on the y-axis is
Ry  R sin t 
x
The power for a resistive AC circuit
and the rms current and voltage
When the AC voltage source
is applied on the resistor, the
voltage over and current
through the resistor are:
vR  Vmax sin t 
iR  I max sin t 
Both average to zero.
But the power over the resistor is
pR  pmax sin2 t 
And it does not average to zero.
The averaged power is:
pav 
1
T
T
2

P
Pmax
Pav
Pmax
pmax sin 2 t dt
2 0
T
2 p  t 1
p

 max   sin  2t    max
T  2 4
2
0
2
x 1
2
sin
xdx

 sin  2 x 

2 4
The power for a resistive AC circuit
and the rms current and voltage
So the averaged power the resistor consumes is
p
1
pav  max  Vmax  I max
2
2
If the power were averaged to zero, like the current
and voltage, could we use AC power source here?
The averaged power can also be written
2
as:
1 Vmax
1 2
P
pav 
 I max
R
2 R
2
Pmax
Define a root mean square for the voltage
Pav
and current:
1
1 2
2
2
2
Vrms
 Vmax
, and I rms
 I max
2
2
or
Vmax  2Vrms , and I max  2I rms
One get back to the DC
formula equivalent:
pav  Vrms  I rms 
2
Vrms
R
2
 I rms
R
Pmax
Resistors in an AC Circuit, summary

Ohm’s Law applies. Te current through and voltage over the
resistor are in phase.
vR  Vmax sin t  iR  I max sin t 

The average power consumed by the resistor is
pav  Vrms  I rms 

2
 I rms
R
R
From this we define the root mean square current and voltage.
AC meters (V or I) read these values.
Vrms 

2
Vrms
Vmax
2
, and I rms
I max

2
The US AC system of 110V/60Hz, here the 110 V is the rms
voltage, and the 60 Hz is the frequency f, so
Vmax  2Vrms  156 V
  2 f  377 sec1
Inductors in an AC circuit, voltage and
current
The voltage over the inductor is
vL  v  Vmax sin t 
To find the current i through the inductor,
we start with Kirchhoff’s loop rule:
v  vL  0
di
0
or Vmax sin t   L
dt
Solve the equation for i
di 
Vmax
sin t  dt
L
V
V


i   di   max sin t dt   max cos t   I max sin  t  
L
L
2

or
V


i  I max sin  t   , with I max  max
2
L

Inductors in an AC circuit, voltage leads
current
Examining the formulas for voltage over and
current through the inductor:


i

I
sin

t

vL  v  Vmax sin t 
max


2

Voltage leads the current by ¼ of a period (T/4
or 90° or π/2) . Or in a phasor diagram, the
rotating current vector is 90° behind the
voltage vector.
PLAY
ACTIVE FIGURE
Inductive Reactance, the “resistance” the
inductor offers in the circuit.
Examining the formulas for voltage over and
current through the inductor again:
vL  Vmax sin t 
V


i  I max sin  t   , with I max  max
2
L

This time pay attention to the relationship between
the maximum values of the current and the voltage:
V
I max  max
L
This could be Ohm’s Law if we define a “resistance” for the inductor to be:
X L  L
And this is called the inductive reactance. Remember, it is the product of the
inductance, and the angular frequency of the AC source. I guess that this is
the reason for it to be called a “reactance” instead of a passive “resistance”.
The following formulas may be useful:
Vmax  X L I max , and Vrms  X L I rms
vL  I max X L sin t 
Capacitors in an AC circuit, voltage and
current
The voltage over the capacitor is
vC  v  Vmax sin t 
To find the current i through the capacitor,
we start with Kirchhoff’s loop rule:
v  vC  0
or
Vmax sin t  
q
dq
 0, with i 
C
dt
Solve the first equation for q, and take the derivative for i
q  CVmax sin t 
dq


i
 C Vmax cos t   I max sin  t  
dt
2

or
Vmax


i  I max sin  t   , with I max 
1
2


C
 
Here I still like to keep the
Ohm’s Law type of formula
for voltage, current and a type
of resistance.
Capacitors in an AC circuit, current leads the
voltage
Examining the formulas for voltage over and
current through the capacitor:

vC  Vmax sin t  i  I max sin  t  
2

Current leads the voltage by ¼ of a period (T/4
or 90° or π/2) . Or in a phasor diagram, the
rotating voltage vector is 90° behind the
current vector.
PLAY
ACTIVE FIGURE
Capacitive Reactance, the “resistance” the
capacitor offers in the circuit.
Examining the formulas for voltage over and
current through the capacitor again:
vC  Vmax sin t 
Vmax


i  I max sin  t   , with I max 
1
2

C 
This time pay attention to the relationship between
the maximum values of the current and the voltage:
I max 
Vmax
1
C 
This could be Ohm’s Law if we define a “resistance” for the capacitor to be:
XC 
1
C
And this is called the capacitive reactance. It is the inverse of the product of
the capacitance, and the angular frequency of the AC source. The following
formulas may be useful:
Vmax  X C I max , and Vrms  X C I rms
vL  I max X C sin t 
The RLC series circuit, current and voltage
The voltage over the RLC is
v  vR  vL  vC
Now let’s find the current. From the this
equation, write out each component:
di q
Vmax sin t   iR  L 
dt C
d
to both sides, and remember
dt
dq
That i 
we have
dt
2
Apply
Vmax cos t   R  L
d i i

2
dt
C
Phase angle between
current and voltage
“Simply” solve for the current i :
i
Vmax
Where:
Z
v  Vmax sin t 
sin t     I max sin t   
Z  R   X L  XC 
2
2
Overall
resistance
X L  XC
tan  
R
Phase
angle
The RLC series circuit, current and voltage,
solved with Phasor Diagrams
The RLC are in serial connection, the current i
is common and must be in phase:
i
v  Vmax sin t 
i  iR  iL  iC
So use this as the base (the x-axis) for the
phasor diagrams:
The RLC series circuit, current and voltage,
solved with Phasor Diagrams
Now overlap the three
phasor diagrams, we
have:
The RLC series circuit, current and voltage,
solved with Phasor Diagrams
Now from final phasor
diagram, we get the
voltage components in xand y-axes:
From:
Vmax  VR 2   VL  VC 
or:
I max  Z 
We have:
2
 I max R    I max X L  I max X C 
2
Z  R2   X L  X C 
2
2
Here Z is the overall “resistance”, called
the impedance.
From the diagram, the phase angle is
We have: tan  
X L  XC
R
tan  
VL  VC I max X L  I max X C

VR
I max R
PLAY
ACTIVE FIGURE
Determining the Nature of the Circuit



If  is positive
V
I max  max
 XL> XC (which occurs at high frequencies)
Z
 The current lags the applied voltage
2
Z  R2   X L  X C 
 The circuit is more inductive than capacitive
X  XC
If  is negative
tan   L
R
 XL< XC (which occurs at low frequencies)
 The current leads the applied voltage
 The circuit is more capacitive than inductive
If  is zero
1
1
2

L

,
or


 XL= XC (which occurs at
)
C
LC
 The circuit is purely resistive and the impedance is minimum, and
current reaches maximum, the circuit resonates.

Often this resonant frequency is called 0 
1
LC