SUMMARY
Circuit
element
Average
Power
Resistor
R
E
 PR   m
2R
R
 PC   0
1
XC 
C
Current leads voltage
by a quarter of a
period
IC
VC  I C X C 
C
X L  L
Current lags behind
voltage by a quarter
of a period
VL  I L X L  I L L
Capacitor
C
Inductor
L
Reactance
2
 PL   0
Phase of current
Current is in phase
with the voltage
Voltage amplitude
VR  I R R
(31 - 17)
(31 - 18)
The series RCL circuit
An ac generator with emf E  Em sin t is connected to
an in series combination of a resistor R, a capacitor C
and an inductor L, as shown in the figure. The phasor
for the ac generator is given in fig.c. The current in
i  I sin t   
this circuit is described by the equation: i  I sin t   
The current i is common for the resistor, the capacitor and the inductor
The phasor for the current is shown in fig.a. In fig.c we show the phasors for the
voltage vR across R, the voltage vC across C , and the voltage vL across L.
The voltage vR is in phase with the current i. The voltage vC lags behind
the current i by 90. The voltage vL leads ahead of the current i by 90.
A
i  I sin t   
B
Z  R2   X L  X C 
O
I
2
Em
Z
Kirchhoff's loop rule (KLR) for the RCL circuit: E  vR  vC  vL . This equation
is represented in phasor form in fig.d. Because VL and VC have opposite directions
we combine the two in a single phasor VL  VC . From triangle OAB we have:
Em2  VR2  VL  VC    IR    IX L  IX C   I 2  R 2   X L  X C   


Em
I
The denominator is known as the "impedance" Z
2
2
R   X L  XC 
2
of the circuit.
I
2
2
2
Z  R 2   X L  X C   The current amplitude I 
2
Em
Z
Em
1 

R2    L 
C 

2
(31 - 19)
i  I sin t   
(31 - 20)
1
XC 
C
A
B
Z  R2   X L  X C 
tan  
O
X L  XC
R
X L  L
From triangle OAB we have: tan  
2
VL  VC IX L  IX C X L  X C


VR
IR
R
We distinguish the following three cases depending on the relative values
of X L and X L .
1. X L  X C    0 The current phasor lags behind the generator phasor.
The circuit is more inductive than capacitive
2. X C  X L    0 The current phasor leads ahead of the generator phasor
The circuit is more capacitive than inductive
3. X C  X L    0 The current phasor and the generator phasor are in phase
1. Fig.a and b: X L  X C    0
The current phasor lags behind
the generator phasor. The circuit is more
inductive than capacitive
2. Fig.c and d: X C  X L    0 The current phasor leads ahead of the generator
phasor. The circuit is more capacitive than inductive
3. Fig.e and f: X C  X L    0 The current phasor and the generator phasor are
(31 - 21)
in phase

1
LC
I res
Em

R
Resonance
In the RCL circuit shown in the figure assume that
the angular frequency  of the ac generator can
be varied continuously. The current amplitude
in the circuit is given by the equation:
Em
I
The current amplitude
2
1 

R2    L 


C


1
has a maximum when the term  L 
0
C
1
This occurs when  
LC
The equation above is the condition for resonance. When its is satisfied I res 
Em
R
A plot of the current amplitude I as function of  is shown in the lower figure.
This plot is known as a "resonance curve"
(31 - 22)
2
Pavg  I rms
R
Pavg  I rms Erms cos 
(31 - 23)
Power in an RCL ciruit
We already have seen that the average power used by
a capacitor and an inductor is equal to zero. The
power on the average is consumed by the resistor.
The instantaneous power P  i 2 R   I sin t     R
2
T
The average power Pavg
1
  Pdt
T 0
1 T
 I 2R
2
2
Pavg  I R   sin t    dt  
 I rms
R
2
T 0

E
R
Pavg  I rms RI rms  I rms R rms  I rms Erms  I rms Erms cos 
Z
Z
The term cos in the equation above is known as
2
the "power factor" of the circuit. The average
power consumed by the circuit is maximum
when   0
Transmission lines
Erms =735 kV , I rms = 500 A
(31 - 24)
home
Step-down
transformer 110 V
Step-up
transformer
T2
T1
R = 220Ω
 1000 km
Power Station
Energy Transmission Requirements
The resistance of the power line R 
Heating of power lines Pheat

.
R is fixed (220  in our example)
A
2
 I rms
R This parameter is also fixed
( 55 MW in our example)
Power transmitted Ptrans  Erms I rms
(368 MW in our example)
In our example Pheat is almost 15 % of Ptrans and is acceptable
To keep Pheat we must keep I rms as low as possible. The only way to accomplish this
is by increasing Erms . In our example Erms  735 kV. To do that we need a device
that can change the amplitude of any ac voltage (either increase or decrease)
The transformer
(31 - 25)
The transformer is a device that can change
the voltage amplitude of any ac signal. It
consists of two coils with different number
of turns wound around a common iron core.
The coil on which we apply the voltage to be changed is called the "primary" and
it has N P turns. The transformer output appears on the second coils which is known
as the "secondary" and has N S turns. The role of the iron core is to insure that the
magnetic field lines from one coil also pass through the second. We assume that
if voltage equal to VP is applied across the primary then a voltage VS appears
on the secondary coil. We also assume that the magnetic field through both coils
is equal to B and that the iron core has cross sectional area A. The magnetic flux
dP
dB
through the primary  P  N P BA  VP  
 NP A
(eqs.1)
dt
dt
dS
dB
The flux through the secondary  S  N S BA  VS  
 NS A
(eqs.2)
dt
dt
VS
V
 P
NS NP
dP
dB
 NP A
(eqs.1)
dt
dt
dS
dB
 S  N S BA  VS  
 NS A
(eqs.2)
dt
dt
If we divide equation 2 by equation 1 we get:
dB
NS A
VS
dt  N S  VS  VP

VP  N A dB
NP
NS NP
P
dt
 P  N P BA  VP  
The voltage on the secondary VS  VP
NS
NP
If N S  N P 
NS
 1  VS  VP We have what is known a "step up" transformer
NP
If N S  N P 
NS
 1  VS  VP We have what is known a "step down" transformer
NP
Both types of transformers are used in the transport of electric power over large
distances.
(31 - 26)
IS
IP
VS
V
 P
NS NP
IS NS  I P NP
VS
VP
We have that:

NS NP
 VS N P  VP N S
(eqs.1)
If we close switch S in the figure we have in addition to the primary current I P
a current I S in the secondary coil. We assume that the transformer is "ideal"
i.e. it suffers no losses due to heating then we have: VP I P  VS I S
If we divide eqs.2 with eqs.1 we get:
IS 
(eqs.2)
VI
VP I P
 S S  IP NP  IS NS
VP N S
VS N P
NP
IP
NS
In a step-up transformer (N S  N P ) we have that I S  I P
In a step-down transformer (N S  N P ) we have that I S  I P
(31 - 27)
Hitt
A generator supplies 100 V to the primary coil of a
transformer. The primary has 50 turns and the
secondary has 500 turns. The secondary voltage is:
A. 1000 V
B. 500 V
C. 250 V
D. 100 V
E. 10V
hitt
The main reason that alternating current replaced
direct current for general use is:
A. ac generators do not need slip rings
B. ac voltages may be conveniently transformed
C. electric clocks do not work on dc
D. a given ac current does not heat a power line as
much as the same dc current
E. ac minimizes magnetic effects