• Introduce phasors and alternating current
• Study voltage, current, and phase angle
• Consider resistance and reactance
• Impedance and its application to the L-R-C series circuit
• Power in AC circuits
• R-L-C and AC power
• Transformers
1
– Thomas Edison invents the light bulb
– Edison favored DC current power distribution
– George Westinghouse favored AC current distribution
– Low voltage power is safe but high power loss in transmission
– High voltage power is not safe but low power loss in transmission
– AC power distribution allows transformers to step down high voltage transmission
– AC current in long metal conductor create electromagnetic waves, radio waves.
2
1

A wave is a disturbance. Unlike water waves, electrical waves cannot be seen directly but they have similar characteristics. All periodic waves can be constructed from sine waves , which is why sine waves are fundamental.
Sine Wave Characteristics
Sine waves are characterized by the amplitude and period.
The amplitude is the maximum value of a voltage or current; the period is the time interval for one complete cycle.
The amplitude ( A of this sine wave is 20 V
)
The period is 50.0
 s
20 V
15 V
10 V
0 V
0
A
25 37.5
50.0
t ( s)
-10 V
-15 V
-20 V T
2

Period and frequency
The period and frequency are reciprocals of each other. f

1
T and T

1 f
Thus, if you know one, you can easily find the other.
(The 1/ x key on your calculator is handy for converting between f and T .)
If the period is 50
 s, the frequency is 0.02 MHz = 20 kHz.
Sine wave voltage and current values
The voltage of a sine wave can also be specified as either the peak-to-peak or the rms value. The peak-topeak is twice the peak value. The rms value is 0.707 times the peak value.
20 V
15 V
The peak-to-peak voltage is 40 V.
10 V
0 V
0
V
PP 25
V rms
37.5
50.0
t ( s)
The rms voltage is 14.1 V.
-10 V
-15 V
-20 V
3

Angular measurement
Angular measurements can be made in degrees ( o ) or radians. The radian (rad) is the angle that is formed when the arc is equal to the radius of a circle. There are 360 o or
2 p radians in one complete revolution.
1.0
0.8
R
R
0.6
0.4
0.2
0
0
-0.2
-0.4
-0.6
-0.8
p
4 p
2
3 p
4 p
5 p
4
3 p
2
-1.0
7 p
4
2 p
Angular measurement
Because there are 2 p radians in one complete revolution and 360 o in a revolution, the conversion between radians and degrees is easy to write. To find the number of radians, given the number of degrees: rad

2 p rad
360

 degrees
0.01745 rad/deg
To find the number of degrees, given the radians: deg

2
360
 p rad
 rad
57.295 deg/rad
4

Lagging Phase shift
-20
-30
- 40
40
30
20
10
0
0

Referenc e
45

90

Example of a wave that lags the reference …and the equation has a negative phase shift
Peak voltage v = 30 V sin ( q -
45 o )
135

180

225

270

315

360

405

Notice that a lagging sine wave is below the axis at 0
Angle (

) o
Leading Phase Shift
40
30
20
10
-45
 0
-10
-20
-30
-40
0

Referenc e
Example of a wave that leads the reference
Notice that a leading sine wave is above the axis at 0 o
Peak voltage v = 30 V sin ( q
+ 45 o )
45

90

135

180

225

…and the equation has a positive phase shift
Angle (

)
270

315

360

5

– Alternating current is sinusoidal v

V cos
 t
– Alternating current can be represented by a phasor or a vector rotating at ω radians per second
– A phasor is not a physical quantity. It is a geometric entity that helps analyze sinusoidal quantities
– The polar form is v

V
– q
Phasors
The sine wave can be represented as the projection of a vector rotating at a constant rate. This rotating vector is called a phasor .
90
180 0
0 90 180 360
11
6

Phasors
The position of a phasor at any instant can be expressed as a positive angle, measured counterclockwise from 0
 or as a negative angle equal to q -
360

. positive angle of q negative angle of q -
360
 phasor
• The values of voltage and current are constantly changing in AC, unlike in DC in which they are steady. We can measure AC voltages in two ways:-
• Measure the peak to peak voltage, easily done on a cathode ray oscilloscope (CRO).
• Measure the root mean square (rms) value, or the effective value.
• We use the rms value because its use allows us to do electrical calculations as if they were direct currents .
7

• Imagine a heater connected to an ac supply.
• It would repeatedly heat up then cool down.
• The power, P, supplied to a heater of resistance R:-
P

I
2
R
• Recall that direction of the current doesn’t matter so the power supplied varies with the square of the current.
15
8

• When I is at its maximum, I
0
, max. power is supplied.
• When I is zero, zero power is supplied.
• The mean power supplied:-
P mean

1
2
I
0
2
R
• This is from the symmetrical shape of the power curve.
• The root mean square value of an alternating current is the value of direct current that would give the same heating effect as the alternating current in the same resistor.
I rms
2
R
P


1
2
I rms
2
R
I
0
2
R
I rms

I
0
2
I rms
2 
1
2
I
0
2 V rms

V
0
2
9

rms
I rms

V rms
R
P

I rms
2
R
10

– A diode conducts current in one direction
21
Household current
• In the US and Canada household power is distributed as AC 120 volts rms
• The peak voltage is V

2 V rms
 
V
 
170 V
• Follow Example 31.1, illustrated by Figure 31.6.
• Personal computer draws 2.7A from 120-V, 60 Hz line
• Average current? Peak current?
22
11

• Receptacles have three holes each
• Lower (rounded) hole is earth ground
– connected to pipes, usu.
– green wire
• Larger slot is “neutral”
– for current “return”
– never far from ground
– white wire
– if wired correctly
• Smaller slot is “hot”
– swings to +170 and -
170
– black wire
– dangerous one
23
In a pure resistance circuit voltage and current are in phase
Figure 31-4 i v
R


I v
R cos v ab

V
R

 t
Ri cos

 t
RI cos
 t

= 2 p f where

= angular frequency in rad/sec
f = frequency in Hertz (cycles/sec)
24
12

AC circuit with pure inductance, the voltage leads the current by 90 i v
L
 v
L
I

 cos v ab
V
L
 t

L di cos( dt
 t
 -
I

L

90

) sin
 t

IX
L cos(
 t

90

)
V
L

IX
L
X
L
 
L

2 p fL (inductive reactance in ohms)
25
In a pure capacitance circuit the voltage lags the current by 90 i v c

 dq dt v ab


I v c

V
C cos
1
C cos(

 t

I
t cos
90

 t dt
) q

Cv c
 
I

I

C sin
 t cos
 tdt
V

C
IX
C
X
C

1

C

1
2 p fC
(capacitive reactance in ohms)
26
13

Comparing AC circuit elements
• Table 31.1 (bottom) provides a valuable summary/comparison for circuit elements.
• Figure 31.11 (below) shows items as a function of the angular frequency.
27
In an L-R-C series circuit the voltage and current depend upon the frequency and values of L-R-C
+ v – i

I cos
 t
X
L
> X
C
X
L
< X
C
V

V
R
2 
( V
L
-
V
C
)
2 
V

IZ
( IR )
2 
( IX
L
-
IX
C
)
2 
I R
2 
( X
L
-
X
C
)
2
Z

X
( R )
2 
( X )
2

X
L
-
X
C
(impedance in ohms)
(total reactance in ohms) tan
 
V
L
-
V
R
V
C

IX
L
-
IX
C
IR

X
L
-
X
C
R

X
R
(phase between V and I)
28
14

The L-R-C circuit Example 31.4
R=300Ώ, L=60 mH, C=0.50µf, ω=10,000rad/s
Find X
L
, X
C
, Z and I
The loudspeaker Crossover Network
• The woofer (low tones) and the tweeter (high tones) connect in parallel through a “crossover.”
• Consider Figure 31.12.
29
30
15

Power in AC Circuits
Arbitrary AC Circuit p
 vi


V cos(
 t
 
)

I cos
 t
 cos(
 t
 
)
 cos
 t cos
 sin
 t sin

Figure 31-12 p

VI cos
 cos
2
 t
-
VI sin
 cos
 t sin
 t p

VI cos

(
1
2

1
2 cos

Avg
2


0
 t )
-
VI sin

(
1 sin 2
 t
 
Avg

0
) cos

= power factor
P av

V rms
I rms cos

P av

I
2 rms
R
31
Power in Pure Resistance AC Circuits
P av

VI
2
V
Figure 31-12

I

R
V
2 v p

V
 vi cos
 t

( V cos i
 t )(

I
I cos
 t cos
 t ) p

VI p

VI
2 cos
2

VI
2
 t

VI 
1
2 cos 2
 t
1
2 cos 2
 t


I
2

V rms
I rms V rms

V
2
I rms

I
2
V rms

I rms
R
2 2
P av

V rms
I rms
 2
I rms
R

2
V rms
R
(31.29)
32
16

Power in a resistor-inductor-capacitor
• Consider current, voltage, and power as functions of time.
• P = vi for all components
• Average power will be different because the phase between current and voltage are different
33
P av

V rms
I rms
  
90
 cos

P av

V rms
I rms cos 90
 
V rms
I rms
( 0 )

0
P av

 
V rms
-
90

I rms cos

P av

V rms
I rms cos(
-
90

)

V rms
I rms
( 0 )

0
P av

 
V rms
0

I rms cos

P av

V rms
I rms cos( 0

)

V rms
I rms
34
17

V

IZ
Z

( R )
2 
( X )
2
X

X
L
-
X
C
(impedance in ohms)
(total reactance in ohms)
At resonance X
L

X
C and X

0
Therefore Z

( R ) 2

R tan
 
X
L
-
X
C
R

X
R

0
2 p f
0
L

1
2 p f
0
C and f
0

2 p
1
LC
 
0

(resonant frequency)
35
Circuit behavior at resonance
At resonance impedance is a minimum and current a maximum http://physci.kennesaw.edu/javamirror/CCP/21-
5/CircuitiE.html
Example of resonant circuit with varying R
L-R_C circuit V=100 V,
L=2.0H, C=0.536µf
36
18

Resonance and tuned circuit application
• Follow Example 31.8.
• Find resonant frequency, impedance, rms current, rms voltages
Crystal set radio uses a tuned circuit
37
• Let’s power our homes with DC power
– DC means direct current: just like what batteries deliver
• But want power plants far from home
– and ability to “ship” electricity across states
• So power lines are long
– resistance no longer negligible long transmission line power plant home appliance
R wire looks like: R load
38
R wire
19

Power Dissipated in an Electricity Distribution System
150 miles
120 Watt
Light bulb
Power Plant on Colorado River
12 Volt
Connection Box
• Estimate resistance of power lines: say 0.001 Ohms per meter, times 200 km =
0.001
W
/m

2

10 5 m = 20 Ohms
• We can figure out the current required by a single bulb using P = VI so I = P / V
= 120 Watts/12 Volts = 10 Amps (!)
• Power in transmission line is P = I 2 R = 10 2

20 = 2,000 Watts! (one wire)
• Efficiency = 120 Watts/4120 Watts = 0.3%!
(P=2000+2000+120)
• What could we change in order to do better?
39
• The thing that kills us most is the high current through the (fixed resistance) transmission lines
• Need less current
– it’s that square in I 2 R that has the most dramatic effect
• But our appliance needs a certain amount of power
– P = VI so less current demands higher voltage
• Solution is high voltage transmission
– Repeating the above calculation with 12,000 Volts delivered to the house draws only
I = 120 Watts/12 kV = 0.01 Amps for one bulb, giving
P = I 2 R = (0.01) 2 20 = 20

10
-4
Watts, so
P = 0.002 Watts of power dissipated in transmission line
Efficiency in this case is e
= 120 Watts/120.004 =
99.996%
40
20

• But having high voltage in each household is a recipe for disaster
– sparks every time you plug something in
– risk of fire
– Electric shock more likely
• Need a way to step-up/step-down voltage at will
– can’t do this with DC, so go to AC
• A step-down power distribution system is needed
41
step-up to 500,000 V step-down, back to 5,000 V
~5,000 Volts step-down to 120 V
High Voltage Transmission Lines, Low Voltage to Consumers
42
21

three-phase “live” wires
500,000 230,000 138,000 69,000 7–13,000 long-distance neighborhood to house
43
Transformers
44
22

Transformers relationships
• Alternating current in primary sets up alternating flux in the core.
• This induces an emf in each winding according to Faraday’s law.
• The induced emf in the secondary produces an alternating current in the secondary
V
1

N
1 d

B dt
V
2

N
2 d

B dt
V
1
V
2

N
1 d

B
N
2 d dt

B dt

N
1
N
2
Transformers can also
“transform” impedance.
This can be used for impedance matching .
V
1
V
2
I
1
I
2
Z
1
Z
2

N
1
N
2
N
2
N
1
 


N
1
N
2


2
V
1

N
1
V N
2 2
I
1

N
I N
2 1
2
45
• Windings i 2 R losses – minimize with low gauge wire and low current
• Hysteresis losses – minimize with soft iron core
• Eddy current losses – minimize with a laminate d core (can produce a
“hum”
46
23

Q31.1
A resistor is connected across an ac source as shown. For this circuit, what is the relationship between the instantaneous current i through the resistor and the instantaneous voltage v ab across the resistor?
A. i is maximum at the same time as v ab
B. i is maximum one-quarter cycle before v ab
C. i is maximum one-quarter cycle after v ab
D. not enough information given to decide
Q31.2
An inductor is connected across an ac source as shown. For this circuit, what is the relationship between the instantaneous current i through the inductor and the instantaneous voltage v ab across the inductor?
A. i is maximum at the same time as v ab
B. i is maximum one-quarter cycle before v ab
C. i is maximum one-quarter cycle after v ab
D. not enough information given to decide
24

Q31.3
A capacitor is connected across an ac source as shown. For this circuit, what is the relationship between the instantaneous current i through the capacitor and the instantaneous voltage v ab across the capacitor?
A. i is maximum at the same time as v ab
B. i is maximum one-quarter cycle before v ab
C. i is maximum one-quarter cycle after v ab
D. not enough information given to decide
Q31.4
An L-R-C series circuit as shown is operating at its resonant frequency.
At this frequency, how are the values of the capacitive reactance X
C inductive reactance X
L
, the
, and the resistance R related to each other?
A. X
L
= R ; X
C
can have any value.
B. X
C
= R ; X
L
can have any value.
C. X
C
= X
L
; R can have any value.
D. X
C
= X
L
= R.
E. none of the above
25

Q31.5
In an L-R-C series circuit as shown, the current has a very small amplitude if the ac source oscillates at a very high frequency. Which circuit element causes this behavior?
A. the resistor R
B. the inductor L
C. the capacitor C
D. misleading question — the current actually has a very large amplitude if the frequency is very high
Q31.6
In an L-R-C series circuit as shown, there is a phase angle between the instantaneous current through the circuit and the instantaneous voltage v ad
across the entire circuit. For what value of the phase angle is the greatest power delivered to the resistor?
A. zero
C. 180° D. 270°
E. none of the above
B. 90°
26

Q31.7
In an L-R-C series circuit as shown, suppose that the angular frequency of the ac source equals the resonance angular frequency.
In this case, the circuit impedance
A. is maximum.
B. is minimum, but not zero.
C. is zero.
D. is neither a maximum nor a minimum.
E. not enough information give to decide
Q31.8
In the transformer shown in the drawing, there are more turns in the secondary than in the primary. In this situation, the voltage amplitude is
A. greater in the primary than in the secondary.
B. smaller in the primary than in the secondary.
C. the same in the primary and in the secondary.
D. not enough information given to decide
27

Q31.9
In the transformer shown in the drawing, there are more turns in the secondary than in the primary. In this situation, the current amplitude is
A. greater in the primary than in the secondary.
B. smaller in the primary than in the secondary.
C. the same in the primary and in the secondary.
D. not enough information given to decide
28