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STEADY-STATE POWER ANALYSIS
LEARNING GOALS
Instantaneous Power
For the special case of steady state sinusoidal signals
Average Power
Power absorbed or supplied during one cycle
Maximum Average Power Transfer
When the circuit is in sinusoidal steady state
Effective or RMS Values
For the case of sinusoidal signals
Power Factor
A measure of the angle between current and voltage phasors
Complex
Power
Power Factor
Correction
Measure
of
power
usingtransfer
phasorsto a load by “aligning” phasors
How to improve
power
Single Phase Three-Wire Circuits
Typical distribution method for households and small loads
INSTANTANEOUS POWER
Instantaneous
Power Supplied
to Impedance
p( t )  v ( t ) i ( t )
In steady State
v (t )  VM cos( t   v )
Assume : v (t )  4 cos( t  60),
Z  230
Find : i (t ), p(t )
V 460
I 
 230( A)
Z 230
i (t )  2 cos( t  30)( A)
VM  4, v  60
I M  2, i  30
p(t )  4 cos 30  4 cos(2 t  90)
i (t )  I M cos( t   i )
p(t )  VM I M cos( t   v ) cos( t   i )
1
cos1 cos2  cos(1  2 )  cos(1  2 )
2
V I
p(t )  M M cos( v   i )  cos(2 t   v   i )
2
constant
LEARNING EXAMPLE
Twice the
frequency
AVERAGE POWER
LEARNING EXAMPLE
For sinusoidal (and other periodic signals)
we compute averages over one period
1
P
T
t 0 T
 p(t )dt
T

VR
2
t0
Find the average
power absorbed
by impedance


VM I M
cos( v   i )  cos(2 t   v   i )
2
V I
P  M M cos( v   i ) It does not matter
who leads
2
p( t ) 
If voltage and current are in phase
1
2
 v   i  P  VM I M Purely
resistive
If voltage and current are in quadrature
v   i  90  P  0 Purely
inductive or
capacitive
I
1060 1060

 3.5315( A)
2  j 2 2 245
VM  10, I M  3.53, v  60, i  15
P  35.3 cos(45)  12.5W
Since inductor does not absorb power
one can use voltages and currents across
the resistive part
VR 
2
1060  7.0615(V )
2  j2
1
P  7.06  3.53W
2
LEARNING EXAMPLE
Determine the average power absorbed by each resistor,
the total average power absorbed and the average power
supplied by the source
Inductors and capacitors do not absorb
power in the average
Ptotal  18  28.7W
Psupplied  Pabsorbed  Psupplied  46.7W
Verification
If voltage and current are in phase
I  I1  I 2  345  5.3671.57
I  8.1562.10( A)
2
1 2
1
1
V
M
 v   i  P  VM I M  RI1M 
2
2
2 R P  VM I M cos(   )
v
i
1245
2
I1 
 345( A)
1
4
Psupplied  12  8.15  cos(45  62.10)
1
2
P4  12  3  18W
2
1245
1245
I2 

 5.3671.57( A)
2  j1
5  26.37
1
P2   2  5.362 (W )  28.7W
2
LEARNING EXTENSION
I
I
Find average power absorbed by each resistor
I2 
1260
Zi
I 2  1.9041.6
I2
Z i  2  (4 ||  j 4)
4( j 4) 8  j8  j16 25.3  71.6


4  j4
4  j4
4 2  45
Zi  4.47  26.6
Zi  2 
1260
 2.6886.6( A)
4.47  26.6
1 2 1
P2  RI M
  2  2.682  7.20W
2
2
I
 j4
4  90
I
 2.6886.6
4  j4
4 2  45
1
P4   4  1.902 (W )
2
LEARNING EXTENSION
I1 
Vs
Find the AVERAGE power absorbed by each PASSIVE
component and the total power supplied by the source
I2
1
P4   4  4.122 (W )
2
Pj 2  0(W )

I1 
4  j2
1030
3  4  j2
4.4726.57
I1 
1030  6.1440.62( A)
7.2815.95
1 2 1
P3  RI M
  3  6.142 (W )
2
2
I 2  1030  6.1440.62
3
3030
1030 
3  4  j2
7.2815.95
 4.1214.05( A)
I2 
Power supplied by source
Method 1. Psupplied  Pabsorbed
Psupplied  P3  P4  90.50W
Method 2: P  1 V I cos(   )
M M
v
i
2
Vs  3I1  18.4240.62
1
P   18.42  10  cos(40.62  30)
2
LEARNING EXAMPLE
Determine average power absorbed or supplied by each
element
1230  60 10.39  j 6  6

 6  j 4.39
j1
j
 7.43  36.19( A)
1
P60   6  7.43 cos(0  36.19)  18W
2
I3 
Passive sign convention
I2 
1230
 630( A)
2
I1  I 2  I 3  5.20  j 3  6  j 4.39  11.2  j1.39( A)
 11.28  7.07
1 2 1
P2  RI M
  2  62  36(W )
2
2
Pj1  0
To determine power absorbed/supplied
by sources we need the currents I1, I2
Average Power
For resistors
2
1
1
1
V
2
M
P  VM I M cos( v   i ) P  RI M 
2
2
2 R
1
P1230   12  11.28  cos(30  7.07)
2
 54(W )  36  18
LEARNING EXTENSION
Determine average power absorbed/supplied by each
element
V4
I2
I1
Loop Equations
I1  120
430   j 2 I 2  4( I 2  120)
I2 
44.58177.43
 9.97204( A)
4.47  26.57
 4( I1  I 2 )  4(12  9.97204)(V )
I2 
V4
430  480 3.46  j 2  48

4  j2
4.47  26.57
1
P120    19.92  12  cos(54.5  0)  69.4(W )
2
1
P430    4  (9.97) cos(30  204)  19.8(W )
2
Check: Power supplied =power absorbed
Alternative Procedure
Node Equations
V
V  430
 120  4  4
0
4
 j2
430  V4
I2 
2j
 4(12  9.108  j 4.055)(V )  19.92  54.5(V )
For resistors
Average Power
1 VM2 1 19.922
P4 
 
 49.6W
2
1
2 R 2
4
1
1
V
2
M
P  VM I M cos( v   i ) P  RI M 
2
P2 j  0(W )
2
2 R
Determine average power absorbed/supplied by each
element
LEARNING EXTENSION
V
I1 
240  V 24  14.77  j1.85

 4.62  j 0.925
2
2
I1  4.71  11.32( A)
I1
I2
I2 
I2 
120  V 12  14.77  j1.85  j


j2
j2
j
 1.85  j 2.77
 0.925  j1.385( A)  1.67123.73( A)
2
1
P2   2  4.712  22.18(W ) For resistors
2
1 2 1 VM2
1 14.882
P4  
 27.67(W ) P  RI M 
2
2 R
2
4
1
2 j (V  24)  2(V  12)  jV  0
P120    12  1.67 cos(0  123.73)  5.565(W )
2
(2  3 j )V  24  j 48
1
P240    24  4.71 cos(0  11.32)  55.42(W )
24  j 48 2  j 3 192  j 24
2
V


Check :
2  j3 2  j3
13 Average Power
Pabsorbed  22.18  27.67  5.565(W )
 14.887.125(V )
1
P  VM I M cos( v   i )
Psupplied  55.42(W )
 14.77  j1.85(V )
2
Node Equation
V  240 V  120 V

  0  j4
2
j2
4
MAXIMUM AVERAGE POWER TRANSFER
ZTH  RTH  jXTH
1 | Z L || VOC |2
PL 
2 | Z L  ZTH |2
RL
RL2  X L2
Z L  ZTH  ( RL  RTH )  j ( X L  X TH )
| Z L  ZTH |2  ( RL  RTH ) 2  ( X L  X TH ) 2
Z L  RL  jX L
1
PL  VLM I LM cos(VL   I L )
2
1
 | VL || I L | cos(VL   I L )
2
ZL
ZL
VL 
VOC | VL |
| VOC |
Z L  ZTH
Z L  ZTH
VL  I L  VL  Z L
|V |
| I L | L
Z L  V   I  Z L
| ZL |
L
L
Z L  RL  jX L  tan(Z L )  X L
RL
IL 
cos 
1
1  tan 2 
 cos(VL   I L ) 
RL
RL2  X L2
1
| VOC |2 RL
PL 
2 ( RL  RTH )2  ( X L  X TH ) 2
PL

 0
X L
  X L   X TH
 
PL
 0   RL  RTH

RL
*
 Z Lopt  ZTH
PLmax
1  | VOC |2 

 
2  4 RTH 
LEARNING EXAMPLE Find Z L for maximum average power transfer.
Compute the maximum average power supplied to the load
 Z Lopt

*
ZTH
PLmax
1  | VOC |2 

 
2  4 RTH 
Remove the load and determine the Thevenin equivalent of remaining circuit
8  j 4 (8  j 4)(6  j1) 52  j16



6  j1
37
37
8  j 4 8.9426.57


 1.4716.93
6  j1 6.089.64
ZTH  4 || (2  j1) 
Z L*  1.47  16.93  1.41  j 0.43
I1
VOC  4 
2
320
40 
 5.26  9.64
6  j1
6.089.64
PLmax
We are asked for the value of the
power. We need the Thevenin voltage
1 5.262
 
 2.45(W )
2 4  1.41
LEARNING EXAMPLE Find Z L for maximum average power transfer.
Compute the maximum average power supplied to the load
 Z Lopt

PLmax
*
ZTH
1  | VOC |2 

 
2  4 RTH 
Circuit with dependent sources!
ZTH 
VOC
I SC
40  Vx'  (2  j 4) I1
KVL
VX'  2 I1
40  (4  j 4) I1  (4 245) I1
I1 
KVL
40
 0.707  45( A)
4 245
VOC  2 I1  40  1  j1  4  3  j1  10  161.5
Next: the short circuit current ...
LEARNING EXAMPLE (continued)...
Original circuit
 Z Lopt

*
ZTH
LOOP EQUATIONS FOR SHORT
PLmax
1  | VOC |2 

 
2  4 RTH 
CIRCUIT CURRENT
 V x"  j 4 I  2( I  I SC )  4  0
40  2( I SC  I )  j 2 I SC  0
CONTROLLING VARIABLE
Vx"  2( I SC  I )
Substitute and rearrange
(4  j 4) I  4 I SC  4
 2 I  (2  j 2) I SC  4  I  (1  j1) I SC  2
4(1  j )(1  j ) I SC  2  4 I SC  4
I SC  1  j 2( A)  5  116.57
VOC  2 I1  40  1  j1  4  3  j1  10  161.57
ZTH  2  45  1  j1  Z Lopt  1  j1
PLmax
1 ( 10 )2
 
 1.25(W )
2
4
LEARNING EXTENSION
Find Z L for maximum average power transfer.
Compute the maximum average power supplied to the load
1  | VOC |2 
max
opt
*

PL  
 Z L  ZTH
2  4 RTH 
ZTH   j 2  (2 || j 2)   j 2 
ZTH 

VOC
I

VOC  120  j 2 I
 12  j 2  9(1  j )
 6  j18
VOC  18.97471.57(V )
| VOC |2  62  182  360
360  (2  j 2) I
36(2  j 2)
I
 9(1  j )  12.73  45
8
4j

2  j2
4
8  j8

 1  j ()
2  j2
8
Z Lopt  1  j ()
1 360
PLmax  
 45(W )
2 4
LEARNING EXTENSION
Find Z L for maximum average power transfer.
Compute the maximum average power supplied to the load
1  | VOC |2 
max
opt
*

PL  
 Z L  ZTH
2  4 RTH 
ZTH
ZTH  j 2 || (2  j 2) 
V j2

VOC

Vj2 
j2
240  2490
j2  j2  2
VOC  120  2490  12  j 24(V )
| VOC |2  122  242  720
KVL
j 2(2  j 2)
2  j2  j2
ZTH  2  j 2()
Z Lopt  2  j 2()
1 720
PLmax  
 45(W )
2 4 2
EFFECTIVE OR RMS VALUES
Instantaneous power
i (t )
p( t )  i 2 ( t ) R
If the current is sinusoidal the average
power is known to be P  1 I 2 R
av
The effective value is the equivalent DC
value that supplies the same average power
If current is periodic with period T
 1 t0 T 2


 p(t )dt  R T  i (t )dt 
t0
 t0

I eff : Pav  Pdc
2
Pdc  RI dc
1
T
For a sinusoidal signal
x (t )  X M cos( t   )
the effective value is
X
X eff  M
2
t 0 T
If current is DC (i (t )  I dc ) then
2
I eff

M
1 2
2
 I eff
 IM
2
R
1
Pav 
T
2
t 0 T
2
i
 (t )dt
t0
I eff 
1
T
t 0 T
i
1
For sinusoidal case Pav  VM I M cos( v   i )
2
Pav  Veff Ieff cos( v   i )
effective  rms (root mean square)
2
(t )dt
t0
Definition is valid for ANY periodic
signal with period T
LEARNING EXAMPLE
Compute the rms value of the voltage waveform
T 3
X rms
One period
v
1
2
3
(t )dt   (4t ) dt   (4(t  2))2 dt
0
2
0
2
1
3
16 3  32
v
(
t
)
dt

2


 3 t   3
0
0
2
Vrms 
t 0 T
x
2
(t )dt
t0
The two integrals have the
same value
 4t 0  t  1

v (t )  
0 1 t  2
  4( t  2) 2  t  3

T
1

T
1 32
  1.89(V )
3 3
LEARNING EXAMPLE
Compute the rms value of the voltage waveform and use it to
determine the average power supplied to the resistor
T  4( s)
i (t )
R  2
R
i (t )  16; 0  t  4
2
X rms
I rms  4( A)
2
Pav  RI rms
 32(W )
1

T
t 0 T
x
t0
2
(t )dt
LEARNING EXTENSION
Compute rms value of the voltage waveform
T 4
X rms
v  2t
Vrms
12
2

(
2
t
)
dt

40
2
1  8
  t 3   (V )
3 0 3
1

T
t 0 T
x
t0
2
(t )dt
LEARNING EXTENSION
Compute the rms value for the current waveforms and use
them to determine average power supplied to the resistor
i (t )
R  4
X rms
R
T 6
2
I rms
4
6

1 2
   4dt  16dt   4dt   8  32  8  8
6 0
6

2
4
T 8
2
I rms
6

1 2
  16dt  16dt   8
8 0

4
P  32(W )
P  8  4  32(W )
1

T
t 0 T
2
x
 (t )dt
t0
2
Pav  I rms
R
THE POWER FACTOR
I M  i

Z L V 
M
v
V

v
V  ZI  V  Z  I
v   z  i
1
P  VM I M cos( v   i )  Vrms I rms cos( v   i )
2
pf 
P
Papparent
 cos( v   i )  cos z
z
I
i
Papparent  Vrms I rms
P  Vrms  I rms  pf
pf
z
0
 90
pure capacitive
0  pf  1  90   z  0 leading or capacitive
1
0
resistive
0  pf  1 0   z  90 lagging or inductive
0
90
pure inductive
 90   z  0
current leads
(capacitiv e)
0   z  90
current lags
(inductive )
V
LEARNING EXAMPLE
Find the power supplied by the power company.
Determine how it changes if the power factor is changed to 0.9
P  Vrms  I rms  pf
 cos z  0.707   z  45
Current lags the voltage
Power company
88  103 (W )
I rms 
 259.3( A) rms
480  0.707
2
Plosses  I rms
R  259.33  0.08  5.378kW
PS  Plosses  88,000(W )  93.378(kW )
If pf=0.9
I rms 
Plosses
88,000
 203.7( A)rms
480  0.9
2
 I rms
R  3.32kW
Losses can be reduced by 2kW!
Examine also the generated voltage
480(V )rms
VSrms
259.3  45( A)rms
 0.08 I rms  VL
 0.08  259.3  45  480
VSrms  0.08  (183.4  j183.4)  480
 494.7  j14.7  495  1.7(V )
If pf=0.9
I rms  203.7  25.8
VS  14.47  j 7.09  480  494  0.82
LEARNING EXTENSION
PL  100kW ,VL  480(V )rms , pf  0.707
Rline  0.1
P  Vrms  I rms  pf
Determine the power savings if the power factor can be increased to 0.94
I rms 
P
Vrms  pf
P 2 Rline
1
Plosses 


2
Vrms
pf 2
1010  0.1
1
Plosses ( pf  0.707) 

(W )
2
2
480
0.707
 2  4.34kW
2
I rms
Rline
1010  0.1
1
Plosses ( pf  0.94) 

(W )
4802
0.942
 1.13  4.34kW
Psaved  0.87  4.34kW  3.77kW
Definition of Complex Power
*
S  Vrms I rms
COMPLEX POWER
The units of apparent
and reactive power are
Volt-Ampere
inductive
| S | Vrms I ms
S  Vrms  v   I rms  i *
| S | P  pf
S  Vrms I rms  v   i
S  Vrms I rms cos( v   i )  jVrms I rms sin( v   i )
P
Active Power
Q
Reactive Power
capacitive
Another useful form
*
Vrms  ZI rms  S  ( ZI rms ) I rms
 Z | I rms |2
 P  R | I rms |2
Z  R  jX  
2
Q  X | I rms |
ANALYSIS OF BASIC COMPONENTS
RESISTORS

Q0
INDUCTORS
CAPACITORS
Supplies reactive power!!
WARNING
 IF
X  0
LEARNING EXAMPLE
Given :
PL  20kW , pf  0.8 lagging,VL  2200rms , Z L  0.09  j 0.3, f  60 Hz
Determine the voltage and power factor at the input to the line
inductive
inductive
P  Re{S} | S | cos( v   i ) | S |  pf
capacitive
Q 2  S L 2  P 2  Q  15kVA S L  20  j15(kVA)  2536.87
VS  248.63  j 21.14  249.534.86
S L  VL I L*
*
 S   25,00036.87 
 IL   L   
 113.64  36.86( A)

2200

 VL  
*
*
 20,000  j15,000 
IL  
 90.91  j 68.18( A)

220


VS  (0.09  j 0.3) I L  2200
VS  (0.09  j 0.3)(90.91  j 68.18)  220(V )
4.86
IL
VS
 36.86
pfsource  cos(41.72)  0.746
lagging
LEARNING EXAMPLE
Compute the average power flow between networks
Determine which is the source
j1
VA  12030(V )rms
I
VB  1200(V )rms
VA  VB 12030  1200

Z
j1
(103.92  j 60)  120
 60  j16.08( A)rms
j
I  62.1215( A)rms
I
S A  VA ( I )*  12030  62.12  195  7,454  165VArms
PA  7,454 cos(165)  7,200(W )
S B  VB ( I )*  1200  62.12  15  7,454  15VArms
PB  7,454 cos(15)  7,200(W )
A supplies 7.2kW average power to B
Passive sign convention.
Power received by A
LEARNING EXTENSION
0.1
Determine real and reactive power losses and
real and reactive power supplied
j 0.25
inductive
40kW
pf  0.84 lagging
P  Re{S} | S | cos( v   i ) | S |  pf
P 40
| S L |

 47.62kVA | QL | | S L |2  P 2  25,839(VA)
pf .84
|S |
S  VI * | I L | L  216.45( A) rms
| VL |
I L  216.45  32.86( A)rms
pf  cos( v   i )   v   i  32.86
Slosses  ( Z line I L ) I L*  Z line | I L |2
Slosses  (0.1  j 0.25)(216.45)2  4,685  j11,713VA
Balance of power
Ssupplied  Slosses  Sload
 4.685  j11.713  40  j 25.839  44.685  j37.552kVA
capacitive
LEARNING EXTENSION
0.12
Determine line voltage and power factor at the supply end
j 0.18
60kW
pf  0.85 lagging
P  Re{S} | S | cos( v   i ) | VL |  | I L |  pf
P
S L  VL I L*
| I L |
 320.86( A) rms
| VL |  pf
 v   i  cos1 ( pf )   v   i  31.79
I L  320.86  31.79( A)rms  272.72  j169.03( A)rms
VS  Zline I L  VL  (0.12  j 0.18)(272.72  j169.03)  220
VS  283.15  j 28.81(V )rms  284.615.81(V )rms
The phasor diagram helps in visualizing
the relationship between voltage and current
IL
 31.79
5.81
VL
VS
pfsource  cos(37.6)  0.792
lagging
POWER FACTOR CORRECTION
Low power factors increase
losses and are penalized by
energy companies
Typical industrial loads
are inductive
VL 
Without capacitor :
Sold  Pold  jQold | Sold |  old
pf old  cos( old )
With capacitor
Snew  S old  Scapacitor
1
I capacitor
jC
Qcapacitor | VL || I capacitor |
| VL |2 C
 Pold  jQold  jQcapacitor
| Snew |  new
pfnew  cos( new )
tan new 
Qold  Qcapacitor
cos 
Simple approach to power factor correction
Pold
1
1  tan 2 
LEARNING EXAMPLE
Economic Impact of Power Factor Correction
Operating Conditions
Current Monthly Utility Bill
Correcting to pf=0.9
New demand charge
Additional energy charge due to capacitor bank is negligible
Monthly savings are approx $1853 per month! A reasonable capacitor bank should
Itself in about a year
f  60Hz.
Determine the capacitor required to increase the power factor
to 0.95 lagging
LEARNING EXAMPLE
Roto-molding
process
50kW ,VL  2200rms
pf  0.8 lagging
P  Re{S} | S | cos( v   i ) | S |  pf
P 50
| Sold |

 62.5kVA | Qold | | Sold |2  P 2  37.5(kVA).
pf .80
cos new  0.95  tan new
2
Q
1  pf new

 0.329  new  Qnew  0.329  P  16.43kVA
P
pf new
Qcapacitor  Qold  Qnew  37.5  16.43  21.07kVA
Qcapacitor | VL || I capacitor |
| VL |2 C
21.07 103
C 

 0.001156( F )  1156 F
 | VL |2 (220)2  (2  60)
Qcapacitor
LEARNING EXTENSION
Determine the capacitor necessary to increase the
power factor to 0.94
PL  100kW ,VL  480(V )rms , pf  0.707
Rline  0.1, f  60 Hz
P  Re{S} | S | cos( v   i ) | S |  pf
P 100
| Sold |

 141.44kVA | Qold | | Sold |2  P 2  100.02(kVA).
pf .707
cos new  0.94  tan new
2
Q
1  pf new

 0.363  new  Qnew  0.363  P  36.3kVA
P
pf new
Qcapacitor  Qold  Qnew  100.02  36.3  63.72kVA
Qcapacitor | VL || I capacitor |
| VL |2 C
63.72 103
C 

 0.000733( F )  733 F
2
2
 | VL |
(480)  (2  60)
Qcapacitor
SINGLE-PHASE THREE-WIRE CIRCUITS
Power circuit normally
used for residencial
supply
General
balanced
case
Line-to-line used
to supply major
appliances (AC, dryer).
Line-to-neutral for lights
and small appliances
An exercise in
symmetry
General case by source
superposition
Basic circuit.
Neutral
current
is zero
Neutral current is zero
LEARNING EXAMPLE
Assume all
resistive
Determine energy use over a 24-hour period and the cost
if the rate is $0.08/kWh
 1Arms
Lights on
 30 Arms
P  Vrms I rms
 0.2 Arms
Stereo on
Energy   p(t )dt  Paverage  Time
Elights  0.12kW  8Hr  0.12kW  7 Hr  1.8kWh
Erange  7.2kW  (2  1  1) Hr  28.8kWh
Estereo  0.024kW  (5  3) Hr  0.192kWh
Edaily  30.792kWh
Cost  $2.46 / day
I aA
31A
1A
KCL
I aA  I L  I R
Outline of
verification
I bB   I S  I R
I nN  I S  I L
Esupplied   psupplied  Vrms  I rms dt
SAFETY CONSIDERATIONS
Average effect of 60Hz current from hand to
hand and passing the heart
Required voltage depends on contact, person
and other factors
Typical residential circuit with ground and
neutral
Ground conductor is not needed for
normal operation
LEARNING EXAMPLE
Increased safety due to grounding
When switched on the tool case is energized
without the ground connector the
user can be exposed to the full
supply voltage!
Conducting due to wet floor
If case is grounded then the supply is shorted and the fuse acts to open
the circuit
More detailed numbers in a related case study
LEARNING EXAMPLE
Wet
skin
150
400
150
Limbs
trunk
Ground prong removed
R(dry skin)
R(wet skin)
R(limb)
R(trunk)
15kOhm
150Ohm
100Ohm
200Ohm
Suggested resistances
for human body
1
I body 
120
 171mA
701
Can cause ventricular fibrillation
LEARNING EXAMPLE
Ground Fault Interrupter (GFI)
In normal operating mode the two currents induce canceling magnetic fluxes
No voltage is induced in the sensing coil
If i1 and i2 become different (e.g., due to a fault)
then there is a voltage induced in the sensing coil
A ground fault scenario
LEARNING EXAMPLE
While boy is alone in the pool
there is no ground connection
x
Ground
fault
Vinyl lining (insulator)
Circuit formed when boy in
water touches boy holding
grounded rail
LEARNING EXAMPLE
150mA
Accidental grounding
Only return path in normal
operation
New path created by the
grounding
Using suggested values of resistance the
secondary path causes a dangerous current
to flow through the body
LEARNING EXAMPLE
A grounding accident
After the boom touches the live line the operator jumps down and starts walking
towards the pole
7200 V
Ground is not a perfect
conductor
10m
720V/m
One step applies 720 Volts to the
operator
LEARNING EXAMPLE
A 7200V power line falls on the car and makes contact with it
7200V
Car body is good conductor
Tires are
insulators
Wet Road
Option 1.
Driver opens door and steps down
7200
Ibody 
Rdry skin  2 Rlimb  Rtrunk
R(dry skin)
R(wet skin)
R(limb)
R(trunk)
Option 2:
Driver stays inside the car
15kOhm
150Ohm
100Ohm
200Ohm
I  460mA Very dangerous! Suggested resistances
for human body
Ibody  0
LEARNING EXAMPLE
Find the maximum cord length
Rcord  13.1A  5V
Rcord  0.382
Minimum voltage for proper
operation
CASE 1: 16-gauge wire
L
Rcord
 95.5 ft
4 m ft
4
m
ft
CASE 2: 14-gauge wire 2.5
L
m
ft
Rcord
 152.8 ft
m

2.5 ft
Working with RMS values the problem is formally the same as a DC problem
LEARNING EXAMPLE
Light dimming when AC starts
 40 A
I light  0.5 Arms
Typical single-phase 3-wire installation
V AN 
AC off
240
 120  119.5Vrms
241
VAN  100Vrms
Circuit at start of AC unit. Current
demand is very high
VAN  115Vrms
AC in normal operation
LEARNING EXAMPLE
DRYER HEATING AND TEMPERATURE CONTROL
Temperature is controlled by disconnecting the heating
element and letting it cool off.
Vary power to thermostat
heater
Safety switch in case control
fails.
If temperature from dryer is high
than the one from thermostat th
open switch and allow heater to
cool off
LEARNING BY DESIGN
Analysis of single phase 3-wire circuit installations
Ia  41.670  41.670  41.67  36.9  119.4  12.1( A)
I m  41.67  36.9
S A  1200  119.4  12.1*
 14.3812.1kVA  14  j 3kVA
S B  1200  41.67  36.9 *
I n  2 41.67
Ib  I m
5000
| I L |
| I H |
120
Option 1
Ia  41.670  41.67  36.9
 536.9kVA  4  j 3kVA
Rlines  0.05


Plosses  0.05  | I a |2  | I b |2  | I n |2  1.147kW
 79.07  18.4( A)
S A  S B  1200  79.0718.4*
 9.518.4kVA  9  j 3kVA
In  0
IL  IH
Ib  Ia
Option 2
Rlines  0.05


Plosses  0.05  | I a |2  | I b |2  0.625kW
Psaved  0.522kW
$ / year  366(@ 0.08$ / kWh)
Steady-state
Power Analysis