Transcript Electric Safety - University of Washington

Electric Safety
Electrical Hazards
Professor Mohamed A. El-Sharkawi
Safety Facts
• More than 1000 people are killed each year in the U.S. due
to electric current, and several thousand more are injured.
• Deaths occurs at lower voltages and currents.
• Current flows inside the body can cause deep burns and
cardiac arrest.
• Individual may not let go of the power circuit due to
involuntary muscle contraction.
• Lungs, brain and heart are the most sensitive organs
affected by current.
(c) M. A. El-Sharkawi, University of Washington
Primary and Secondary Shocks
• Secondary shock: Does not produce direct
physical harm, but may cause involuntary muscle
reactions.
• Primary shock: A shock of a magnitude such that
it may produce direct physical harm. Fibrillation,
respiratory tetanus and muscle contractions are
all due to primary shocks
(c) M. A. El-Sharkawi, University of Washington
Safety Facts (ac current)
• For less than 10 mA at the skin level, the person
merely feels a "funny" sensation
• For currents above 10 mA, the person freezes to
the circuit and is unable to let go
• For currents of 100 mA to 1 A, the likelihood of
sudden death is very high
• More than 1 A, the heart experience a single
contraction, and internal heating is significant.
(c) M. A. El-Sharkawi, University of Washington
Effects of ac and dc currents
Ref: IEEE Standard 524a-1993
Current (mA)
Reaction
dc
ac
Men
Women
Men
Women
No sensation on hand
1.0
0.6
0.4
0.3
Tingling (Threshold of perception)
5.2
3.5
1.1
0.7
Shock: Uncomfortable, muscular control
not lost
9.0
6.0
1.8
1.2
Painful shock, muscular control is not lost
62.0
41.0
9.0
6.0
(c) M. A. El-Sharkawi, University of Washington
Threshold Limit
Ref: IEEE Standard 1048-1990
Current in mA
Threshold
Let-go: Worker cannot release wire
0.5% of population
Men
Women
Men
Women
9
6
16
10.5
23
15
Respiratory Tetanus: Breathing is
arrested
Ventricular Fibrillation: Weak and out of
synch heart pulses
50% of population
100
(c) M. A. El-Sharkawi, University of Washington
67
Biological effect of current
• Tingling: When a person lightly touches a
charged object and the current within the
touch perception threshold. If the person
grip to the object, the current spread out
over a wide contact area and the person
may not feel anything.
(c) M. A. El-Sharkawi, University of Washington
Biological effect of current
• Let go: The maximum current a person can
tolerate and still manage to release a
gripped conductor.
(c) M. A. El-Sharkawi, University of Washington
Let-go level
Early Days!
• “Muscular reaction at
the let-go current value
is increasingly severe
and painful, as shown by
the subject during
laboratory tests”
• IEEE Spectrum. Feb,
1972
(c) M. A. El-Sharkawi, University of Washington
Biological effect of current
• Respiratory tetanus: A current above the
let-go threshold passes through the chest
can cause involuntary contraction of the
muscles, which will arrest breathing as long
as the current continues to flow.
(c) M. A. El-Sharkawi, University of Washington
Biological effect of current
• Ventricular fibrillation: A current passes
through the chest can disturb the heart’s
own electrical stimulation and cause it to
assume an uncontrolled vibration. The
heart may cease to beat.
(c) M. A. El-Sharkawi, University of Washington
Factor affecting human Safety
•
•
•
•
•
•
Voltage level
Current flowing in person
Resistance of body
Frequency of source
Duration of shock
Pathway of current
(c) M. A. El-Sharkawi, University of Washington
1. Effect of voltage
• The higher the voltage the higher the current!
• 100-400 V ac is the most lethal voltage
– High enough to cause significant current flow in the
body
– Can cause muscles to contract tightly on the
energized equipment.
• At higher voltages, fierce involuntary muscle
contractions may throw the victim away from
the hazard.
(c) M. A. El-Sharkawi, University of Washington
2. Effect of Current
• High current causes heating damage to tissues.
• 10 A passing directly through the heart can
cause cardiac arrest. Heart muscle fibers beat
out of sync, so no blood is pumped
• The spinal cord may also be affected, altering
respiration control. 100-1000 mA is sufficient to
induce respiratory arrest and/or cardiac arrest.
• Thermal heating of tissues increases with the
square of the current (I2R).
(c) M. A. El-Sharkawi, University of Washington
3. Effect of Body Resistance
• A palm resistance can range from 100  to
1 M.
• Nerves, arteries and muscle are low in
resistance.
• Bone, fat and tendon are relatively high in
resistance.
• Across the chest of an average adult, the
resistance is about 70-100 .
(c) M. A. El-Sharkawi, University of Washington
Body Resistance
(c) M. A. El-Sharkawi, University of Washington
Body Resistance in Ohms
Ref: IEEE Standard 1048-1990
Hand-to-hand
Hand-to-feet
Resistance
Dry condition
Wet condition
Wet condition
Maximum
13,500
1,260
1,950
Minimum
1,500
610
820
Average
4,838
865
1221
(c) M. A. El-Sharkawi, University of Washington
4. Effect of Source Frequency
• 50-60 Hz current has a much greater ability to
cause ventricular fibrillation than D.C. current.
• At 50-60 Hz, involuntary muscle contractions
may be so severe that the individual cannot let
go of the power source.
• As the frequency gets above about 500 kHz, little
energy passes through the internal organs.
(c) M. A. El-Sharkawi, University of Washington
5. Effect of Duration
• The longer the duration, the more severe
the internal heating of tissues.
• Keep in mind that with 110-240 V, the
individual may become incapable of
letting go.
(c) M. A. El-Sharkawi, University of Washington
Dalziel Formula
• Charles Dalziel carried out research on the time-current
relationship for the primary shock.
K
I
t
• I: ventricular fibrillation current in mA
• t: time duration of the current in seconds
• K: is a constant that depends on the
weight of the test subject
– for people weighing less than 70 kg (154 lb),
K=116
– for people weighing more than 70 kg, K=157.
(c) M. A. El-Sharkawi, University of Washington
1600
Less than 70 kg
More than 70 kg
fibrillation current in mA
1400
1200
1000
800
600
400
200
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
time in mS
(c) M. A. El-Sharkawi, University of Washington
0.8
0.9
1
6. Effect of Pathway
• If the current passes through the brain or
heart, the likelihood of death increases
significantly.
(c) M. A. El-Sharkawi, University of Washington
Example
• A child climbs a tree to retrieve his kite that is
tangled on a power line. The voltage of the
power line is 240 V. The resistance of the wire
from the source to the location of the kite is 0.2
ohms. The tree has a resistance of 500 ohms.
The soil resistance is 300 ohms. Assume the child
resistance is 2000 ohms. If the child touches the
power line with his left hand, estimate the
current through his body. Also, estimate the time
it takes to induce ventricular fibrillation.
(c) M. A. El-Sharkawi, University of Washington
Line resistance Rl
Solution
Child resistance
Rb
Tree resistance
Rt
Source
Ground resistance
Rg
True Ground
V
240
I

 85.7 mA
Rl  Rb  Rt  Rg 0.2  2000  500  300
2
2
 K   116 
t   
 1.83 s
 I   85.7 
(c) M. A. El-Sharkawi, University of Washington
Ground Resistance
(c) M. A. El-Sharkawi, University of Washington
Definition of
ground resistance
• Ground
Resistance:
Determines
the amount
of current
flown through
an object.
(c) M. A. El-Sharkawi, University of Washington
I
Object
R
Center of
Earth
Ground Resistance of Hemisphere
I
I
J

area of hemisphere 2 r 2
I
Conductor
J ( x) 
Surface of earth
r
I
2 x
2
;
xr
Field intensity
E ( x)   J ( x);
 is ground resistivity
(c) M. A. El-Sharkawi, University of Washington
xr
Ground Resistance of Hemisphere
x b
x b
 I 1
1
Vab   E ( x) d x    J ( x) d x 
 

2  a
b
x a
x a
Vab   1
1
Rab 

 

I
2  a
b
I
Conductor
Surface of earth
r
Equipotential
surface
For a=r, b=
a
Vab
b
Va
Vb

Rg 
2 r
(c) M. A. El-Sharkawi, University of Washington
Ground Resistance
Ref: IEEE Standard 1048-1990
IEEE Standard 524a-1993
Soil Composition
Resistivity  (Ohm
meter)
Wet
Organic
Moist
Dry
Bedrock
10
100
1000
10,000
(c) M. A. El-Sharkawi, University of Washington
Example
• Compute the ground resistance of a
hemisphere with 2m diameter buried in a
wet organic soil.
• Also compute the ground resistance at 2m,
10m and 100m away from the center of the
hemisphere.
(c) M. A. El-Sharkawi, University of Washington

10
Rg 

1.6 
2 r 21
Solution
At 2m
 1
1  10 1
1
Rab1 
 
   0.8 


2  a
b  2 1
2
1.8
Resistance (Ohm)
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
50
100
Distance (meter)
150
(c) M. A. El-Sharkawi, University of Washington
Voltage Source
I
Measuring
Ground
Resistance
I
Voltmeter
V
Surface of earth
x
Measured Voltage (V)
Object under test
Current electrode
Potential probe
Vf
xf
(c) M. A. El-Sharkawi, University
of Washington
Distance
of the voltage
probe from object (x)
Ground Resistance of Objects
Object
Rod
Ground Resistance

 2l  r 
ln 

2 l  r 
l is the length of the rod
r is the radius of the rod

4r
r is the radius of the disk
Circular plate
(disk) at the
surface
Buried wire
Parameters

2 l
 l
 l 
 ln    ln   
 2d  
 r
l is the length of the wire
r is the radius of the wire
d is the depth at which the
wire is buried
(c) M. A. El-Sharkawi, University of Washington
Ground Resistance of people
Assume the foot is a circular plate
Rf 

4r
A  r

Rf 
4
Rf
Rf
Remote earth
A

2


4
For standing person
Rg 
Rf  Rf
Rf  Rf
(c) M. A. El-Sharkawi, University of Washington
0.02
 3

 0.5 R f 1.5 
Touch and Step Potential
• Step Potential: Potential between the two
feet.
• Touch Potential: Potential between a hand
and any other part of the body.
(c) M. A. El-Sharkawi, University of Washington
(c) M. A. El-Sharkawi, University of Washington
Static Wire
Ground Wire
Cross arm
High voltage
wire
Insulators
High voltage
wire
Steel Tower
Tower ground or local
(c) M. A. El-Sharkawi, University ofground
Washington
Flashover
Flashover
Energized line
(c) M. A. El-Sharkawi, University of Washington
Touch Potential
I
I
Rman
Itg
Iman
Rg
0.5 Rf
Itg
Rg
Iman
0.5 Rf
I man  I
(c) M. A. El-Sharkawi, University of Washington
Rg
Rg  Rman  0.5 R f
Example
• A power line insulator is partially failed and 10A
passes through the tower structure to the
ground. Assume that the tower ground is a
hemisphere with a radius of 0.5 meter, and the
soil surrounding the hemisphere is moist.
– Compute the voltage of the tower.
– Assume that a man with a body resistance of 3k
touches the tower while standing on the ground.
Compute the current passing through the man.
– Use Dalziel formula and compute the man’s survival
time.
(c) M. A. El-Sharkawi, University of Washington
Solution
Compute the ground resistance of the hemisphere

100
Rg 

 32 
2 r 2  0.5
V  I Rg  10  32  320 V
To compute the current through the man, first compute Rf
R f  3  300 
(c) M. A. El-Sharkawi, University of Washington
Solution
The current through the man is given in Equation (8.12)
I man  I
Rg
Rg  Rman
32
 10
 100 mA
 0.5 R f
32  3000  150
According to Dalziel formula, the man can survive for
2
 K   157 
  
t  
  2.5 s
 I man   100 
2
(c) M. A. El-Sharkawi, University of Washington
Step Potential
a
b
Iman
(c) M. A. El-Sharkawi, University of Washington
Step Potential
Rman
Rth
Vth
b
a
Rf
Rf
(c) M. A. El-Sharkawi, University of Washington
Iman
Example
• During a weather storm, an atmospheric discharge
hits a lightning pole. The pole is grounded through a
hemisphere, and the maximum lightning current
through the pole is 20,000A.
– A person is playing golf 30 meters away from the center of
the hemisphere. The distance between his feet is 0.3m,
and his leg-to-leg resistance is 2k. Assume the soil
surrounding the hemisphere is moist. Compute the
current through the person and his step potential.
– If that person is 3m away from the center of the
hemisphere, repeat the solution.
(c) M. A. El-Sharkawi, University of Washington
Solution
At 30m.
For moist soil,  = 100 m.
I  1
1  20,000* 100  1
1 
Vth 

 105 V
  


2  ra
rb 
2
30.3 
 30
R f  3   300 
Vth
105
I man 

 40.4 mA
2 R f  Rman 600  2000
Vstep  I man * Rman  40.4* 2000  80.8 V
(c) M. A. El-Sharkawi, University of Washington
Solution
At 3m
I  1
1  20,000* 100  1
1 
Vth 

 9.646 kV
  


2  ra
rb 
2
3.3 
3
Vth
9646
I man 

 3.71 A
2R f  Rman 600  2000
Vstep  I man * Rman  3.71* 2000  7.42 kV
(c) M. A. El-Sharkawi, University of Washington
Prevention and Protection from
Electric Shock
Problem: Microshock due to Capacitive
Coupling:
Internal
Electric
Circuit
V
Internal
Electric
Circuit
Equipment
chassis
Leakage
capacitance
Leakage
capacitance
(c) M. A. El-Sharkawi, University of Washington
V
Problem: Macroshock due to Faulted
Equipment
Conductive
enclosure
V
If 
Rman  Rg man  Rg
Fault
If
If
Energized
Electric
Circuit
CB
V
In=0
a
If
Example:
Rg man=100, Rg=20, Rman=1000
If 
V
120

 107 mA
Rman  Rg man  Rg 1000  100  20
Rg man
Circuit breaker will not interrupt this hazardous current
(c) M. A. El-Sharkawi, University of Washington
If
Rg
Solution: Grounding Chassis through the
Neutral Wire
Internal
Electric
Circuit
V
(c) M. A. El-Sharkawi, University of Washington
Problem: Stray Voltage due to Resistance of
Neutral Wire
I
Internal
Electric
Circuit
Rh
V
Rn
In
Rn
I man  I
Rn  Rman
Iman
(c) M. A. El-Sharkawi, University of Washington
Example
• The resistance of the neutral wire is 0.1 Ohm, and
the equipment current is 10 A. Assume the
resistance of the man plus its ground resistance is
2000 Ohm.
• Compute the current through the man assuming
that the service transformer is farther away from
the house, where Rn = 2 Ohm.
• Repeat the problem assuming the appliance is
about 3 kW.
(c) M. A. El-Sharkawi, University of Washington
Solution
Rn
0.1
I man  I
 10
 0.5 mA
Rn  Rman
0.1  2000
3 kW produce about 30A
Rn
2
I man  I
 30
 30 mA
Rn  Rman
2  2000
Circuit breaker will not interrupt this hazardous current
(c) M. A. El-Sharkawi, University of Washington
Solution: Three-prong: Separated
Neutral and Ground
I
Internal
Electric
Circuit
Ileak
V
Ileak
Ileak
In
Equipment Grounding
Conductor (EGC)
(c) M. A. El-Sharkawi, University of Washington
Problem: Macroshock due to Faulted Equipment
Conductive
enclosure
Panel circuit
breaker
If
If
Internal
Electric
Circuit
If
V
If
EGC
In=0
If
If
Example:
Rg1=Rg2=10.
If 
V
120

6 A
Rg1  Rg 2 20
VEGC  I f Rg1  6 *10  60 V
Rg2
Rg1
If
6A is very small current and within the level
of normal loads.
The smallest circuit breaker in most premises
would not open and the fault will not be
cleared.
(c) M. A. El-Sharkawi, University of Washington
earth is not an
effective path for
fault current
Solution: NEC 250.24
Conductive
enclosure
Neutral and ground
bonding
If
Internal
Electric
Circuit
Service Panel
If
If
V
Rwire
If
EGC
If2
Rg2
If
EGC
VEGC  I f 1 Rg1  I f
Rwire Rg1
Rg1
Rg1  Rg 2  Rwire
(c) M. A. El-Sharkawi, University of Washington
If1
Example
Example:
Rg1=Rg2=10.
I f V
Rg1  Rg 2  Rwire
R
g1
 Rg 2 * Rwire
 120
20.5
 246 A
10
This is large fault current and will be easily detected
and cleared by the panel circuit breaker.
(c) M. A. El-Sharkawi, University of Washington
NEC 250.24
House
Circuit Breaker
120V
House
meter
Neutral
EGC
120V
120V
Neutral
Ground rod
Conductive enclosure bonded
to ground wire
(c) M. A. El-Sharkawi, University of Washington
Appliance
Distribution Panel
(c) M. A. El-Sharkawi, University of Washington
(c) M. A. El-Sharkawi, University of Washington
3-prongs receptacles
Neutral (N)
Hot (H)
Ground (G)
(c) M. A. El-Sharkawi, University of Washington
3-prongs receptacles and Plug (Europe)
(c) M. A. El-Sharkawi, University of Washington
(c) M. A. El-Sharkawi, University of Washington
(c) M. A. El-Sharkawi, University of Washington
Ground Fault Circuit-Interrupters (GFCI)
(c) M. A. El-Sharkawi, University of Washington
Ground Fault Circuit-Interrupters
(GFCI)
• Accidents and several deaths
involving electric appliances and
water occur every year in USA.
• Changes were made in the
National Electric Code which
require the use of (GFCI).
(c) M. A. El-Sharkawi, University of Washington
Without GFCI (touch potential)
House breaker
I
Internal
Electric
Circuit
Rh
Rn
In
Water is inside
Iman
(c) M. A. El-Sharkawi, University of Washington
V
Main idea of GFCI
Ihot
core
winding
Ineutral
(c) M. A. El-Sharkawi, University of Washington
Stray Voltage
(c) M. A. El-Sharkawi, University of Washington
What is Stray Voltage
• Elevated voltage on neutral wires
• The stray voltage problems are often
associated with low voltage magnitudes
– nuisance instead of hazard.
• However, in some scenarios, the stray
voltage could be lethal.
(c) M. A. El-Sharkawi, University of Washington
Problem: Voltage Drop on Utility Neutral
Tower 1
Tower 2
Tower 3
I
Hot Wire
I
V1
V2
Tower 4
Neutral
V3
V4
V2  V1  V ; V3  V1  2V ; V4  V1  3V ;...
V4  V3  V2  V1
(c) M. A. El-Sharkawi, University of Washington
3.72
What is Stray Voltage?
• When a grounded object carries current, the
voltage of the object is non-zero.
• This voltage is called Stray voltage
• The current that produce the stray voltage is
called stray current.
(c) M. A. El-Sharkawi, University of Washington
Causes of Stray Voltage
• The neutral and ground wires are not adequately
separated
• The neutral wire is deteriorated
• The neutral is poorly grounded.
• The neutral wire carries excessive currents at the
service transformers due to unbalanced loads
• Electromagnetic field coupling with metallic
object (fence, pole, etc.)
(c) M. A. El-Sharkawi, University of Washington
Stray Voltage
I
I
Internal
Electric
Circuit
In
V
Ig
In
Stray Current
(c) M. A. El-Sharkawi, University of Washington
Ig
What is the Problem?
• Stray voltage can cause behavioral problems in farm
animals due to their nerve stimulation
• May cause sensitive hospital equipment to
malfunction.
• Swimming pools and outdoor shower shocks
• In severe cases, the stray voltage could reach lethal
levels if the neutral wire is broken
(c) M. A. El-Sharkawi, University of Washington
Stray
Current
(mA)
1-3
3-4
5-6
>6
Effect of Stray Current on
Livestock (USDA)
Effect on Livestock
Signs of awareness by livestock, but no milk
production is lost
Animal may become more difficult to manage
Short term changes in feed/water consumption or
milk production
Long term changes in feed/water consumption or
milk production
(c) M. A. El-Sharkawi, University of Washington
Case Study
10A
Internal
Electric
Circuit
Service
panel
I=10A
In
10A
Rn=1Ω
Rcow=500Ω
Ig
Rg1=20Ω
Rg2=30Ω
Ig
Rg3=20Ω
Icow
Icow
(c) M. A. El-Sharkawi, University of Washington
V
Ig+Icow
Analysis of Case Study
I g  I cow  I
Rn
1
 10
 0.203 A
Rn  Rg1  Rg 2 //( Rcow  Rg 3 )
1  20  30 //( 500  20)
I cow  0.203
Rg 2
Rg 2  Rg 3  Rcow
30
 0.203
 11 mA
30  20  500
According to the table, this level of stray current will have long term changes in
animal’s feed and water consumption as well as milk production.
(c) M. A. El-Sharkawi, University of Washington
Equipotential Grid in Farms
Equipotential grid
(c) M. A. El-Sharkawi, University of Washington
Service Transformer: Auto
Connection to Detect Faults
If
Hot
If
vs
N
y
If
Neutral (N)
If
x
(c) M. A. El-Sharkawi, University of Washington
3.8
Stray Voltage in Swimming Pools
I1
Service
Box
I2
Rw
N2
Vs
N1
Rw
I2
I5
I3
Earth surface
EGC
I4
I7
I9
I6
Rg1
Iman
Rg2
Rg3
Insulated lining
of pool
I6
Iman
(c) M. A. El-Sharkawi, University of Washington
I3  I5  I9
Elimination of Stray Voltage in
Swimming Pools
Double
bushing or
isolation xfm
Service
Box
I2
I2
I2
a
Earth
surface
I4=0
I5=0
Iman=0
Rg2
(c) M. A. El-Sharkawi, University of Washington
Rg4
EG for Swimming
Pool
3 ft
Swimming Pool
(c) M. A. El-Sharkawi, University of Washington
NEC 680.26 (Conductive Shell)
Pool’s reinforcing
steel
Light fixture
EGC
Barrier
Pool’s reinforcing
steel
EGC
Light fixture
(c) M. A. El-Sharkawi, University of Washington
NEC 680.27.A3 (Insulated Shell)
Barrier
Light fixture
EGC
EGC
Intentional bond
Ground the water? Huh?
(c) M. A. El-Sharkawi, University of Washington
Stray Voltage in Outdoor
Showers
I1
Service
Box
Rw
I2
Vs
I2
Rw
I4
Rman
I2
I3
I6
Iman
I5
Rg1
Rg2
I5
Iman
(c) M. A. El-Sharkawi, University of Washington
Iman
Rg4
Stray Voltage in Hospitals
Ground lead of
a catheter
Conductive
Chassis
Rwire
I
Internal
Circuit
Rwire
Iman
Rwire
Iman
C2
C1
Ic1
Iman
Rgman
Rg
Iman
(c) M. A. El-Sharkawi, University of Washington
Equipotential Bonding
Conductive Chassis of
medical equipment
Ground lead of
a catheter
Rwire
I
Internal
Circuit
Rwire
Rwire
Ic
Bed frame
Rg
• For the isolated or the non-isolated systems, equipotential
grounding must be established by bonding the patient to the
chassis of the equipment and the bed frame
(c) M. A. El-Sharkawi, University of Washington
Equipotential Grounding
Grounding strip of
equipotential area
Outlet
Water pipes
and faucet
Conductive
window
frame
Patient’s bed
Building
ground
(c) M. A. El-Sharkawi, University of Washington
Conductive
door frame
Severe Conditions
• Most stray voltages due to the bonding of
the ground and neutral wires at the
customers site are small to cause
hazardous conditions
• However, if the neutral wire is damaged,
lethal voltages could exist on objects
frames
– Not uncommon on lighting poles, fences and
side walk fixtures.
(c) M. A. El-Sharkawi, University of Washington
Hazard of Neutral Damage
I2
Hot
I=I1+I2
In2
Neutral
Rn2
I2
In2
I1
Rg2
Ig2
I
In1
I2
Electric
load 2
In1+In2
Rn1
I1
Electric
load 1
Ig1+Ig2
Ig1
Rg1
Rg
Ig1
Ig2
(c) M. A. El-Sharkawi, University of Washington
I2
Hot
I=I2
In2=0
Neutral
I2
Rn1
In1=Is1
I
In1=Is1
I2
Electric
load 2
Rg2
Electric
load 1
Ig2=I2
Rg1
Is2
Is1
Is1
Is2
(c) M. A. El-Sharkawi, University of Washington
Rg
Example
the load current I2=5A, Rg=20, Rg1=Rg2=30
and Rn1=1.
Assume that the neutral current between the
two loads is broken and load 1 is not energized
Compute the stray voltage at both loads.
(c) M. A. El-Sharkawi, University of Washington
Solution
I 2  I s1  I s 2
Rg
20
I s1  I 2
5
2 A
Rn1  Rg  Rg1
1  20  30
I s 2  I 2  I s1  3 A
Vstray1  I s1 * Rg1  2 * 30  60 V
Vstray2  I 2 * Rg 2  5 * 30  150 V
Notice that although the first load is de-energized, its stray
voltage is 60 V due to the broken neutral wire.
(c) M. A. El-Sharkawi, University of Washington
How to Control Stray Voltage?
• The most fundamental requirement in
controlling stray voltage is to separate the
Ground and Neutral conductors.
– The Neutral wire provides the return path from
the load back to the power source.
– The neutral is grounded only once at the
service entrance.
– The neutral wire must never be grounded at a
second place in the system.
(c) M. A. El-Sharkawi, University of Washington