SINGLE-PHASE THREE-WIRE CIRCUITS
Power circuit normally
used for residencial
supply
Line-to-line used
to supply major
appliances (AC, dryer).
Line-to-neutral for lights
and small appliances
General
balanced
case
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
I aA
31A
1A
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
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
L
2.5
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.
Safety switch in case control
Vary power to thermostat
fails.
heater
If temperature from dryer is higher
than the one from thermostat then
open switch and allow heater to
cool off
Analysis of single phase 3-wire circuit installations
LEARNING BY DESIGN
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
 536.9kVA  4  j 3kVA
Rlines  0.05


Plosses  0.05  | I a |2  | I b |2  | I n |2  1.147 kW
Option 1
Ia  41.670  41.67  36.9
 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
Rlines  0.05


Plosses  0.05  | I a |2  | I b |2  0.625kW
Psaved  0.522kW
Ib  Ia
$ / year  366(@ 0.08$ / kWh)
Option 2
Steady-state
Power Analysis