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October 19, 2003

                             ET NEWS 
Issue No. 121           10-19-2003

- News
- ET Journal
- NICET Test Dates
- AFAA Class Schedule
- Comments & Acknowledgements


STUDENTS alarmed, voices sound
ASU Ram Page, TX
By Aleshia Claunch. Senior Shavon Jackson is tired of being woken
up in the middle of the night by a buzzing fire alarm—and she is
not alone. ...


BANGKOK skyscraper on fire, all safe
Reuters AlertNet, UK
... Several people complained no fire alarm went off and police
said one woman trapped in a lift on the 15th floor called her
friend on a mobile phone to ask for ...


VOLUNTEER fire department conducts fire safety session
News 24 Houston, TX
... supposed to change them. And then when the fire alarm goes
off, we teach them that they're immediately supposed to leave,"
said Faber. ...


WORRIES arise over safety of charter schools
South Florida Sun-Sentinel, FL
By Karla D. Shores. A new charter school near Boca Raton pays a
county employee to monitor its building all day because it has no
fire alarm system. ...


I'll see you in Anchorage this week

Have fun!




When someone asks me to suggest a book or program that they could
use to learn more about electricity I offer these two suggestions:

1) Indenture yourself in an electrical apprenticeship program
that includes on-the-job training AND classroom training.

2) If this isn't available you can start with Delmar's Standard
Textbook of Electricity:

There's no denying that this is a great book but it *is* rather
expensive (~$100 new), so you might want to look for a used copy
in your local bookstore or browse ebay:

UGLY's Electrical References is also handy to have:

To prepare for a NICET examination you'll have to:

1 - Understand DC circuits
2 - Use Ohm's Law
3 - Manipulate series and parallel circuits
4 - Determine the resistance of wire
5 - Perform voltage drop calculations
6 - Use a VOM (volt-ohm-milliammeter)
7 - Understand AC circuits.

1 - Understand DC circuits

Direct Current is current that always flows in the same
direction. The voltage and current can fluctuate up and down, but
the polarity of the voltage remains the same.

Ohm's law states that it takes one volt to push one amp through
one ohm. In a DC circuit the current is directly proportional to
the voltage and inversely proportional to the resistance.

2 - Use Ohm's Law

The five basic units of electricity are:
[q] Charge, expressed in Coulombs (C)
[V] Voltage, expressed in Volts (V)
[I] Current, expressed in Amperes (A)
[R] Resistance, expressed in Ohms (Greek letter Omega)
[P] Power, expressed in Watts (W)

Charge (q) -> Coulomb q=NE. Charges are the basis of electricity.
A charge exerts forces on other charges. The smallest charge is
the charge of an electron, 0.000000000000000016 Coulombs.  The
magnitude of the charge q = the number of electrons times the
charge of one electron.

Voltage (V) -> joules/Coulomb V=E/q. Voltage is an energy
measure, the energy carried by one coulomb of electrical charge.
The voltage between two points in a circuit is the amount of
energy available for pushing each coulomb of charge from one of
these points to the other.

Current (I) -> Coulomb/second. Current is the rate of flow of
electrical charge.

Resistance (R). The characteristic of materials to oppose the
flow of electricity in an electric circuit.

Power (P) = I x V. Power is the rate of working. When power is
transmitted electrically, the amount is the product of voltage
(V) and current (I).

Ohm's Law and the Power Law:

E=IxR  E=P/I  E=sqrt(PxR)

I=E/R  I=P/E  I=sqrt(P/R)

R=E/I  R=P/I^2  R=E^2/P

P=IxE  P=I^2xR  P=E^2/R

3 - Manipulate series and parallel circuits

A series circuit can be identified by the following
 o There is only one path for current
 o The voltage will drop across each resistor
 o The total resistance is the sum of the individual resistances
 o The total power is the sum of the individual powers

A parallel circuit can be identified by the following
 o Voltage is the same throughout the circuit
 o Current splits into each individual branch
 o Total current is the sum of the branch currents
 o Total resistance is smaller than the smallest branch resistance
 o Total power is the sum of the individual powers

4 - Determine the resistance of wires

The size of a wire is determined by its cross-sectional area.
The unit of measure is the mil. One mil is equal to .001 inch.
The cross-sectional area of a wire is measured in circular-mils
or cmil.

The diameter of a round wire is measured in Circular Mils (CM).
One mil equals .001 inch. A Circular Mil is the diameter of the
wire in mils squared. For example: assume a wire has a diameter
or 0.064 inch. Sixty-four thousandths should be written as a
whole number, not as a decimal or a fraction
(64^2 [64 X 64] = 4096 CM).

The resistivity of wire, expressed in ohms, is the resistance of
a one-foot length of wire that has a cross-sectional area of one

Resistivity of materials:
 Silver - 9.8
 Copper - 10.37
 Gold - 14.7
 Aluminum - 17.02

R=(Greek letter Rho)x(L/A)

R = resistance in ohms
Rho = resistivity
L = length in feet
A = area in circular mils


If I pull wire on a job site and need to know the length of wire
(d) that I just installed, I can twist the pair together at one
end, attach my trusty Simpson(r) Model 260 volt-ohm-milliammeter
to the open pair and read the resistance in ohms represented by
the pair of wires.

Determine the length of wire in this example using NFPA 70 and a

            |<----------  d = ? feet --------->|

|+----+   + +--------------------------------.        
||VOM |     |                                 `-x-<-Twisted
|+----+   - +---------------------------------'     together
Model 260 VOM

R = 4 ohm
d = wire run distance
L = length of copper wire (two times d)

Given that the wire is 10AWG 7-strand uncoated copper.

Step 1)
NFPA 70, Chapter 9, Table 8 "Conductor Properties" lists 10AWG
7-strand uncoated copper conductor with 1.24 ohm per 1000 feet
(at 167 degrees F).

Stated another way:

1.24 ohm
1000 ft 

Step 2)
Determine how long this wire run is by substituting the
information we've gathered into the formula;

R=(Greek letter Rho)x(L/A)

L = 4 ohm x 1000 ft  = 4000 ohm-ft = 3226 ft
            --------   -----------
            1.24 ohm   1.24 ohm

Step 3)
Remember that L is the total length of copper and d is the wire
run distance or L/2;

    3226 ft
L = ------- = 1613 ft

5 - Determine voltage drop calculations

What is the minimum wire size in AWG permitted by NFPA 70-1999
in the following example:

    |<------------- 2000' ------------->|

+---+                                   +-------+
| + +-----------------------------------+ +++++ |
|   |                                   | +++++ |
| - +-----------------------------------+ +++++ |
+---+                                   +-------+
NAC Power Supply                        (10) Horn
Class 2 Power Limited                   .050A ea.
20.4 VDC                                16-33 VDC

 o The average ambient temperature is 75 degrees centigrade.
 o The wire type is 7-strand, uncoated copper.
 o The horn nameplate reads "operating range 16-33 VDC".
 o The horn nameplate reads "current requirements .050 A".
 o There are ten (10) horns attached to the end of the circuit.
 o The distance from the NAC power supply to the horns is 2000'.

Step 1)
The minimum wire size is dependent upon the maximum permitted
voltage drop.

The manufacturer specifies that the horn will operate properly
at a minimum of 16 VDC and at a maximum of 33 VDC. Comparing the
NAC power supply voltage (20.4 VDC) with the minimum voltage at
which the horn will operate (16 VDC), we can determine the
maximum voltage-drop permitted.

20.4 VDC - 16 VDC = 4.4 VDC = Ed (E sub d, the maximum volt-drop

The manufacturer specifies a current requirement of .050 A per
horn times 10 horns.

I = .050 A x 10 = .5 A

Step 2)
Use Ohm's Law and solve for R, which is the maximum resistance
allowed under the maximum voltage drop allowed, which is 4.4 V:

R = Ed/I
R = (20.4 VDC - 16 VDC) / .5 A
R = 4.4 VDC / .5 A
R = 8.8 OHM

Otherwise stated, the maximum amount of resistance permitted by
the circuit conductors is 8.8 OHM.

The example is a 2000' long 2-conductor circuit or a 4000'
1-conductor or;

The maximum amount of resistance permitted by the 4000'
round-trip circuit is 8.8 OHM or;

8.8 OHM / 4000' = 2.2 OHM / 1000'

Step 3)
NFPA 70 Chapter 9, Table 8 "Conductor Properties" is where you
will find that the smallest conductor size permitted, which will
not exceed 2.2 OHM per 1000' of 7-strand, uncoated copper is 12
AWG, which has 1.98 OHM per 1000'.

6 - Use a VOM (volt-ohm-milliammeter)

Learn how to use and read an analog VOM. I suggest the Simpson(r)
Model 260. No, it's not an antique and yes you can still buy them.

7 - Understand AC circuits

 o Alternating Current is current that changes polarity
 periodically due to voltage changing polarity. Probably the
 biggest advantage of AC is the fact that AC current can be
 transformed and DC current cannot. A transformer permits voltage
 to be stepped up or down.

 o Sine wave values of AC include peak-to-peak, peak, RMS, and
    PEAK-TO-PEAK is measured from the maximum value in the
		positive direction to the maximum value in the negative
 o The PEAK value is measured from zero to the highest value
 obtained in either the positive or negative direction. The peak
 value is one-half of the peak-to-peak value.
 o RMS stands for root-mean-square, which is an abbreviation for
 the square root of the mean of the square of the instantaneous
 currents. The RMS value can be found by dividing the peak value
 by the square root of 2 (1.414) or by multiplying the peak value
 by 0.707 (the reciprocal of 1.414).
   RMS = peak x 0.707
   Peak = RMS x 1.414

 o AVERAGE values of voltage and current are actually direct
 current values. The average value must be found when a sine wave
 AC voltage is changed into DC with a rectifier. The average
 value of voltage will produce the same amount of power as a
 nonpulsating source of voltage such as a batter. For a sine wave,
 the average value of voltage is found by multiplying the peak
 value by 0.637 or by multiplying the RMS value by 0.9.
   Average = peak x 0.637
   Average = RMS x 0.9



OR1 PCC Sylvania, Portland;
Test 11/15/03. Postmark deadline 9/27/03.
Test ??/??/04. Postmark deadline 12/1/03.

OR2 Clackamas Community College, Oregon City;
Test 11/15/03. Postmark deadline 9/27/03.
Test ??/??/04. Postmark deadline 12/1/03.

"It is anticipated that the 2004 schedule will be very similar
to the 2003 schedule. The dates for the first test cycle of 2004
will be January 24, February 22 and March 22. The first
application postmark deadline will be December 1, 2003. This
postmark deadline must be used by all examinees testing in the
first test cycle of 2004 until such time as the schedule for your
preferred sessions is established."

For a complete list of all test centers and test dates, visit:


October 21-23, 2003 Anchorage, AK
Intermediate Fire Alarm Seminar

November 3-6, 2003 Anaheim, CA - Sponsored by CAFAA
*NEW* -> Plan Review Seminar <- *NEW*
Advanced Fire Alarm Seminar

November 4-7, 2003 Boston, MA - Sponsored by New England AFAA
*NEW* -> Plan Review Seminar <- *NEW*
*NEW* -> Understanding Code Requirements for Fire Alarm Systems 
More information will be available soon.

November 21, 2003 Reno, NV
*NEW* -> Plan Review Seminar <- *NEW*

December 2-4, 2003 Phoenix, AZ - Co-sponsored by AZ AFAA
*NEW* -> Plan Review Seminar <- *NEW* 
*NEW* -> Advanced Fire Alarm Seminar (NICET III/IV)  <- *NEW*

December 9-11, 2003 San Antonio, TX - Co-sponsored by TX AFAA
Advanced Fire Alarm Seminar


Engineering Technician info

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Code(r), and NFPA 101(r) Life Safety Code(r), Copyright(c) NFPA,
Quincy, MA 02269. This reprinted material is not the complete and
official position of the NFPA on the referenced subject, which is
represented only by the standard in its entirety. 
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This reprinted material is not the complete and official position of the NFPA on the referenced subject, which is represented only by the standard in its entirety.