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July 11, 2004

                               ET NEWS                                
Issue No. 159   7-11-2004 

- News
- ET News Journal
- Job Opportunities
- Fire Alarm Class Schedule
- Comments & Acknowledgements


DAY care building code amended
Arizona Republic - Phoenix,AZ,USA
Offsite fire alarm monitoring systems will be required for providers who take in
more than five children per session, according to a building code amendment ...

UL Launches Online Search Engine for Finding UL Listed Alarm ...
TMCnet - USA
... launch of UL Alarm Finder, an online search engine enabling commercial
business owners and residential dwellers to easily find the UL-Listed fire and
security ...


PERMIT Needed to Paint Your House
WPVI - Philadelphia,PA,USA
... to think of things like the difference between what is UL approved and UL
listed really ... O'Clocks are over the regulations say they'll be at the Fire
Hall on ...


LOOP fire panel wants sprinkler law
Chicago Sun Times - Chicago,IL,USA
... Chairman Bernard Stone (50th) put his foot down in opposition to the mayor's
plan to allow 'bare wires' to be used to install fire alarm notification and ...


I'll see you in Chicago 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 three 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:

3) If you have a working knowledge of basic electricity but need a little review,
I suggest that you buy UGLY's Electrical References. It's only about $10 - $15
plus it contains additional material useful for a NICET exam.

To prepare for a NICET examination in Basic Electricity 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 Coulomb.  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 = I x R  E = P/I      E = sqrt(P x R)

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

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

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

3 - Manipulate series and parallel circuits

A series circuit can be identified by the following characteristics:
 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 characteristics:
 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 mil.

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 distance from the source to the
load (d), 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.

Example: Determine the distance d 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/2)
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     3226 ft
d = --- =  ------- = 1613 ft
     2        2

5 - Determine voltage drop calculations

What is the minimum wire size in AWG permitted by NFPA 70-2002 in the following

    |<------------- 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" is the maximum volt-drop permitted)

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

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 12AWG, 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 average:
   PEAK-TO-PEAK is measured from the maximum value in the positive direction to
   the maximum value in the negative direction.
 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


Place your "job opportunity" or "seeking employment" ad here.


July 13-15, 2004 Chicago, IL
Fire Alarm System Testing and Inspection Seminar
Advanced Fire Alarm Seminar (NICET III/IV)

July 20-22, 2004 Salt Lake City, UT
Fire Alarm System Testing and Inspection Seminar
Intermediate Fire Alarm Seminar (NICET I/II)

August 17-19, 2004 Cincinnati, OH
Basic Fire Alarm Seminar
Fire Alarm System Testing & Inspections Seminar
Fire Alarm Plan Review Seminar

August 24-26, 2004 Indianapolis, IN
Fire Alarm Plan Review Seminar
Understanding IBC Fire Alarm Requirements

August 31 through September 2, 2004 Dallas, TX
Fire Alarm Plan Review Seminar
Understanding IBC Fire Alarm Requirements


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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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|               Fire Alarm System Design & Plan Review               |
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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.