justin59
Member
- Location
- loma linda, ca
does anybody know of a way to test a circuit with flourescent fixtures on it for a dead short before you energize it for the first time? I know a simple continuity test won't work. Thanks.
testing flourescents
- Thread starter justin59
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gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090429-0851 EST
justin59:
A dead short would mean you should see the wire resistance out to the point of the short. #12 wire is 1.588 ohms/1000 ft. So a short 500 ft away from the point at which you make the measurement will produce a reading of about 1.6 ohms.
You could probably apply 6 V plus a series 1 ohm resistor to the circuit to be tested and measure the voltage across the 1 ohm resistor and the circuit under test and from these voltages predict whether a short exists or not. You also want to know what the impedances of your fluorescent fixtures are.
Do you believe you have a short? If so, then is your next question how to find the short?
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justin59:
A dead short would mean you should see the wire resistance out to the point of the short. #12 wire is 1.588 ohms/1000 ft. So a short 500 ft away from the point at which you make the measurement will produce a reading of about 1.6 ohms.
You could probably apply 6 V plus a series 1 ohm resistor to the circuit to be tested and measure the voltage across the 1 ohm resistor and the circuit under test and from these voltages predict whether a short exists or not. You also want to know what the impedances of your fluorescent fixtures are.
Do you believe you have a short? If so, then is your next question how to find the short?
.
justin59
Member
- Location
- loma linda, ca
One of my co-workers energized a 277v lighting circuit with a dead short in one of the fixtures. It was really easy to tell wich one it was (the black one.)
it was a 2x2 drop in fixture and the whole thing had to be replaced because the burn mark was visible. I am just wondering if anybody knew of a trick to tell if there is a short before energizing the circuit.charlietuna
Senior Member
Sounds like "standard operating proceedure" to me !
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090429-2019 EST
You can probably use any low power means to measure low resistance. The 6 V source with 1 ohm internal impedance would probably work. An 8 ft 120 V Slimline standard ballast has a DC resistance of about 4.2 ohms. The 60 Hz AC impedance will be much higher. If you use 6 V at 60 Hz for the test voltage it probably will be easy to detect a dead short.
At 7.4 V 60 Hz my 8 ft Slimline shows a current of 43 MA. In other words an impedance of 172 ohms. You can parallel quite a number of these and not interfere with a dead short test using the low voltage 60 Hz approach.
.
You can probably use any low power means to measure low resistance. The 6 V source with 1 ohm internal impedance would probably work. An 8 ft 120 V Slimline standard ballast has a DC resistance of about 4.2 ohms. The 60 Hz AC impedance will be much higher. If you use 6 V at 60 Hz for the test voltage it probably will be easy to detect a dead short.
At 7.4 V 60 Hz my 8 ft Slimline shows a current of 43 MA. In other words an impedance of 172 ohms. You can parallel quite a number of these and not interfere with a dead short test using the low voltage 60 Hz approach.
.
justin59
Member
- Location
- loma linda, ca
would a 6v battery work or a "wall wart" power supply with an output voltage of 6v be better? and would i connect one side of the resistor to the hot and the other side to the neutral and measure impedance between the two? Im just trying to get a better understanding of what you are saying. BTW thanks for your time and help. I think im going to try this after work one day.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090429-2230 EST
justin59:
I think I favor using using a 6.3 V 2 A 60 Hz filament transformer and a 5 ohm 10 W series resistor.
One transformer lead goes to one end of the 5 ohm resistor. The other end of the resistor and the second end of the transformer secondary are your current limited voltage source. Connect these two output leads to whatever you want to test. One output connection to your hot lead to the fixtures, and the other to either EGC or neutral depending upon what you are testing. The circuit you are testing must be unenergized from any other source. Your load impedance is Zl. The open circuit voltage is Vs and the load voltage is Vl. The constant 5 in the following equation is the 5 ohm resistor and needs to be slightly higher by the amount of equivalent internal resistance of the transfomer. But ignore this because you only need a rough estimate.
Someone else needs to check my math.
Zl = 5/(-1+(Vs/Vl) )
As Vl approaches 0 Zl gets small and approaches 0.
As Vl approaches Vs Zl approaches infinity.
If Zl was 50 ohms and Vs was 6 V, then Vl would equal 5.4545 .
If Zl was 0.5 ohms and Vs was 6 V, then Vl would equal 0.54545 .
.
justin59:
I think I favor using using a 6.3 V 2 A 60 Hz filament transformer and a 5 ohm 10 W series resistor.
One transformer lead goes to one end of the 5 ohm resistor. The other end of the resistor and the second end of the transformer secondary are your current limited voltage source. Connect these two output leads to whatever you want to test. One output connection to your hot lead to the fixtures, and the other to either EGC or neutral depending upon what you are testing. The circuit you are testing must be unenergized from any other source. Your load impedance is Zl. The open circuit voltage is Vs and the load voltage is Vl. The constant 5 in the following equation is the 5 ohm resistor and needs to be slightly higher by the amount of equivalent internal resistance of the transfomer. But ignore this because you only need a rough estimate.
Someone else needs to check my math.
Zl = 5/(-1+(Vs/Vl) )
As Vl approaches 0 Zl gets small and approaches 0.
As Vl approaches Vs Zl approaches infinity.
If Zl was 50 ohms and Vs was 6 V, then Vl would equal 5.4545 .
If Zl was 0.5 ohms and Vs was 6 V, then Vl would equal 0.54545 .
.
You could use the method in THIS VIDEO to locate the problem.
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