Some wires in the circuit are existing. Some are new.
I don't have a megger, but if I did or can get one, If the wires are fine, Would not the inductive kick back cause the breaker to trip any way?
If there is a damaged wire, would not, should not it trip with no loads or the A lamp installed?
gar.. sounds good, but I don't think they want additional lights coming on with the in the pool pool lights. Especially, A light in the panel... :- )
Thank you for the calculator links eHunter.... why did you pick 2.7k ??? just wondering.
And I can see the calculations.......but how or why did in the calculation of 5 w's jump to 25 w for the suggested resistor?
Rich
You are welcome.
I picked 2.7K because it it a common resistor size that would have a relatively low current draw and produce minimal heat.
Wherever you place/mount the resistor, you will need to be cognizant that the
resistor will get hot when energized.
However, I did have difficulty locating a 2.7K 25W unit, but I did find a close value of 3.0K that is readily available. See Below...
The power(W) handling capacity of a resistor should always be at least 2x the calculated power dissipation for longevity.
Your situation has the added element of the transformer inductive kick that is of an undetrmined peak voltage, current and duration, so a SWAG from exprience ~5x the calculated power dissipation was used to be safe and assure the long term life of the resistor.
Voltage (E) = 120 Volts
Current (I) = 0.04 Amps
Resistance (R) = 3000 Ohm
Power (P) = 4.8 Watts
Voltage (E) of 120 known
Standard Resistance (R) of 3.0K Ohm is selected
I = E/R = 120/3000 = 0.04
P = E*I = 120 * 0.04 = 4.8
Digikey Link for 3.0K 25W wirewound resistor:
http://www.digikey.com/product-detail/en/TMC0253K000FE02/TMC25-3.0K-ND/269955
Digikey 2.5K-3.5K 25W search:
http://www.digikey.com/scripts/dkse...t=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25