View Full Version : How to replace burned out switch?
jab751
May 14, 2014, 01:46 PM
Cutting the wattage of a 300 watt bulb to 150 watts using a 3 way pull switch and device to drop the wattage.
ma0641
May 14, 2014, 01:50 PM
Why not a 150 watt bulb? Or a 3 way, 50-100-150?
donf
May 14, 2014, 01:56 PM
What is your question?
A switch will not change either the applied voltage, resistance or amperage so the wattage of the lamp will stay the same
jab751
May 14, 2014, 02:10 PM
I have workbench light with 2 300watt halogen bulbs. Before it burned out, it operated with a SPDT switch @ 150/300watts. The device that was inline with the 1st pull of the switch got burned out beyond recognition, now I do not have drop in wattage. I could not locate the MFG to determine the device that failed. A RESISTOR?
jab751
May 15, 2014, 09:28 AM
I have workbench light with 2 300watt halogen bulbs. Before it burned out, it operated with a SPDT switch @ 150/300watts. The device that was inline with the 1st pull of the switch got burned out beyond recognition, now I do not have drop in wattage. I could not locate the MFG to determine the device that failed. A RESISTOR?
donf
May 15, 2014, 12:56 PM
Okay, most likely it was a resistor that burned out. Wattage is a statement of heat.
Is the lamp connected to a 120 V ac circuit or a 240 V ac circuit?
To calculate the resistance for 300 watts, its is E X E / P or (120 * 120) / 300 = 48 Ohms. For 150 Watts, (120 * 120) /150 = 96 Ohms. So if my math is correct, you would need to place an additional 48 Ohm resistor off the pull switch 1 position to create the necessary reduction to create a 150 Watt output.