Question for Speedy-Petey

[HayZee518]

If you have a lamp rated 60 watts that normally operates on 240 volts, what would be the current draw if it operates on 120 volts?

[Speedy Petey]

Whatever math I did was wrong.

This is the best I could come up with.
I'll stick to doing things how I know.

[LazyPup]

Unless I am doing something wrong, I get .5A

[HayZee518]

The POWER represented by P never changes. P=EI - The E or voltage is a variable and I the current is variable.
60 divided by 240 equals a quarter amp. 60 divided by 120 equals a half amp. Because you are operating the 240 bulb at half its rated voltage the amperage is double its normal operating amperage. Therefore the bulb operates at a half amp.

[Speedy Petey]

SO then why did you ask?
Was this some sort of test?

I always knew a lamp worked at half wattage at half voltage, so I did the math to see if I my thinking was correct and came up different. I was wrong because I did the wrong math. I thought into it too much.
I should have stuck to my instinct.

[pushkins]

Just to make sure I have this straight, the 60 watt lamp at 240 volts glows at 60 watts (naturally), so at 120 volts it would glow at only 15 watts a 1/4 of it's rated output with 1/2 of the power supply ?

[HayZee518]

No, no test. Just a problem for discussion. All the incandescent nightlights I have in my house are 240 volt 6S6 bulbs. I put these in - in the year 2000 and they are still working. The 6S6 bulb has a base like the old Xmas tree bulbs. The 60 watt bulb operating at 240 volts draws a quarter amp. If you operate it at 120 volts the wattage reduces by one-half. The current is the same a quarter amp. 60 divided by 240 = .25. 30 divided by 120 = .25.

[Speedy Petey]

No Push.

60w@ 240v = 30w@120v

I over though it and confused myself over something extremely simple.

[HayZee518]

Alot of people do that. They don't think "out of the box" Don't you find many people complicating something simple? Engineers are good for that. Take a simple switch and add a bunch of relays - Boom! complicated circuits.

[pushkins]

Thanks speedy....Now I'm unconfused.

[HayZee518]

Nuther question........
Say you have a common residential dryer operating at 150 feet from the panelbox. Assuming the dryer is a 2.7Kw. What is the size of the feeder (voltage drop no more than 3% per hundred feet)

[Speedy Petey]

#10cu @ 150 feet @ 240v @ 11.25 amps = 1.74375% voltage drop

Even with an imbalance from the 120v items it is well below 3%

[HayZee518]

1 conductor(s) per phase utilizing a 12 AWG Copper conductor will limit the voltage drop to 2.43% or less when supplying 11.2 amps for 150 feet on a 240 volt system.

For Engineering Information:

* 20 Amps Rated ampacity of selected conductor
* 1.9029 Ohms Resistance (Ohms per 1000 feet)
* 0.054 Ohms Reactance (Ohms per 1000 feet)
* 7.2 volts maximum allowable voltage drop at 3%
* 5.83 Actual voltage drop loss at 2.43% for the circuit
* 0.9 Power Factor

[Speedy Petey]

It's an electric clothes dryer. I'd still use #10.

[Speedy Petey]

For a regular circuit of: #12 @ 150' @ 240v with 11.25 amps I get 6.6825 volt drop equaling 2.784375%