I have a home that receives single phase 240 AC from the utility co.
Is there a way to wire and electric heater rated at 208 volts?
thanx
gary
Is there a way to wire and electric heater rated at 208 volts?
thanx
gary
Ahem,.... the voltage across the "lightbulb" is 32 V, not 240 V and the power would be about 1.5 KW. I doubt such a lightbulb exists but this resistor could be of any type and could be custom built.The light bulb idea is not so bright, I think...
P = E x I, so 10000 / 208 = 48 Amps (I hope you have a spare circuit in your breaker box.)
240V - 208V = 32V across the lightbulb
E = I x R, so R = E / I. Light bulb resistance needed is 32 / 48 = 0.67 Ohm
P = E^2 / R, so light bulb rating = 240 x 240 / 0.67 = 86 kiloWatts.
You are not confused, Recca, I simply expressed myself poorly. Required resistance is indeed 0.67 Ohm.i'm confused why is the rating not 32*48 = 33^2/0.67
Most bulbs in 240 Volt nations are rated for 240 Volts, not 32 Volts. Most bulbs are rated by wattage, not resistance. To find bulb resistance, one must use the given ratings.Ahem,.... the voltage across the "lightbulb" is 32 V, not 240 V
The heater is for industrial use. 208 Vac is one leg of a 240 Vac 3-phase wye system taken to ground.'tis strange that a heater would be built for 208 V. --- I still don't understand where such a heater came from.
actually that answers my doubtMost bulbs in 240 Volt nations are rated for 240 Volts, not 32 Volts. Most bulbs are rated by wattage, not resistance. To find bulb resistance, one must use the given ratings.
Even after reading this several times I was confused but I think I now understand what you are trying to communicate which is that if we could find a bulb rated at 240 V, 86 KW then the resistance would be 0.67 ohm and we could use it for the intended purpose. But that is not the case because the resistance varies greatly with the power being generated and such a bulb connected in such a way would have a *much* lower resistance. Way much lower. It would only have a resistance of 0.67 ohm when connected to 240 V and dissipating 86 KW. What you *really* can use is a lightbulb (or other resistance) rated 32 V, 1.5 KW.Most bulbs in 240 Volt nations are rated for 240 Volts, not 32 Volts. Most bulbs are rated by wattage, not resistance. To find bulb resistance, one must use the given ratings.
Can you explain that? According to my calculations if you have 240 V between two phases then you have 240 / SQRT(3) = 139 V between a phase and neutral. Am I missing something?The heater is for industrial use. 208 Vac is one leg of a 240 Vac 3-phase wye system taken to ground.
true,Well, on a three phase system the line to line voltage is always the line to neutral voltage multiplied by sqrt(3)
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