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I'm assuming this is standard residential single phase. Simple calculation as noted below:

Watts / Volts = Amps

So: 200 Watts / 120 Volts = 1.666~ Amps

If you needed to calculate for a 220 volt run with the same 200 Watts

200 Watts / 220 Volts = 0.909~ Amps

Remember 80% load per circuit breaker so a 15 amp breaker you should only load to 12 amps or less. Using Watt / Volts = Amps is the same as Amps x Volts = Watts.

15 Amps X .8 (80%) = 12 Amps max per circuit (for a 15 amp breaker/fuse)

So 12 Amps x 120 Volts = 1440 Watts max for a 15 amp circuit (typical 14 gauge wire)

***************************************************************

If a 20 amp circuit and 12 gauge wire (smaller gauge = larger dia wire).

20 Amps X .8 (80%) = 16 Amps max per circuit

So 16 Amps x 120 Volts = 1920 Watts max for a 20 amp circuit (typical 12 gauge wire)

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14y ago
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11y ago

Since P = V2/R, you can easily manipulate this equation and find the answer for yourself. Bear in mind that this will give you the 'hot' resistance; the resistance will be very much lower when the lamp is cold.

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13y ago

Power = V2 / Resistance

Resistance = V2 / Power = (120)2 / 100 = 14,400/100 = 144 ohms

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13y ago

P=I*E and I=E/R so P=E^2/R.

So R=(E^2)/P and R=(120v)^2/200w = 72 ohms

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13y ago

About 144 ohms. This assumes the bulb is rated at 120 volts. You might find it rated anywhere from 110v to 130v, which in normal usage means nothing. But it may mean a different resistance.

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Anonymous

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3y ago

1.666

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Q: How many amps does a 200 watt light bulb draw at 120 volts?
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