# A 2KW iron is used on a 100V AC mains. The current in the iron is?

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• Anonymous
1 month ago

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• Anonymous
4 months ago

The gurrent iz twenny ambs.

• 20 amps if i remember correctly

• "A 2KW iron is used on a 100V AC mains. The current in the iron is?"

P=VI

I=P/I

I = 2 000/100

I= 20 Amps

• P=IE thus I = P/E

I = 2,000w/100v = 20a ←

But I question where you would ever find 100vac mains?? Did you mean 120vac as is standard in the US? then I = 2000/120 = 16.7a

• P=E*I

2000=100*I

20=I

20A

• "A 2KW iron is used on a 100V AC mains. The current in the iron is?"

Assuming the supply mains are single phase, and the iron is purely resistive, then this can be answered by simple Watt's Law.

P=VI

I=P/I

I = 2 000/100

I= 20 Amps

.

• P = EI

2000 watts = 100 v x I

I = 20 amps

that assumes the device is designed to draw 2 kW at 100 volts.

• 2,000/100 = 20 amps. P divided by E equals I. P divided by I equals E. P is the measure of power when you multiply E x I. E is electromotive force. I is current flow.