Salman asked in Science & MathematicsEngineering · 5 months ago

# 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.

• garry
Lv 5
4 months ago

20 amps if i remember correctly

• 5 months ago

"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

• Jim
Lv 7
5 months ago

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

• 5 months ago

P=E*I

2000=100*I

20=I

20A

• 5 months ago

"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

.

• 5 months ago

P = EI

2000 watts = 100 v x I

I = 20 amps

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

• 5 months ago

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.