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

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  • 3 weeks ago

    20 amps if i remember correctly

  • 4 weeks 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
    4 weeks 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

  • 4 weeks ago

    P=E*I

    2000=100*I

    20=I

    20A

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  • 4 weeks 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

    .

  • 4 weeks 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.

  • 4 weeks 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.

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