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A 2.2 kW electric iron operates normally on a voltage of 220 V.

Fill in the blank.

The current in the iron when it is operating normally is _____ A.

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Answer:

Current equals Power divided by Voltage (I=P/V), Power equals Current times Voltage (P=VxI), and Voltage equals Power divided by Current (V=P/I).

P = V I

2.2*10³ = 220 * I

I = 10 amps

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