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Question

An electric kettle for use in a 230 V supply is rated at 3000 W. For safe working what should be the current carried by a cable connected to it?

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Solution

Step 1, Given data

Potential difference , v = 230 v

Power = 3000 W

Step 2, Finding the current
We know,

Power = V×I

Putting all the values

3000=230×I

Or,I=3000230=13.04A

Hence the current flowing is 13.04 A



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