Question

If a TV of rating 100 W is operated for 6 hours per day, find the amount of energy consumed in any leap year.

Answer 1

Power of the TV = 100 W = 0.1 kW.
Energy consumed per day = Power * Time
= 0.1 * 6 = 0.6 kWh.

Energy consumed in the leap year
= 0.6 * 366 days
= 219.6 kWh

Answer 2

Hello friend
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Let's write the given..

So,

Power (P) = 100 W

which will be converted into kilowatt.

1 kW = 1000W

= 100/1000

= 0.1 kW

Time,(t) = 6 x 366 ......(leap year)

= 2196 hours.

We have to find,

Electric energy consumed

So,we know that,

Electric energy consumed = power x time

So now put the given values...

Electric energy consumed = 0.1 x 2196

= 219.6 kWh

Therefore,the amount of energy consumed is 219.6 kWh.

Thanks.