## Calculate the total cost of running the following electrical devices in the month of September, if the rate of 1 unit of electricity is ` 6.00.

## (i) Electric heater of 1000 W for 5 hours daily.

## (ii) Electric refrigerator of 400 W for 10 hours daily.

**
Answer:
**

(i) P
_{
1
}
= 1000 W = 1000 / 1000 kW = 1 kW

t
_{
1
}
= 5 hours

Energy, E
_{
1
}
= P
_{
1
}
* t
_{
1
}
* number of days

= 1 * 5 * 30

= 150 kWh

(ii) P 2 = 400 W = 400 / 1000 kW = 0.4 kW

t
_{
2
}
= 10 hours

Energy, E
_{
2
}
= P
_{
2
}
* t
_{
2
}
* number of days

= 0.4 * 10 * 30

= 120 kWh

Therefore, Total energy = E
_{
1
}
+ E
_{
2
}

= 150 + 120 kWh

= 270 kWh

Given,

cost of 1 kWh of energy = ₹ 6

So,

**
Total Cost
**
= Total energy * cost of 1 kWh of energy

= 270 * 6

**
=
**
**
₹
**
**
1620
**