Question

A refrigerator having a power of 350 W operates for 10 hours a day. Calculate the cost of electrical energy to operate it for a month of 30 daus. The rate of electrical energy Rs. 3.40 per kWh.​

Answer 1

$\bf{\pink{Solution:}}$

Electrical energy,

$\tt{\purple{E = P \times t}} \\ \\ \tt{\green{Power, \: P = 350\:W}} \\ \\ \tt{\blue{ \frac{350}{1000}\: kW}} \\ \\ \tt{\orange{ = 0.35\:kW}} \\ \\ \tt{\red{And, \: \: \: \: \: Time, \: T = 10 \times 30 \: hours}} \\ \\ \tt{\green{ = 300\:h}}$

Now, putting these values of P and t in the formula,

$\tt{\orange{E = P \times t}} \\ \\ \tt{\purple{We \: get: \: \: \: \: \: E = 0.35 \times 300 \: kWh}} \\ \\ \tt{\green{ = 105 \: kWh}}$

Thus, the electrical energy consumed by the refrigerator in a month of 30 days is 105 kilowatt-hours.

$\tt{\orange{Now, \: \: \: Cost \: of \: 1 \: kWh \: of \: electricity}}\\ \tt{\pink{ = Rs. \: 3.40}} \\ \\ \tt{\blue{So, \: \: \: Cost \: of \: 105 \: kWh \: of \: electricity}} \\ \tt{\purple{= Rs. \: 3.40 \times 105}}$

$\large{\mathcal{\boxed{\red{Rs.\:357}}}}$

Answer 2

Explanation:

Electrical energy,

E=P×t

Power,P=350W = 350 / 1000 = 0.35kW

And,

Time,T=10×30hours = 300h

Now, put the values of P and t in the formula, we get ,

E=P×t

E = 0.35×300kWh = 105kWh

Thus, the electrical energy consumed by the refrigerator in a month of 30 days is 105 kilowatt-hours.

Now,

==> Cost of 1kWh of electricity = Rs.3.40

So,

==> Cost of 105kWh of electricity = Rs3.40×105 = Rs 357