Question

A refrigerator having a power rating of 350W operates for 12 hours a day. calculate the cost of electrical energy to operate it for 30 days. the rate of electrical energy is rs4.50 per kWh. calculate the minimum fuse rating to be put in the refrigerator circuit if the supply voltage is 240V.​

Answer 1

Answer:

See the answers below.

Explanation:

The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.

$Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]$

The fuse can be calculated by knowing the amperage.

$P=V*I$

where:

P = power = 350 [W]

V = voltage = 240 [V]

I = amperage [amp]

Now clearing I from the equation above:

$I=P/V\\I=350/240\\I=1.458[amp]$

The fuse should be larger than the current of the circuit, i.e. about 2 [amp]