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Question
A house is provided with 15 bulbs of 40W, 5 bulbs of 100W, 5 fans of 80 W, and one heater of 1.0 kW. Each day bulbs are used for 4h, fans for lOh, and heater for 2h. The voltage of mains is 220 V. Calculate:
(i) Maximum power of the circuit in the house,
(ii) maximum current capacity of the main fuse in the house,
(iii) the electrical energy consumed in a week,
(iv) cost of electricity consumed at 1.25 Rs. per kWh.
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Solution
(i) Maximum power of the circuit = (40 × 15) + (100 × 5) + (80 × 5) + (1000 × 1)
= 600 + 500 + 400 + 1000
= 2500 W = 2.5 kW.
(ii) Maximum current capacity = `"Total wattage"/"Voltage of mains" = 2500/220 = 11.36` (or 11.4) A.
(iii) Electrical energy consumed in a week
In 40 W bulbs = `((40 xx 15) xx (4 xx 7))/1000 = 16.8` kWh
In 100 W bulbs = `((100 xx 5) xx (4 xx 7))/1000` = 14 kWh
In 80 W fans = `((80 xx 5) xx (10 xx 7))/1000` = 28 kWh
In 1 kW heater = `((1000 xx 1) xx (2 xx 7))/1000 = 14` kWh
∴ Total electrical energy consumed = 16.8 + 14 + 28 + 14 = 72.8 kWh
(iv) Cost of electricity = 72.8 × 1.25 = Rs. 91
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Unit of electric power may also be expressed as:
