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प्रश्न
An electric motor of 2 H.P. and two coolers of 500 W each operate on 250 V mains for 4 hours a day. If the energy costs ₹ 1.80 per kWh, calculate (a) weekly bill (b) minimum fuse rating. [Take 1 HP = 750 W]
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उत्तर
(a) Power of electric motor = \[2 \times 750 = 1500\text{ W}\]
Power of 2 coolers = \[2 \times 500 = 1000\text{ W}\]
∴ Total power of motor and coolers = \[1500 + 1000 = 2500\text{ W}\]
∴ Energy consumed by motor and coolers in a day = \[2500 \times 4 = 10000\text{ Wh} = 10\text{ kWh}\]
∴ Energy consumed by motor and coolers in a week = \[10 \times 7 = 70\text{ kWh}\]
\[\text{Weekly bill} = 70\text{ kWh} \times 1.80\frac{\text{₹}}{\text{kWh}}\]
\[\therefore \text{Weekly bill} = \text{₹}126\]
(b) Total power of motor and coolers = \[1500 + 1000 = 2500\text{ W}\]
\[\therefore \text{Minimum fuse rating} = \text{Current drawn} = \frac{\mathrm{P}}{\mathrm{V}}\]
\[= 2500 \div 250\]
\[= 10\text{ A}\]
