Question

Two bulbs of 60 W each and two coolers of 200 W each work on average for 4 hours a day. If the energy costs ₹ 3.50 per kWh, calculate the monthly bill and the minimum fuse rating when power is supplied at 200 V.

Answer 1

(a) Power of 2 bulbs = 2 × 60 = 120 W

∴ Energy consumed by 2 bulbs in a day = P × t = 120 × 4 = 480 Wh

Power of 2 coolers = 2 × 200 = 400 W

∴ Energy consumed by 2 coolers in a day = 400 × 4 = 1600 Wh

∴ Energy consumed by bulbs and coolers in a day = (480 + 1600) = 2080 Wh

∴ Energy consumed by bulbs and coolers in a month = 2080 × 30 = 62400 Wh = 62.40 kWh

\[\text{Monthly bill} = 62.40\text{kWh} \times 3.50\frac{\text{₹}}{\text{kWh}}\]

∴ Monthly bill = ₹ 218.40

(b) Total power of bulbs and coolers = 120 + 400 = 520 W

\[\therefore \text{Minimum fuse rating} = \text{Current drawn} = \frac{\mathrm{P}}{\mathrm{V}}\]

\[= 520 \div 200\]

\[= 2.6\text{A}\]