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Question: A house is fitted with seven tubelights of rating 220 V, 40 W each, two bulbs of rating 220 V, 60 W ...

A house is fitted with seven tubelights of rating 220 V, 40 W each, two bulbs of rating 220 V, 60 W each, five fans each drawing a current of 0.4 A at 220 V, and a heater of resistance 48.4 Ω\Omega. The main line power supplied to the house is at 220 V. Calculate the bill for the month of January if tubelights and bulbs are used for 6 h daily, fans for 1 h daily, and heater for 10 h daily. The electricity is to cost ₹2 per unit.

Answer

₹796.08

Explanation

Solution

  1. Calculate the power of each appliance:

    • Tubelights: 7×40 W=280 W7 \times 40 \text{ W} = 280 \text{ W}
    • Bulbs: 2×60 W=120 W2 \times 60 \text{ W} = 120 \text{ W}
    • Fans: 5×(220 V×0.4 A)=5×88 W=440 W5 \times (220 \text{ V} \times 0.4 \text{ A}) = 5 \times 88 \text{ W} = 440 \text{ W}
    • Heater: (220 V)248.4Ω=4840048.4 W=1000 W\frac{(220 \text{ V})^2}{48.4 \Omega} = \frac{48400}{48.4} \text{ W} = 1000 \text{ W}
  2. Calculate daily energy consumption (in kWh):

    • Tubelights: 280 W×6 h1000=1.68 kWh\frac{280 \text{ W} \times 6 \text{ h}}{1000} = 1.68 \text{ kWh}
    • Bulbs: 120 W×6 h1000=0.72 kWh\frac{120 \text{ W} \times 6 \text{ h}}{1000} = 0.72 \text{ kWh}
    • Fans: 440 W×1 h1000=0.44 kWh\frac{440 \text{ W} \times 1 \text{ h}}{1000} = 0.44 \text{ kWh}
    • Heater: 1000 W×10 h1000=10 kWh\frac{1000 \text{ W} \times 10 \text{ h}}{1000} = 10 \text{ kWh}
  3. Total daily energy consumption:

    • 1.68 kWh+0.72 kWh+0.44 kWh+10 kWh=12.84 kWh1.68 \text{ kWh} + 0.72 \text{ kWh} + 0.44 \text{ kWh} + 10 \text{ kWh} = 12.84 \text{ kWh}
  4. Total monthly energy consumption (January = 31 days):

    • 12.84 kWh/day×31 days=398.04 kWh12.84 \text{ kWh/day} \times 31 \text{ days} = 398.04 \text{ kWh}
  5. Total electricity bill:

    • 398.04 kWh×2/kWh=796.08398.04 \text{ kWh} \times ₹2/\text{kWh} = ₹796.08