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How should two resistors with resistances R1 ohm and R2 ohm be connected to a battery of emf V volts so that the electric power consumed is maximum? In a house, 3 bulbs of 60 watt each are lighted for 3 hours daily, 4 fans of 100 watt each are used for 8 hours daily and an electric heater of 2 kwh used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 4/kwh.

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Final answer:

To maximize electric power consumption, two resistors should be connected in parallel to the battery. The total energy consumed by 3 bulbs, 4 fans, and an electric heater in a household for a month of 31 days is 147.18 kWh, costing Rs 588.72 at the rate of Rs 4/kWh.

Step-by-step explanation:

To maximize the electric power consumed by two resistors with resistances R1 ohm and R2 ohm when connected to a battery of emf V volts, the resistors should be connected in parallel. In this configuration, the total resistance is decreased, allowing more current to flow from the battery, which increases the power consumed according to P=IV, where I is current and V is voltage.

For the energy consumption in a household, we can calculate the total amount used by various appliances and their duration of use over a period of 31 days. The formula for energy consumed in kilowatt-hours (kWh) is Energy (kWh) = Power (Watts) × Time (hours) × Number of Days × (1/1000) to convert Watts to Kilowatts.


  1. Bulbs: 3 × 60W each × 3 hours/day × 31 days × (1/1000) = 16.98 kWh

  2. Fans: 4 × 100W each × 8 hours/day × 31 days × (1/1000) = 99.2 kWh

  3. Heater: 2 kWh × 0.5 hours/day × 31 days = 31 kWh

Total energy consumed = 16.98 kWh + 99.2 kWh + 31 kWh = 147.18 kWh

The cost of energy consumption at Rs 4 per kWh is:

Cost = 147.18 kWh × Rs 4/kWh = Rs 588.72

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