2.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? b) 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.
PLEASE DO ANSWER IT...Class 10--Physics
Answers
Answer:
To obtain maximum current ,we will arrange the two given resistors in parallel .Only then,the equivalent resistance will be minimum and current maximum. R(eq)=R1*R2/(R1+R2) Therefore,V=I*R(eq) gives maximum current. (b)3bulb of 100 w for 5 hrs. 3 ×100×5 = 1500 w = 3/2 kwh 2 fans of 50 watt each used for 10 hours daily. 2 × 50 × 10 = 1000 w = 1 kwh. an electric heater of 1.00 kW is used for half an hour daily 1 kwh × 1/2 hr = 1/2 kwh. for 31 days. 3/2 × 31 = 93/2 1 × 31 = 31 1/2 × 31 = 31 / 2. now adding these 93/2 + 31/2 + 62/2 = 196/2. = 98. now cost is 98 × 3.6 = Rs. 352.8
Resistors should be connected in parallel to get get maximum electric power consumed
Explanation:
Power = V²/R
Battery of emf V
hence V is constant
so to get maximum power
R should be minimum
in Series Resistance get added
& in Parallel Resistance reduces
Hence Resistors should be connected in parallel
3 bulbs of 60 watt each are lighted for 3 hours daily
= 3 * 60W * 3 = 540 Wh = 0.54 kWh
4 fans of 100 watt each are used for 8 hours daily
= 4 * 100W * 8 = 3200 Wh = 3.2 kWh
electric heater of 2 kwh used for half an hour daily.
= 2kW * 1/2 = 1 kWh
Total Consumption daily = 0.54 + 3.2 + 1 = 4.74 kWh
in 31 Days consumption = 31 * 4.74 = 146.94 kWh
Cost per kWh = Rs 4
Total Cost = 4 * 146.94 = Rs 587.76
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