An electric iron draws a current of 4 Ampere on a 240 V supply
1.what is its power
2.what is its resistance
3.what is the cost of using the iron for the month of January 10 hours a day if 1 KWH costs Rs.3.40?
Answers
Answered by
20
1. P = IV
=> P = 4A × 240V
Therefore, P = 960W
2. V = IR
=> R = V/I
=> R = 240V ÷ 4A
Therefore, R = 60 Ohms
3. 1KW = 1000W
1000W = Rs.3.40
=> 1W = Rs. (3.40÷1000)
=> 960W = Rs. {(3.40 × 960) ÷ 1000}
Therefore, Rs. 3.264
Ps. Please recheck the answer for 3. (It might not be correct)
=> P = 4A × 240V
Therefore, P = 960W
2. V = IR
=> R = V/I
=> R = 240V ÷ 4A
Therefore, R = 60 Ohms
3. 1KW = 1000W
1000W = Rs.3.40
=> 1W = Rs. (3.40÷1000)
=> 960W = Rs. {(3.40 × 960) ÷ 1000}
Therefore, Rs. 3.264
Ps. Please recheck the answer for 3. (It might not be correct)
Answered by
3
Answer:
Explanation:Volt =240v,and current(I)=4A
So,the power(p)=volt *current=240*4=960w then the total energy consumed by the iron in 31 days. 960w*10 hour*31 day=297600wh. We covert this into kWh=297600/1000=297.6kwh. 1kwh=3.40rupees for 297.6kwh=297.6*3.40=1011.84 rupees hence,the total cost of using the iron is 1011.84 rupees
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