A. 532 k cal.B. 735 k cal.C. 746 k cal.D. 860 k cal.Answer: D. 860 k cal.
Explanation: Concept:Electric power (P) is the rate at which work is done by the e.m.f. a source in maintaining current in the circuit. That is, Electric Energy is the total work done by an e.m.f. a source in maintaining current in the circuit for a given time. Its SI unit is the joule (J) or (kWh).
That is given by, Electrical energy = Electrical power × time1 unit = 1 kWh = Power (in kW) × time (hours).
Calculation:Given that, electrical energy = 1 kWhWhere as 1 Joule = 1 Watt-sec
And1 kW = 1000 W1 h = 60 × 60 secSo, in terms of joule 1 kWh can be calculated as
1 kWh = 1 × 103× 60 × 60 = 3600 ×103 = 3600 Killo joule
1 killo Joule = 0.239006 Killo Calorie
1KWh = 3600 * 0.239006 = 860 k Cal
A. 532 k cal.
B. 735 k cal.
C. 746 k cal.
D. 860 k cal.
Answer: D. 860 k cal.
Explanation:
Concept:
Electric power (P) is the rate at which work is done by the e.m.f. a source in maintaining current in the circuit. That is,
Electric Energy is the total work done by an e.m.f. a source in maintaining current in the circuit for a given time. Its SI unit is the joule (J) or (kWh).
That is given by,
Electrical energy = Electrical power × time
1 unit = 1 kWh = Power (in kW) × time (hours).
Calculation:
Given that, electrical energy = 1 kWh
Where as 1 Joule = 1 Watt-sec
And
1 kW = 1000 W
1 h = 60 × 60 sec
So, in terms of joule 1 kWh can be calculated as
1 kWh = 1 × 103× 60 × 60 = 3600 ×103 = 3600 Killo joule
1 killo Joule = 0.239006 Killo Calorie
1KWh = 3600 * 0.239006 = 860 k Cal
