Asked by mike
a 230v/30v with 50Hz transformer that has a primary current of 0.6A which supply's number of 30v and 30 watts halogen light bulb. what is the efficiency?
Answers
Answered by
Henry2
Pi = Ep * Ip = 230 * 0.6 = 138 Watts = Power in.
Is = Ep/Vs * Ip = 230/30 * 0.6 = 4.6A. = Current in secondary or output.
Po = Vs * Is = 30 * 4.6 = 138 Watts. = Power out.
Assuming the given information is correct, Po = Pi = 138 Watts; but 4.6 light bulbs are required to consume 138 Watts. So we'll have to use 4 light
bulbs which consumes only 120 Watts(Po).
Efficiency = (Po/Pi) * 100% = (120/138) * 100% =
Is = Ep/Vs * Ip = 230/30 * 0.6 = 4.6A. = Current in secondary or output.
Po = Vs * Is = 30 * 4.6 = 138 Watts. = Power out.
Assuming the given information is correct, Po = Pi = 138 Watts; but 4.6 light bulbs are required to consume 138 Watts. So we'll have to use 4 light
bulbs which consumes only 120 Watts(Po).
Efficiency = (Po/Pi) * 100% = (120/138) * 100% =
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.