... 0.35A.calculate.
a. the efficiency of the transformer
b. the current in the lamp
A transformer is use to light a 240v, 70w lamp from a 240 a.c main current in the main cable is?
- Posted:
- 3+ months ago by Nature
- Topics:
- lamp, light, cable, current, transformer, transformers
Answers (1)
Get a ruler in your hands. Measure things until you start to understand how a ruler works. Measure some stuff and figure out where the center is. Say you measure a book and it's 7/8" thick. You look at your ruler and see that every eighth is divided into two sixteenths, so obviously half of 7/8" is going to be 7/16". If you write that out you have 1/2 x 7/8 = 7/16. And you notice that 1/2 is divided into 2/4 and then into 4/8 and so on, so you can convert anything to anything by multiplying all the numbers on top and then all the numbers on bottom.
Other rulers are divided into 10 and 100 parts. But an inch is still an inch, so anything on one ruler can be translated to the other ruler. A half inch on one ruler is 5/10 or 50/100 on the other. An eighth inch is just 12.5 marks when you have 100 marks per inch. A metric ruler divides an inch into 25.4 parts, so a half inch would be 12.7 of those parts. Pretty simple, isn't it? Practice this a bit and people will think you went to wizard school.
Percent is simply a ruler with 100 marks. The only confusion is trying to keep track of what the marks represent, since that changes from time to time.
For the main cable power=240 x 0.35 = 84 watts. Efficiency is expressed in percent, so we have a ruler with 100 marks worth 84 watts and each mark is worth 0.84 watts. Then 70watts/0.84watts = 83.333% efficiency. In practice you would figure 70watts/84watts x 100% = 83.333%.
You have to assume that the transformer output is exactly equal to the input at the current being drawn. If the transformer is simply 1:1 The voltage would be reduced as part of the efficiency loss, the output would be less than the input, and the lamp would not draw 70 watts. The question does not mention any such assumptions. But if the lamp operates at stated capacity, with 240 volts applied, then the current is 70/240 = 0.291666 amp.