E and M : About power

很可惜樓上的朋友錯了。我是電機工程畢業的，希望能夠作出更正及給出正確的答案=)

(1)

First of all, Power is always the product of the voltage and current. This is always true. When you calculate the power consumption of a device, you can use P=VI for sure. Remember we CAN ALSO USE P=VI

However, in most cases, either V or I is known. Hence we derive another two formula just for the sake of easy calculation. THey are P=(I^2)R and (V^2)/R.

Note that the derivation is based on Ohm's law. Wherever Ohm's law does not hold, P=(I^2)R and (V^2)/R are NO LONGER VALID. On the other hand, P=VI is still TRUE.

Normally, when the current I changes, voltage changes too. So sometimes it is hard to use P=VI. In normal cases (which appear in secondary school Phy), Resistance R is a constant.

So whenever comparing the power losses of two device of different currents and same voltage, we use P=(I^2)R. On the other hand, whenever comparing the power losses of two device of different voltages and same current, we use (V^2)/R.

(2)
Assuming the transformer is ideal and there is no power loss, i.e. input electrical power equals output electrical power

(Vp)(Ip) = (Vs)(Is)
Hence, Vs/Vp = Ip/Is ------------ (1)

Also from Faraday's law, we can derive Vs/Vp=Ns/Np------(2)
(***The one you typed in the question is WRONG)

From (1) and (2), we can calculate the current in the secondary winding (here we use the term WINDING instead of COIL)

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Why not V=IR?

This is an interesting and good question and I am quite sure you are a CE student. I will try to explain to you by easy words. there are a few points you need to know:

a) the voltage developed at the secondary winding is NOT 100% taken by the secondary winding resistance. In fact, the secondary winding resistance only take a very little amount of developed voltage. But where does the rest of developed voltage go? please see the next paragraph.

b) the voltage and current in the transformer are ac, not dc (@steady state). the winding not only has resistance, but also has inductance. Inductance causes significant voltage drop for high frequency ac. Even now it's 50Hz AC, the inductance is still great. Hence, Vs mainly consists of two parts --- the voltage drop due to resistance + voltage drop due to inductance

Because most of the things in CE are using dc, like batteries, cells, etc., you seldom hear using ac source. So you are puzzled by V=IR

The above explainations are made of easy words. I simplified the explainations for easy understanding. If I try to tell the whole story about transformer, it will make you further not understand.

I hope I can help you. Thank you for reading

2007-02-28 20:35:30 補充：
to 樓上的朋友: I don't mean to be offensive. I just want to correct your mistakes1) all three formulas are equivalent unless Ohm's law does not hold. There is nothing dealing with the accuracy of the answer.. We cannot simply have a smaller error by choosing a special formula.

2007-02-28 20:35:44 補充：
2) when a practical transformer is produced, normally the number of turns is UNKNOWN. Instead, the turns ratio is provided. Also, the reason why we do not use V=IR is not because the turns-ratio method is more convenient. in fact, V=IR just cannot be used in this way

for the first question, because when considering the transformer, voltage will become very large and the current would become very small, therefore, when calculating the power loss, if we use two terms which the order of magnitudes are too far away, the answer will be less accurate. Therefore, we would like to use I^2R so as to make the error become smallest.

For the secord question, it is only becuase of teaching purposes in HK. The main reason is the HKEAA want us to learn about the principle of transformer -- the voltage can be step up or step down by number of turns of coil, so nearly all the text book would use the equation you mentioned above. Another consideration is about the manufacturing factor, when we produce an transformer, the number of turn would be known and no need to use additional equipment to measure about it, therefore, it should be the most concenient method to calculate the stepped voltage in the secondary coil.