Hello friends,
I am completely newbie in basic electronics and i really don't know how to answer this assignment i have. I don't know how to approach the solution, please help me get the right answers for this questions or point in the direction how to approach the solutions.....

http://img33.picoodle.com/img/img33/...5m_0ef3e59.jpg

questionnaires:

1) Look at the circuit. An ammeter is being used to measure the current in the primary winding. The ammeter displays a reading of 2 A. Which is the following statements about the switch positions is correct?
a) only sw S1 is closed b) switch s1 and sw s2 are both closed c) only sw s3 is closed d) all the three switches are closed.

2) Look again at the circuit. Assuming that all switches are closed, which of the ff. statements are correct.
a) resistor r1 will reach highest temp.
b) R2 will reach the highest temp.
c) R3 will reach the highest temp.
d) all 3 resistors have the same temp.

3) In the same circuit. when all 3 switches are closed, the power in the primary circuits is .
a) at maximum b) 200 W c) less than that any single secondary winding d) reduced by eddy current losses.

4) compared to a single-phase alternator, a three phase alternator will generate?

a) a higher output voltage b) three sine waves 120 degrees displaced c) a higher output current d) three sine waves delta-connected.

Please give me an idea which is the correct answer or how to solve the problem. Thank you in advance....

 

Hi Randolfo,

WATTS the problem ?

on1aag.

 

Just in case on1aag's answer confounds you, I'll say a bit more, but won't spell things out.
Looking at your penciled in marks, you have correctly calculated the voltage across all of the resistors, however your current calculations are not correct, save the last one, for 200V.
Ohm's Law: I = V/R

I = 200V/200ohm = 1A

apply this to all of the secondaries and you will have your answers for the first 3 questions.

For the 3phase question think of a circle. How many degrees are there in a circle? Divide that by 3, and you will have an answer, but you should google this to learn more.

 

Hi Beepop,

Thanks for your reply, i'll take it from there. Just asking is it not possible that three phase alternator produced higher current than single phase alternator, i am more inclined to tick this one, do you have additional idea on this...?

 

You should base your answer only on the evidence given. So do you see any evidence that supports higher current? I think they are asking for the fundamental difference between a single and triple phase alternator.

 

Just want to make sure, is it not the formula to find the temperature of resistor is (I squared * Resistance) ? Am I correct? It mentioned in my textbook that if the ratio of the voltage in the primary is lets say 1:3 of the secondary that means the current in the secondary will be 3:1 just to equalize the power (E * I) in both primary and secondary. That's why I am thinking the current can not be all the same in this circuit because the voltages ratio are all different . Please clarify, I am confused... The current can not be all 1 A ..

 

Hi BeeBop,

I see you are online, thanks . Yeah you are right no evidence to supports higher current, but it is one of the difference right? I think they want the fundamental difference, ok.

 

Well, the Power dissipated by the resistor will be:
P = IV, but V = IR, so P = I x I x R so yes, but this is power, measured in watts.

The more power the resistor dissipates, the hotter it will get.

Perhaps the ratio thing is confusing you. Not regarding losses, the power drawn by the secondary will be the same as the power in the primary, and it is measured in watts, thus on1aag gave you the answer, but I wondered if you would catch it.

Now I'll use just the one you had correct
There is 200V across R3
R3 is 200ohms
so there is 1A flowing through it.
Since the secondary here is 2x the number of turns of the primary, there will be twice the current, or 2A drawn in the primary winding.

You work out the others, and get back.

 

hi beebop,

I am still confused. If i work out the other two i'll get 1 A current for each secondaries. The rules of ratio is not followed only in the secondary 3 the ratio is correct, on the other two secondaries the ratio are not correct. See , in the first secondary there is voltage ratio of 10:1 in the primary, the second one is 5:1 . So if applied the rules of ratio, I should get 10 x 2A = 20A in the first secondary, 5 x 2A =10A in the second secondary. Why is that? what i am missing here...

 

Yes, I get 1A in each Secondary. If you get 1A for each secondary, why are you using 2A in your math? Perhaps you are also looking at the ratios in reverse? Try 1:10

 

Each secondary has it's own current based upon the primary to secondary turns ratio and load. For a perfect transformer each secondary it completely independent and has no effect on the other secondaries.

All secondary load currents are reflected into the primary and added together based upon the turns ratio of each secondary. If you calculate it all correctly, the primary power will equal the sums of the secondary powers.

__________________
Carl