PE Exam Sample Question Wrong?

[steve66]

I agree with Charlie (and others) that the right half of the circuit is an open if V2 is "disconnected", and since V2 is open, there is on current flow on the right hand side, the values of the mutual inductance, L2 and the 75 ohms don't matter.

That gives simply L1 in series with R1, which gives answer B, so I believe it is an incorrect answer. I also think you should check the errata at the site given earlier.

The Camara text (at least in my copy) gives the equivalent circuit for mutual inductance (with the secondary open) as R1 in series with (L1 - Lmutual) and in series with Lmutual. Not as R1 in series with L1 and Lmutual as shown in the solution.

(The equivalent circuit is a "T" with R1 in series with L1 - Lm on the left side, Lm in the center, and R2 in series with L2-Lm on the right side.)

 

[GoldDigger]

Note that (L1 - Lm) in series with Lm is just L1. Why go to the extra trouble?

 

[Ingenieur]

Wouldn't the equivilent be a T circuit with the mutual being the middle leg?
then if the sec is open Z = Rp + Zlp + Zm
edit:?Zm must be subtracted from each side also

 

[K8MHZ]

deleted

never mind...

 

[steve66]

Note that (L1 - Lm) in series with Lm is just L1. Why go to the extra trouble?

The equivalent circuit also works for when the V2 is something other than "disconnected".

I'm a little surprised question #46 isn't something like "V2 is 10 volts. Find the current through L1."

What's the point of having a practice problem with mutual inductance if it doesn't matter what the mutual inductance is?

 

[GoldDigger]

The equivalent circuit also works for when the V2 is something other than "disconnected".

I'm a little surprised question #46 isn't something like "V2 is 10 volts. Find the current through L1."

What's the point of having a practice problem with mutual inductance if it doesn't matter what the mutual inductance is?

It seems pretty clear from the solution guide that the writer of the problem thought that it did matter. <sigh>.

 

[gar]

160316-2015 EDT

The "T" network when reflected to the primary side is:

Rp in series with Lp*(1-k) on the left,
Lp*k in the middle, and
Rs*(Lp/Ls) in series with Lp*(1-k) on the right.
Clearly the assumption was made in the derivation of the equations that the secondary leakage inductance when reflected to the primary was equal to the primary leakage inductance.

Clearly if there is no secondary load, then the input impedance is Rp in series with Lp*(1-k)+k*Lp or simply the inductive component is Lp.

Furthermore, no matter what the input impedance is it can not be less than Rp. Thus, if the secondary is shorted the input impedance has to be between these two values.

It is interesting that in the design of the problem k = M / sq root (Lp*LS) = 15 / sq root (40*65) = 15 / 51 = 0.2941 or 0.3 which is a fairly nice number.

I believe that it is very clear that the problem was to be solved for a 60 Hz frequency.

It also seems clear that the possible answers included one that results from adding Lp to Lm, answer C. And this is not a correct answer. Answer D is even bigger and thus not possible.

I don't believe that the person that wrote the question wrote the sample analysis of the question. You don't have answer B in the list unless you know how to analyze the circuit.

.

 

[Ingenieur]

They give us V sin (377 t)

w = 2 Pi f = 377
f = 60