PE Exam Sample Question Wrong?

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gray.one

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I call bs for 2 reasons.

1). The wording doesn't say if v2 is open or shorted. Measuring the impedance of the whole circuit requires a shorted v2 to be of much value since mutual inductance was mentioned.

2). If v2 is open, mutual inductance will not be in play. Put the system of equations together and the mutual inductance is either multiplied by I2 or is in the second (open v2 equation) and does feed back into the first equation.


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GoldDigger

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I agree that the analysis is wrong.
You could interpret "voltage sources deenergized to mean that they are shorted out.
But even with V2 shorted out the 75 ohm resistance is in series with the L2 and so is coupled by mutual inductance. That cannot be ignored. The result is that the equivalent circuit containing only L1 and Lmutual along with the 50 ohm resistor is wrong.
 
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Ingenieur

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It's confusing but we may be confusing an actual measurement of the primary winding with an equivilent ckt

the n is 5:1 so the sec Z reflected to the primart is /25

they also state the mutual is a factor when energized?

de-energizing the v sources implies open ckt or removal from the ckt
 
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dkarst

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I don't believe (but may be incorrect) this question is from the "official" typical exam questions book provided by NCEES. There are a few other providers where the sample questions are not as clear as on the actual exam.
 
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gar

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If V2 is assumed to be an open circuit, the answer is close to B. With the question as worded my normal assumption would be that V2 should be considered a short which won't match any of the answers. That assumption puts the value between answers A and B.

In the limit if L1 was shorted the answer would be A, but whether V2 is a shorted or open in the actual circuit the answer can not be A.

To get answer B refer to https://en.wikipedia.org/wiki/Leakage_inductance for an equivalent circuit analysis. See the equivalent circuit labeled "Real transformer equivalent circuit in terms of coupling coefficient k".

With V2 an open circuit ignore everything to the right of kLp. My quick calculation gave me about 52.2 ohms @ 16.6 deg.

.
 
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gray.one

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dkarst said:
I don't believe (but may be incorrect) this question is from the "official" typical exam questions book provided by NCEES. There are a few other providers where the sample questions are not as clear as on the actual exam.
Click to expand...

It is indeed.
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gray.one

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PE Exam Sample Question Wrong?

Gar, we agree. If the are going to work this as meaning the secondary is open, then mutual inductance doesn't come into play.


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Phil Corso

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Gentlepeople...

The fact that the circuit is de-energized rules out the effect of "polarity" (Subtractive or Additive), rules out the "dot" rule, and rules out "turns-ratio"! More importantly, it rules out any effect of "mutual inductance"! Hence, the problem is treated as a determining the impedance of a simple reactor, containing just R and L1... making 'B' the correct answer!

However, one could argue that the Impedance Meter must contain a frequency source, thus the effective or equivalent inductance should include mutual impedance, thus the effective impedance includes R1 and the combination of L1 and Lm... making 'C' the correct answer!

Regards, Phil Corso
 
charlie b

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I haven't done this sort of calculation in a long time, but I am inclined to go with answer "A." With source V2 de-energized, and with source V1 disconnected and replaced by a meter, the entire right hand side of the circuit becomes irrelevant. There is no mutual inductance, unless there first is inductance. No power source means no frequency of a power source, and that means no inductance. That makes L1 essentially a wire. It will have some resistance, but that will be negligible compared with the 50 ohm resistor. So the "most nearly" measured impedance will be 50 ohms at an angle of 0 degrees.

 
K8MHZ

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charlie b said:
I haven't done this sort of calculation in a long time, but I am inclined to go with answer "A." With source V2 de-energized, and with source V1 disconnected and replaced by a meter, the entire right hand side of the circuit becomes irrelevant. There is no mutual inductance, unless there first is inductance. No power source means no frequency of a power source, and that means no inductance. That makes L1 essentially a wire. It will have some resistance, but that will be negligible compared with the 50 ohm resistor. So the "most nearly" measured impedance will be 50 ohms at an angle of 0 degrees.
Click to expand...

Please note that the question clearly states 'impedance meter'. A real impedance meter has a setting (that must be chosen) for frequency. The question also contained enough information to figure out what the frequency needed to be applied to the circuit would be so the meter could be set for the correct reading.
 
jumper

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K8MHZ said:
Please note that the question clearly states 'impedance meter'. A real impedance meter has a setting (that must be chosen) for frequency. The question also contained enough information to figure out what the frequency needed to be applied to the circuit would be so the meter could be set for the correct reading.
Click to expand...

I get 60 Hz.
 
charlie b

charlie b

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Phil Corso said:
Charlie...You're wrong.
Click to expand...
Well, there is a first time for everything. :lol:
Phil Corso said:
L1 is specified as a a coil, therefore it has self-inductance, commonly referred to as 'Inductance'!
Click to expand...
Not if the meter imposes a DC voltage, in order to take its measurements. That was the basis of my answer. However, not having any experience as a lab technician, I would not know exactly how an "impedance meter" functions. If it imposes an AC signal of a frequency that the operator must select, then L1 will behave as the inductor that it is.
 
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GoldDigger

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charlie b said:
Well, there is a first time for everything. :lol:
Not if the meter imposes a DC voltage, in order to take its measurements. That was the basis of my answer. However, not having any experience as a lab technician, I would not know exactly how an "impedance meter" functions. If it imposes an AC signal of a frequency that the operator must select, then L1 will behave as the inductor that it is.
Click to expand...

An impedance meter does have a (usually variable, or at least switchable) frequency source for the measurement and provides two analog or two digital outputs, usually for the equivalent resistance and inductance/capacitance rather than the complex Z at the selected frequency.
An impedance bridge, much more common before the days of digital electronics, used mechanically variable resistance and capacitance to balance a bridge network in which the unknown two terminal network formed one leg.
By changing the bridge leg on which the variable capacitance was located you could avoid the need for a variable inductor in the test set.
Some impedance bridges would accept an external oscillator for the test frequency in addition to its built in source.
 
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