Transformer Maximum Power Transfer Quesiton

[tankfarms]

I'm studying for my PE exam and encountered a sample question like below. I don't quite understand the third solution provided. How come we are not using impedance (Z) for calculation but instead we only use resistance (R). Also, to calculate the load current, I thought it would be just to divide the secondary voltage (1240/4 = 30V) by the impedance of the secondary circuit, but using such thinking got me the wrong answer.

Can anyone help explain the third question and its provided solution?

 

[T.M.Haja Sahib]

I'm studying for my PE exam and encountered a sample question like below. I don't quite understand the third solution provided. How come we are not using impedance (Z) for calculation but instead we only use resistance (R). Also, to calculate the load current, I thought it would be just to divide the secondary voltage (1240/4 = 30V) by the impedance of the secondary circuit, but using such thinking got me the wrong answer.

Can anyone help explain the third question and its provided solution?

Maximum power transfer when source impedance is equal to load impedance. Here capacitance on primary and inductance on secondary and so on maximum power transfer, they cancel, leaving the resistances.

Applying transformation ratio to find out the secondary voltage does not work here because of presence of large resistance on the primary side.

 

[LMAO]

I'm studying for my PE exam and encountered a sample question like below. I don't quite understand the third solution provided. How come we are not using impedance (Z) for calculation but instead we only use resistance (R). Also, to calculate the load current, I thought it would be just to divide the secondary voltage (1240/4 = 30V) by the impedance of the secondary circuit, but using such thinking got me the wrong answer.

Can anyone help explain the third question and its provided solution?

First of all, there is a forum dedicated to FE/PE candidates. I used it when I was studying for my PE. It is here:
http://engineerboards.com/index.php?showforum=4

About your question: I must say it is a very interesting problem and I'd not seen one like this before. The way I understand it, they are looking for the frequency at which the primary capacitor impedance cancels out secondary inductor impedance (Z=jlw -j/(cw)). So jlw=j/(cw) and w^2*l*c=1 where w=2pi*f, l is inductance and c is capacitance. But l is in the secondary side so that's where a^2 comes from.

 

[drbond24]

Maximum power transfer occurs when source and load impedances match.

When source and load impedances match, you have a resonant circuit (circuit operating at resonant frequency).

The very definition of resonant frequency is: the frequency at which a circuit becomes purely resistive.

They got you with a trick question. At max power transfer you are at resonance and therefore you have a resistive circuit. You can ignore the L and C.

By the way, I got all of that out of the Electrical Engineering Reference Manual by John A. Camara. If you don't have that book, I cannot emphasize strongly enough how much you need it to take the PE. It is worth every penny of the $100+ they charge for it.

 

[steve66]

drbond is on the right track. Here are my solutions:

1) They are looking for the turns ratio for max power transfer (although they seem to forget to say that in the problem.) The resistor and inductors on the secondary can be redraw in series with the primary circuit by multiplying by alfa^2. See the attached PDF. Then you set the load resistance equal to the source resistance for maximum power transfer. So Rs = alfa^2 * Rl.

2) The inductance and capacitance cancel out when Xl = Xc. Again, the L becomes alfa^2 * L to move it to the primary.

3) Again, working with everything on the primary, as mentioned, the Xc and Xl cancel, so you don't have to worry about them. You have 120V/ (Rs + alfa^2 * Rl) = 120/(256+256). That gives 0.234 amps. That is the current in the primary of the transformer. To get the final answer, you change back to the original circuit, and know that the current in the secondary is going to be 4 times the current in the primary (the turns ratio = 4).

4* 0.234 = 0.9375 amps.

View attachment Equivalent Circuit.pdf



Your mistake in part C was forgetting some voltage is dropped across Rs.

(Edit: One more gripe: They should have said the answer to part C is calculated using the frequency in answer B).

 

[steve66]

You also could have worked part C by moving the primary resistance on the secondary. Then, you have 16 ohms in series with 16 ohms. And the answer is 30 volts/32 ohms = .9375 amps. Here, the 30 volts comes from 120V/4.

Steve

 

[drbond24]

(Edit: One more gripe: They should have said the answer to part C is calculated using the frequency in answer B).

They did. B) says to calculate f0. C) says to calculate the load current at f0.

 

[steve66]

They did. B) says to calculate f0. C) says to calculate the load current at f0.

OK, I'll give you that.

But for part A, I still don't see where it says the transformer is operating at maximum power transfer, which you would need to find alfa.

 

[gar]

120530-1724 EDT

When I saw the first post of this thread I did not know what the original question was? Was the question as presented in the original test ever presented in this thread?

The question is a very good question to determine if one understands transformers, resonance, and the conditions for maximum power transfer. This is not really a trick question, but a question to get the reader to think about the circuit.

If I wrote the question I might present it as follows ---
What is the transformer turns ratio to obtain maximum power transfer to the load resistor R_L?
What is the input frequency to obtain this maximum power transfer?
What are the primary and secondary currents when maximum power transfer occurs?
At this turns ratio what is the approximate secondary current when the frequency is 60 Hz?

It is assumed the transformer is ideal, the resistances, capacitor, inductor, and the source voltage are all constant.

.

 

[tankfarms]

Thanks guys for all the information. I like the simple explanaiton of what a true "resonant circuit" is (resistive), that well explains lots of my questions earlier!

 

[tankfarms]

You also could have worked part C by moving the primary resistance on the secondary. Then, you have 16 ohms in series with 16 ohms. And the answer is 30 volts/32 ohms = .9375 amps. Here, the 30 volts comes from 120V/4.

Steve

How do you "transform" the primary circuit's 256 ohm resistance to the secondary circuit's equivalent of 16 ohms?

 

[Besoeker]

OK, I'll give you that.

But for part A, I still don't see where it says the transformer is operating at maximum power transfer, which you would need to find alfa.

I'm inclined to agree.For the solution given in A you have to assume the circuit is operating at resonant frequency and, like you, I can't where that assumption comes from.

 

[Besoeker]

How do you "transform" the primary circuit's 256 ohm resistance to the secondary circuit's equivalent of 16 ohms?

Turns ratio squared.

 

[gar]

120631-1344 EDT

Anything other than resonance will produce a lower series current.

Then anything other than having the reflected resistance being equal to the other resistance will produce less than maximum power to the load resistance.

In this problem both the source and load resistances were kept constant and the variable was the transformer ratio to control the relationship of the load resistance to the primary source resistance.

Tankfarms:

To see why turns ratio squared is the transfer factor in referring the secondary resistance to the primary do the following:

Let R_{viewed at the primary} be equal to V_pri / I_pri, and assume no resistance in the primary, and N is the turns ratio of primary to secondary, then what are the values of the secondary voltage and current? From this you should be able to calculate the R_{on secondary} and show yourself the relationship.

.

 

[steve66]

How do you "transform" the primary circuit's 256 ohm resistance to the secondary circuit's equivalent of 16 ohms?

Its basically the same "equivalent circuit" concept that is used when we combine 2 parallel resistors into a single resistor, or 2 series resistors using formulas like Rt = R1+R2. We simplify the circuit to find some particular value (like the current in a series resistor circuit, or the current in the primary of your transformer circuit.) Then we work back toward the original circuit finding individual currents and voltage across individual components.

A transformer basically steps the current up by the turns ratio, while it steps the voltage down by the turns ration. So impedences are multiplied by the square of the turns ratio.

So an X or an R on the secondary can be moved to the primary by multiplying each value by alfa^2. Then, you work the problem with the applied 120 volts. That allows you to find the primary current. Then you go back to the original circuit, and use the primary current to find the secondary current.

You can also move X and R to the secondary by dividing each by alfa^2. Then you work the problem with the 30 volts the ideal transformer would provide on the secondary.

 

[Besoeker]

120631-1344 EDT

Anything other than resonance will produce a lower series current.

But that's not the first question asked.
To come up with the 4:1 ratio in the solution given you have to assume resonance.

 

[gar]

1205331-2122 EDT

By inspection of the circuit you know that it is a series resonant circuit. Resonant frequency at the start is unknown. We do have to assume that the resistances are constant with respect to frequency, but there was no information to the contrary.

Once it is observed that there is a series resonant circuit, then from common knowledge of the characteristics of such circuits it is known that maximum current flows when the series resonant circuit is excited at resonance. Thus, the first question to ask is how to adjust the transformer ratio to obtain maximum power transfer to R_L.

When the transformer ratio has been determined, then is is possible to determine the resonant frequency.

This question addresses several areas of knowledge all in one question. Actually a very good question.

.

 

[topgone]

MY CALCS

MY CALCS

I tried playing around with different frequencies to find the resonant frequency and I got a very different answer on item 3 ---> amps at Fo = 0.47 A (b).
Please see pasted figures from my work sheet:

C =	0.0000033 Farads	Cpri
R1 =	256	Rpri
L =	0.003 Henry	Lsec
R2 =	16	Rsec
alpha =	4.272869019	= SQRT(1/((2*PI()*Fo)^2*Cpri*Lsec))-->answer is (b) on item 2.
Resonance Frequency =	374.3544106	answer is (d) on item 1.
Left side of circuit:
R1 =	256	Rpri
Xc1 =	128.8318483	= 1/((2*PI()*Fo*Cpri))
Z1 =	286.5896808	= SQRT(C8^2+C9^2)
Right side(referred to primary):
R2 (pri) =	0.876356521	= Rsec/alpha^2
XL2 (pri) =	128.8319744	= (2*PI()*Fo)*(1/alpha^2)
Z2(pri) =	128.834955	= SQRT(Rsecpri^2+Lsecpri^2)
Check:	-0.00000097864	= (Xc1-Lsecpri)/Lsecpri
Total impedance =	256.8763565	= SQRT((C8+Rsecpri)^2+(C9-Lsecpri)^2
Current, amps =	0.467150818	= amps ---> answer is (b)on item 3.
Power transferred to RL =	0.191247184	= power_transferred

 

[Besoeker]

1205331-2122 EDT

By inspection of the circuit you know that it is a series resonant circuit.

With series L and C in any circuit there will be a frequency at which these will cancel but, it seems to me that, for part A, you have to assume that the circuit is operating at that frequency in order to calculate turns ratio.

 

[gar]

120601-0920 EDT

Besoeker:

It is common knowledge, based on prior analysis, that the minimum impedance of a series resonant circuit is at resonance. Thus, maximum current from a constant excitation voltage is at resonance, and maximum power in the series resistance of the equivalent circuit is at resonance. You do not need to assume resonance, you know resonance is required.

.