Transformer Maximum Power Transfer Quesiton

[Besoeker]

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.

Yes, I know all that. I'm simply pointing out that part A appears to ask the question without any prior reference to maximum power.
Maybe it was somewhere in part of the question not posted.

 

[gar]

120601-1203 EDT

Besoeker:

That was effectively the question I asked in post #9. I do not know that we ever saw a direct statement of the question that was presented in the test. The title of the thread implied the question was about maximum power transfer. Was the thread title the actual question, or was there a more complete statement for the question?

.

 

[Besoeker]

120601-1203 EDT

Besoeker:

That was effectively the question I asked in post #9. I do not know that we ever saw a direct statement of the question that was presented in the test. The title of the thread implied the question was about maximum power transfer. Was the thread title the actual question, or was there a more complete statement for the question?

Good points. And put much more succinctly than I did.
Maybe the OP can clarify.

 

[steve066]

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

I'm not following your table. Line 5 has 4.2 for alfa, but from part (a), alfa is exactly 4.0. There is also an Fo in the formula for alfa, but we can't find Fo until we have found alfa.

 

[jim dungar]

I'm not following your table. Line 5 has 4.2 for alfa, but from part (a), alfa is exactly 4.0.

Actually, the question (a) used the phrase "most nearly".

 

[topgone]

I'm not following your table. Line 5 has 4.2 for alfa, but from part (a), alfa is exactly 4.0. There is also an Fo in the formula for alfa, but we can't find Fo until we have found alfa.

As mentioned, the question asked for a range of numbers by saying "most nearly". In the same way, you can set your iterative solution to "guess" what the alpha is and re-compute for the Fo.

The idea is to make sure the inductive reactance (referred to the primary side of the ideal transformer) cancels the capacitive reactance in series:

Xc1 = XL x 1/alfa²

1/(w x C) = w x L x 1/alfa²

solving for w:

w² = 1/ (L x C x alfa²,

And w = sqrt(1/(L x C x alfa²))

Then we get Fo:

Fo = w/(2 x pi)

Conversely, we can solve for alfa given an Fo:

alfa = sqrt(1/ (L x C x w²))

 

[gar]

120602-0914 EDT

For maximum power transfer the turns ratio is exactly 4.

At maximum power transfer the secondary resistance reflected to the primary is 256. Thus, total load on the 120 V supply is 256 + 256 = 512 ohms. The primary current is 120/512, or on the secondary side 480/512 or 30/32 = 0.9375 . Notice that integers based on 2^N have been used in this problem.

My calculation for the resonant frequency is 399.89 Hz. I suspect the question was originally designed with L = 3.0 mH, f = 400 Hz, and then C rounded to 3.3 ufd.

.

 

[topgone]

120602-0914 EDT

For maximum power transfer the turns ratio is exactly 4.

At maximum power transfer the secondary resistance reflected to the primary is 256. Thus, total load on the 120 V supply is 256 + 256 = 512 ohms. The primary current is 120/512, or on the secondary side 480/512 or 30/32 = 0.9375 . Notice that integers based on 2^N have been used in this problem.

My calculation for the resonant frequency is 399.89 Hz. I suspect the question was originally designed with L = 3.0 mH, f = 400 Hz, and then C rounded to 3.3 ufd.

.

I followed your calcs and it looks correct! I must have missed something in my worksheet. Will review more where the fault lies in there.

 

[steve066]

As mentioned, the question asked for a range of numbers by saying "most nearly".

Yes, but alfa = sqrt(256/16) which is exactly 4.0. Not 4.29. Your frequency seems to be low by about the same amount your alfa is high.

It looks like you somehow miscalculated alfa. But the formula you list next to alfa doesn't seem right- i didn't look at it in detail - but it looks more like the equation for Fo. So I'm not sure how you got alfa??

 

[topgone]

Yes, but alfa = sqrt(256/16) which is exactly 4.0. Not 4.29. Your frequency seems to be low by about the same amount your alfa is high.

It looks like you somehow miscalculated alfa. But the formula you list next to alfa doesn't seem right- i didn't look at it in detail - but it looks more like the equation for Fo. So I'm not sure how you got alfa??

Alfa formula in my worksheet is derived by equating the absolute primary reactance with the secondary inductance (referred to the primary) for a certain frequency.
|Xc| = |XL x (alfa)²|
1/wC = wL x (alfa)²
1 = wC x wL x (alfa)²
Solving for alfa:
alfa = sqrt[(1/(w² x LC)]

Using your Fo = 400 Hz, we get:
alfa = sqrt(1/({2 x pi() x 400}² x 3.3 x 10^-06 x 3 x 10^-13)
alfa = 3.999, say 4!

BTW, I found the error in my worksheet: truncation errors accumulating into significant levels due to the circuitous path of the calculations!