# Solve This Problem and Get Paid



## Cram For The PE (Jul 19, 2020)

Note- This problem has been solved.

Reward has doubled!!!!! I will give the first person to solve this problem my complete set of books for free OR can get $100 cash. See link for complete details.

http://cramforthepe.com/index.php/2020/04/27/circuit-problem-solve-the-problem-and-get-a-discount/

An AC analog amp-meter with a range of 0 to 1000 amps is placed in the primary of the circuit to determine the primary current as shown below. If both transformers are subtractive polarity, what would the amp meter read?




Answer to nearest hundredth. Please show details about answer.


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## DLD PE (Jul 22, 2020)

5A


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## matt267 PE (Jul 23, 2020)

42


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## Cuseman17 (Jul 24, 2020)

1. Equivalent Z = 50.225
2. Transferring to the primary side, Zp= 50.225 * 7/2 * 7/2 since transformers have same polarity, the ratio is 7/2, while transferring impedance, we multiply by square of the ratio
3. I = 240/Zp = 0.3900 A


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## DoctorWho-PE (Jul 24, 2020)

C

(Just kidding... but how I would answer, since not electrical)


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## Cram For The PE (Jul 24, 2020)

Cuseman17 said:


> 1. Equivalent Z = 50.225
> 2. Transferring to the primary side, Zp= 50.225 * 7/2 * 7/2 since transformers have same polarity, the ratio is 7/2, while transferring impedance, we multiply by square of the ratio
> 3. I = 240/Zp = 0.3900 A


This is incorrect.


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## Dude99 (Sep 3, 2020)

0


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## Warrior PE (Sep 8, 2020)

0.351 A?


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## Cram For The PE (Sep 9, 2020)

Warrior said:


> 0.351 A?


You have to detail how you got the answer when you give one please.


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## Cram For The PE (Sep 9, 2020)

Dude99 said:


> 0


You have to detail how you got the answer when you give one please.


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## Cram For The PE (Sep 9, 2020)

matt267 PE said:


> 42


You have to detail how you got the answer when you give one please.


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## Orchid PE (Sep 9, 2020)

Cram For The PE said:


> matt267 PE said:
> 
> 
> > 42
> ...


I think I can answer this one.

https://en.wikipedia.org/wiki/Phrases_from_The_Hitchhiker's_Guide_to_the_Galaxy#The_Answer_to_the_Ultimate_Question_of_Life,_the_Universe,_and_Everything_is_42


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## Cram For The PE (Sep 9, 2020)

Chattaneer PE said:


> I think I can answer this one.
> 
> https://en.wikipedia.org/wiki/Phrases_from_The_Hitchhiker's_Guide_to_the_Galaxy#The_Answer_to_the_Ultimate_Question_of_Life,_the_Universe,_and_Everything_is_42


I will give a hint. The answer is NOT 42! . Narrows it down for you guys a little.


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## Orchid PE (Sep 9, 2020)

Cram For The PE said:


> I will give a hint. The answer is NOT 42! . Narrows it down for you guys a little.


I've never covered a topic like this before. I'm not super refreshed on the specifics of transformers, but it seems like the magnetic fields inside the cores of the transformers would be conflicting against each other. E.g. If we assume the same current is flowing in the primaries, and assume the correct ratio of current is produced on the secondary of the 4:1 transformer, then that secondary current would be producing a magnetic field in the 3:1 transformer that is fighting against the magnetic field produced from its primary. So I don't really know where to go from here.

Looking forward to seeing the explanation!


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## Dude99 (Sep 11, 2020)

VpT1 = 4/7 • V                              VsT1 = 1/4 •4/7 • V = V/7

VpT2 = 3/7 x V                              VsT2 = 1/3 • 3/7 • V = V/7

subtractive
VsT1 - VsT2 = V/7- V/7 = 0

Is = 0/Z = 0 therefore Ip = 0


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## Orchid PE (Sep 11, 2020)

Since both transformers are subtractive, their secondary voltages should add together. If one transformer was subtractive and the other was additive, then the difference would be used.


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## JayD (Sep 12, 2020)

Due to subtractive polarity, combined turns ratio is 4-3:1 i.e. 1:1

Equivalent impedance referred to primary is thus same i.e. calculated 50&lt;-2.2 or 50.

I=240/50,-2.2=4.8&lt;2.2 or 5A.


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## Cram For The PE (Sep 12, 2020)

JayD said:


> Due to subtractive polarity, combined turns ratio is 4-3:1 i.e. 1:1
> 
> Equivalent impedance referred to primary is thus same i.e. calculated 50&lt;-2.2 or 50.
> 
> I=240/50,-2.2=4.8&lt;2.2 or 5A.


This is incorrect.


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## Cram For The PE (Sep 12, 2020)

Dude99 said:


> VpT1 = 4/7 • V                              VsT1 = 1/4 •4/7 • V = V/7
> 
> VpT2 = 3/7 x V                              VsT2 = 1/3 • 3/7 • V = V/7
> 
> ...


This is incorrect.


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## Cram For The PE (Sep 13, 2020)

Reward has been doubled. Solve it and get all my books for free or $100 cash.


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## Orchid PE (Sep 14, 2020)

Based on my limited knowledge, I'm going with I_pri = 0.

Since we're assuming ideal transformers (transformers a &amp; b):

I_pri = I_pri_a = I_pri_b

I_sec = I_sec_a = I_sec_b

4 * I_pri_a = I_sec_a

3 * I_pri_b = I_sec_b

The only number that satisfies all of these equations is I_pri = 0.

We're assuming these are two individual transformers and _not _two windings on the same core, correct?


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## Cram For The PE (Sep 14, 2020)

Chattaneer PE said:


> Based on my limited knowledge, I'm going with I_pri = 0.
> 
> Since we're assuming ideal transformers (transformers a &amp; b):
> 
> ...


You Sir have solved the problem. Which is good because I been waiting to post other problems and was sick of this one. Email me your information and tell me what you would like.

MMF balance is impossible. The network draws no current and the input impedance is infinity (or the same as an open circuit). It is completely irrelevant what is connected on the secondary. As Mr Chattaneer PE has shown above, the only number that fits all of it is zero. Dennis Karst posted at my website and realized something was wrong. Chattaneer PE also realized something was wrong about MMF balance when he posted this:



Chattaneer PE said:


> I've never covered a topic like this before. I'm not super refreshed on the specifics of transformers, but it seems like the magnetic fields inside the cores of the transformers would be conflicting against each other. E.g. If we assume the same current is flowing in the primaries, and assume the correct ratio of current is produced on the secondary of the 4:1 transformer, then that secondary current would be producing a magnetic field in the 3:1 transformer that is fighting against the magnetic field produced from its primary. So I don't really know where to go from here.
> 
> Looking forward to seeing the explanation!




Another way to think about this- If both primaries where connected in parallel we would have a short circuit. When you connect them in series you have an open circuit.

Congrats!


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## Orchid PE (Sep 14, 2020)

Woohoo!


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## Dude99 (Sep 14, 2020)

imho my solution is correct using V.  I don't care about the $.
 

As noted the I's in each xfmr prim and sec are equal in magnitude and opposite in sign summing to 0 at the node between each xfmr.  This makes the V's also opposite and equal summing to 0. If you did have a resultant V you would have a current since you have a load.


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## Cram For The PE (Sep 14, 2020)

Dude99 said:


> imho my solution is correct using V.  I don't care about the $.
> 
> 
> As noted the I's in each xfmr prim and sec are equal in magnitude and opposite in sign summing to 0 at the node between each xfmr.  This makes the V's also opposite and equal summing to 0. If you did have a resultant V you would have a current since you have a load.


Chattaneer PE pointed out the problem with that solution. You can't solve the problem in the way you did. If your method was used correctly, you would have added the voltages as Chattaneer PE pointed out. It was really a combination of both his answers that made him correct. First, he recognized something was wrong with the MMF balance. Second, he wrote out the current and realized only zero would work to make the formulas correct. At that point, it was enough for him to get credit for the solution.


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## Orchid PE (Sep 14, 2020)

Dude99 said:


> VpT1 = 4/7 • V                              VsT1 = 1/4 •4/7 • V = V/7
> 
> VpT2 = 3/7 x V                              VsT2 = 1/3 • 3/7 • V = V/7


You can't assume that the voltage on the 4:1 transformer is 4/7, and on the 3:1 is 3/7. Since the transformers have separate cores this ratio cannot be assumed. 



Dude99 said:


> subtractive
> VsT1 - VsT2 = V/7- V/7 = 0


This isn't what subtractive polarity means. Since both transformers have subtractive polarity, their secondary voltages will add.



Dude99 said:


> As noted the I's in each xfmr prim and sec are equal in magnitude and opposite in sign summing to 0 at the node between each xfmr.  This makes the V's also opposite and equal summing to 0.


This is incorrect. The currents in the primaries and secondaries are not opposite in sign. The currents in the primaries flow in the same direction, and the currents in the secondaries flow in the same direction. The voltages on the primaries add, and the voltages on the secondaries add.


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## akyip (Sep 15, 2020)

(Accidental double-post, please ignore.)


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## akyip (Sep 15, 2020)

Chattaneer PE said:


> Based on my limited knowledge, I'm going with I_pri = 0.
> 
> Since we're assuming ideal transformers (transformers a &amp; b):
> 
> ...


Wow, I'm blown away by the answer. Congrats, Chattaneer!!!

I'm hoping a question of this caliber doesn't pop up on the actual exam too much haha!

Adding this information onto my list of "Things to know about Transformers" on my Bible binder... which won't be used for the CBT, but I can still read it over and now know this as useful information!


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