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[Cram For The PE]

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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.

 

[DLD PE]

5A

 

[matt267 PE]

42

 

[Cuseman17]

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

 

[DoctorWho-PE]

C

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

 

[Cram For The PE]

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.

 

[Dude99]

0

 

[Warrior PE]

0.351 A?

 

[Cram For The PE]

0.351 A?

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

 

[Cram For The PE]

0

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

 

[Cram For The PE]

42

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

 

[Orchid PE]

42

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

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

 

[Cram For The PE]

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.

 

[Orchid PE]

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!

 

[Dude99]

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

 

[Orchid PE]

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.

 

[JayD]

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<-2.2 or 50.

I=240/50,-2.2=4.8<2.2 or 5A. 

 

[Cram For The PE]

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<-2.2 or 50.

I=240/50,-2.2=4.8<2.2 or 5A. 

This is incorrect.

 

[Cram For The PE]

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

This is incorrect.

 

[Cram For The PE]

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