Solve This Problem and Get Paid

[Orchid PE]

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

Since we're assuming ideal transformers (transformers a & 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?

 

[Cram For The PE]

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

Since we're assuming ideal transformers (transformers a & 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?

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:

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!

 

[Orchid PE]

Woohoo!

 

[Dude99]

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.

 

[Cram For The PE]

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. 

 

[Orchid 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

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. 

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.

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.

 

[akyip]

(Accidental double-post, please ignore.)

 

[akyip]

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

Since we're assuming ideal transformers (transformers a & 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?

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!