I would go by basic MMF balance in an ideal transformer. If assuming high mu_r core, then 3 MMF sources in an ideal transformer will be in parallel. So that N*Is = N*I1 = N*I2 => Is = I1 = I2. Also with energy conservation power input must be total power output, so Ps = P1+P2 or Vs*Is = V1*I1 + V2*I2 => Vs = V1+V2 = I1*R1+I2*R2. Now : 1) Vs = Is*R1+Is*R2 = Is*(R1+R2) = Is*5 => Is = Vs/5 = 10/5 = 2A = I1 = I2 and V1 = 2*4 = 8V and V2 = 2*1 = 2V 2 & 3) if R1 is shorted => V1 = 0 but I1 != 0 but it also holds that Is = I1 = I2 due to MMF balance so Vs = V1+V2 = V1 + 0 = 10V. I1 = V1/4 = 2.5A = Is = I2
If the coupling is ideal (infinite Ur core), then because the magnetic circuit is of a 3 MMFs connected in parallel, the two resistors will reftect to the voltage source side as a single series resistor of 4 + 1 = 5Ohms, therefore the source current as well as the two resistors' currents would be 10/5 = 2A (1:1:1 reflection). Thus, V1 would be 2*4 = 8V, and V2 would be 2*1 = 2V. When R1 is shortet (=0Ohms) then the total reflected resistance would be 4+0 = 4 Ohms. and the 3 currents would be 10/4 = 2.5A, thus. V2 would be 2.5*4 = 10V, which makes the magnetic circuit behave as a two-winding Xformer with a 1:1 ratio. The 3rd shorted winding will cancel out (opposite) it's leg's flux (shorted winding leg flux will be zero, thus circuit will become a simple ideal transformer).
This is a real headscratcher (especially the last two questions). Here are my answers, happy to hear of the problem behind my reasonings, as I am really not sure: 1) V1 and V2 will be 5 Vac with a 180-degree phase shift. This is because the flux splits up from the middle excitation winding, giving half the excitation to each winding. 2) The currents should match in a sub-loop (right side) so allow for a closed magnetic loop path. However, the flux swing will be minimal in this path due to the loss of resistance. So 2.5 A (ac) is based on the exciting winding's current. 3) Because voltage can't be present on V1 (as this requires the resistance), the flux swing has to appear at the left sub-loop. Therefore, it's now a fully coupled circuit, i.e., V2 = 10 Vac.
I'm not sure my sign conventions for the fluxes are the right way round but here's my take anyway.. Let: V0 = 10V Then: (0) V0 = -NAdB0/dt (1) V0 = -N(A/2)dB0/dt -N(A/2)dB0/dt Assume direction of fluxes of each section are such that from (1) we get: (2) V2 = N(A/2)dB1/dt +N(A/2)dB0/dt (3) V1 = -N(A/2)dB1/dt +N(A/2)dB0/dt where there is a common flux - Φ0 flowing between each half of the centre section and each of the two side sections - Φ1 flowing between the two side sections (2)-(3) => (4) V2-V1 = -NAdB1/dt (2)+(3) & (0) => (5) V2+V1 = V0 (4)+(5) => (6) V0 = 2V2 +NAdB1/dt (6)+(0) => V0 = V2 +N(A/2)dB1/dt -N(A/2)dB0/dt & (3) => V0 = V2 -V1 & (5) => V2+V1 = V2 -V1 => V1 = 0 & (5) => V2 = V0 = 10V. But my flux sign convention can't be correct as it introduces an asymmetry in a symmetrical circuit, right?
Only from reading the first couple of comments i learned a lot totally new insights. Flux splitting and the summing of outputvoltages and so on. The drawing did me think of 3fase transformers from wich i never understand how the different fases can use the same core. I look so much out for the answervideolesson, in fact as always
1) V1 = 2 V and V2 = 8 V The voltage of each secondary winding is related to the impedance reflected in the primary winding. 2) I1= 2.5 A The current of secondary 1 is equalised to the current delivered by secondary 2. 3) V2 = 10 V Since secondary 1 is short-circuited, the impedance seen from the primary is the load of secondary 2. Due to the 1:1 ratio, the voltage is the same as that of the primary.
The more loaded a coil is, the more it will deflect the magnetic fields away from itself by creating an opposing field. 1 - 8V+2V. both coils will see half the magnetic flux of the middle one when open circuit. When loaded the more loaded one deflects the field away, so less voltage is induced. 2 - (10V/4Ohm)/2=1.75A. I1=Isupp/2, as it has to run just enough current to match the open circuit magnetic field magnitude. 3 - 10V. The magnetic path towards coil1(the right one) is effectively blocked, leaving out a 2 coil transformer.
If R1 & R2 are infinite (I1=I2=0), then the answer is easy: the magnetic flux (Φ = B.A) generated by the centre winding splits equally between the other two core legs, and from Faradays law (V = NdΦ/dt) each of these windings sees half the voltage, ie: V1 = V2 = 5Vac. However, this flux distribution is not maintained under the loaded condition, since I1 and I2 are not zero. In the reluctance diagram, the total MMF generated by centre leg sees two other MMFs in series. Since the nmbr of turns are the same, the currents must be the same, and the two voltages have to add to 10V. So we solve two simultaneous equations: (1) V1 + V2 = 10. (2) I1 = I2 (3) V1 / R1 = V2 / R2, where R1=1Ω, R2=4Ω. Writing V2 in terms of V1 from (1), and subbing into (3) we get: V1 / 1Ω = (10 - V1) / 4Ω --> V1= (2.5 - V1 /4Ω) --> 1.25x V1 = 2.5 , So V1 = 2, and V2 = 8. And I1 = I2 = 2A. Power: P1 = V1.I1 = 2V x 2A = 4W. P2 = V2.I2 = 8V x 2A = 16W. Total power = 20W, so 2A in the centre winding (ignoring magnetising current). If R1=0, then this situation **defines** the winding voltage to be zero , so V1=0, this means zero magnetic flux, so all the flux (and hence all of the primary voltage) is moved over to V2, so: V2=10V, and I2 = 10V / 4Ω = 2.5A. Power = 10V x 2.5A = 25W. Same for R2=0, V2 is forced to 0, so: V1=10V, and I1 = 10V / 1Ω = 10A. Power = 10V x 10A = 100W. Another way to look at it: let's leave R1=1Ω, and let R2 vary from infinite to zero. For R2 open ckt (infinite), then all the voltage appears at V2, & V1=0. Total load power is zero. As R2 reduces, then V2 voltage starts to reduce, and V1 starts to increase, and load power increases. At R2=R1=1Ω, both the voltage and the currents will be equal, the voltages will sum to 10V, so 5V each, and 5A each, so 25W per winding, 50W total. As R2 reduces toward a 0, V2 reduces while V1 increases, until at R2=0 we get V2=0, V1=10V, I1=10A, and Power = 100W. So power increases quite rapidly as R2 reduces from 1Ω to 0Ω. We can think of this as a way of controlling how magnetic flux is split up from one path to another path.
Assuming unity coupling factor (otherwise it is incorrect) : 1. The load on V1 is 4 times greater, so 4*V1 = V2, as well as V = V1 + V2, yielding V1 = 2Vac, V2 = 8Vac. 2. If R1 = 0, V1 = 0, so V2 = 10Vac, therefore I2 = 2.5A. We must therefore have I1 = 2.5A. 3. V2 = 10Vac.
I would have assumed the flux divides evenly until considering short circuit cases, which tells me flux does not divide evenly. Lower resistance should get less flux. 1. V1=7.2 V2=2.8 (Thus P1=13W, P2=5.2W. Hmm.) 2. Any flux down this leg is resisted by the superconducting shorted turn. Seems undefined? The current has to be nonzero to counteract the input flux. Power is zero. I don’t know! 3. 10V, its a normal transformer on the left half since the right half is effectively disconnected.
1) 10V 2) 2.5A 3) 10V The reluctance imposed by R1 is n^2.s/R1, where s=j.2.pi.f The exact overall formula using non zero leg reluctances is tedious but straightforward.
I do not know, but here is my 2 cents (PS: I see I'm completely wrong by other's answers... Therefore I look forward for you video) -- If the core was ideal ( *no saturation* ): 1) V1 = V2 = 10/2 Vac (b/c the flux of the center portion splits 50/50 onto the other legs) 2) I1 → ∞ 3) V2 would still be 10/2 Vac -- With real ferrite core (i.e. *with saturation* ): 1) V1 = V2 = 10/2 Vac (if R1 and R2 are high enough for the core not to saturate) 2) As R1 → 0 the current in the "primary" increases. Eventually the core saturates and I think that then I1 will reach a maximum I1 = Isat and it will remain that way. V1 will drop to 0 as R1 → 0. 3) When the core saturates the magnetic flux will be Φsat and will stop increasing. V2 therefore will clamp also to V2_sat. Bonus: when the core saturates I_primary → ∞ (i.e. will become a short and bad things will happen)
1. Since the ratios are 1:1:1, V1 & V2 are same as the driving source or 10V. 2. Assuming an ideal transformer and windings, i.e no other resistances than the values of R1 & R2, and that all the core limbs are equal x-section shorting out R1 will completely short out the flux and draw unlimited current. 3. Because of a short on R1 (all flux diverted to R1) and therefore same on the driving source there can't possibly be a voltage across R2, so V2 = 0V. I'm now ready for the bad news as I've never considered this scenario before.
Completely wrong above. Just did some actual measurements - did not have access to a suitable ferrite core but had a pair of steel laminated UNICOREs from AEM Cores on which I wound up 3 coils as in the video, 23 turns each. also the signal source I have was incapable of driving such low resistances shown in the video so I scaled the values to 1k for R1 and 4k for R2 keeping the same ratio of loads. in addition had to keep the drive signal to the middle coil at 1V rms to prevent overloading of the drive circuit when R1 shorted. The input frequency is 10kHz. 1. Measured V1 at 433.2 mV rms and V1 at 417.5 mV rms. 2. R1 shorted: I1 measured at 17.4 mA rms. 3. R1 shorted: V2 measured at 787mV rms. As these are real values with real lossy magnetics etc, I'd be curious to see how they compare to the theoretical values using the resistors and input I used. PS sorry for the James T. Kirk method of solving problems.😊 Cheers Mark G
@@sambenyaakov Thanks for the response, look forward to the answers video. Just out of curiosity, other than a possibility of errors in measurements is there a problem of scaling by using different values in my attempt? EDIT: Further investigation revealed some issues with my ac voltmeter, I also recalibrated my Rigol scope. Using only the scope to do all the measurements I found that if the two secondary windings are either unloaded or loaded with equal resistances, the outputs end up being equal but only half the value of the applied voltage. this makes sense as the total applied flux is split in two equal values, one left and the other right of center limb - this is only true if the L & R magnetic path areas are identical - my cores seem to have a small difference between the two halves. Unbalancing the two loads unbalances the voltages in the direction where the lower resistor develops a lower voltage and the opposite for the higher value resistor. The sum of the two voltages add up to the value on the primary winding. If either of the outside windings is shorted out the opposite winding ends up with almost full primary voltage across it, an ideal transformer without any losses I suspect would get a full primary voltage on the winding with resistive load whose current would be V2 / R2. so in case of R2 = 4R and Vmax across it = 10V, this would end up at 2.5A Not sure what the current I1 would end up being, don't have a clamp meter to do the measurement, replacing R1 with a much lower resistor and measuring the voltage drop across it suggests that the current increases. Will be quite interesting to see the math behind it, hopefully it's not beyond my level of understanding. Thanks for a nice little head scratcher leading up to a weekend. Cheers Mark G
Since the windings are on opposite sides of the middle core, I *think* it's easier to imagine this as two seperate transformers. . Assuming stable power supply and no saturation, Initial voltage drop across each resistor Voltage drop across R1 should be 10Vac, with I1 10A & I2 2.5A. This already feels like I'm wrong. Shorting either resistor would immediately saturate the ferrite core on that side. But it would also saturate the primary core, dropping the impedance to the DC resistance (0 if ideal). Without a voltage drop on the primary you also can't have a secondary voltage, so.... Well, 0 makes me sure I'm even more wrong. By my calculations, the Professor has just invented the worse current-sense transformer ever conceived. I wouldn't normally post such an obviously wrong answer, but since there's only one other comment so far and I already told that guy that he was totally wrong, it only seems fair to admit that I don't have a clue either.
@@sambenyaakov But that was the only idea I could come up with! 😁 At this point I either need to find a big ferrite core in my parts bin or wait for the answer video!
@@SkippiiKai Or, if you are familiar with, build a magnetic circuit diagram with reluctances etc. from which the problem and solution can be better envisioned. OR, wait to my video.
@@sambenyaakovIs it possible to see it as two normal transformers with instead of the middle leg the two primaries in series? Is that really an equivalent circuit?
I would go by basic MMF balance in an ideal transformer. If assuming high mu_r core, then 3 MMF sources in an ideal transformer will be in parallel. So that N*Is = N*I1 = N*I2 => Is = I1 = I2. Also with energy conservation power input must be total power output, so Ps = P1+P2 or Vs*Is = V1*I1 + V2*I2 => Vs = V1+V2 = I1*R1+I2*R2.
Now :
1) Vs = Is*R1+Is*R2 = Is*(R1+R2) = Is*5 => Is = Vs/5 = 10/5 = 2A = I1 = I2
and V1 = 2*4 = 8V and V2 = 2*1 = 2V
2 & 3) if R1 is shorted => V1 = 0 but I1 != 0 but it also holds that Is = I1 = I2 due to MMF balance
so Vs = V1+V2 = V1 + 0 = 10V.
I1 = V1/4 = 2.5A = Is = I2
👍
If the coupling is ideal (infinite Ur core), then because the magnetic circuit is of a 3 MMFs connected in parallel, the two resistors will reftect to the voltage source side as a single series resistor of 4 + 1 = 5Ohms, therefore the source current as well as the two resistors' currents would be 10/5 = 2A (1:1:1 reflection). Thus, V1 would be 2*4 = 8V, and V2 would be 2*1 = 2V.
When R1 is shortet (=0Ohms) then the total reflected resistance would be 4+0 = 4 Ohms. and the 3 currents would be 10/4 = 2.5A, thus. V2 would be 2.5*4 = 10V, which makes the magnetic circuit behave as a two-winding Xformer with a 1:1 ratio.
The 3rd shorted winding will cancel out (opposite) it's leg's flux (shorted winding leg flux will be zero, thus circuit will become a simple ideal transformer).
Good answer! thanks for participating.
This is a real headscratcher (especially the last two questions). Here are my answers, happy to hear of the problem behind my reasonings, as I am really not sure:
1) V1 and V2 will be 5 Vac with a 180-degree phase shift. This is because the flux splits up from the middle excitation winding, giving half the excitation to each winding.
2) The currents should match in a sub-loop (right side) so allow for a closed magnetic loop path. However, the flux swing will be minimal in this path due to the loss of resistance. So 2.5 A (ac) is based on the exciting winding's current.
3) Because voltage can't be present on V1 (as this requires the resistance), the flux swing has to appear at the left sub-loop. Therefore, it's now a fully coupled circuit, i.e., V2 = 10 Vac.
I think the same way as you. So I agree that V1 = V2 = 5V due to flux splitting and we would have a V2 = 10V if we shorted R1.
1 wrong. 2, 3 OK.
@@SergiuCosminViorel Think in term of reluctances.
@@sambenyaakov reluctances not good. give another terms. reluctances rejected, are meaningless. give me something real
@@SergiuCosminViorel Hold on to m answer video
I'm not sure my sign conventions for the fluxes are the right way round but here's my take anyway..
Let:
V0 = 10V
Then:
(0) V0 = -NAdB0/dt
(1) V0 = -N(A/2)dB0/dt -N(A/2)dB0/dt
Assume direction of fluxes of each section are such that from (1) we get:
(2) V2 = N(A/2)dB1/dt +N(A/2)dB0/dt
(3) V1 = -N(A/2)dB1/dt +N(A/2)dB0/dt
where there is a common flux
- Φ0 flowing between each half of the centre section and each of the two side sections
- Φ1 flowing between the two side sections
(2)-(3) =>
(4) V2-V1 = -NAdB1/dt
(2)+(3) & (0) =>
(5) V2+V1 = V0
(4)+(5) =>
(6) V0 = 2V2 +NAdB1/dt
(6)+(0) =>
V0 = V2 +N(A/2)dB1/dt -N(A/2)dB0/dt
& (3) =>
V0 = V2 -V1
& (5) =>
V2+V1 = V2 -V1 =>
V1 = 0
& (5) =>
V2 = V0 = 10V.
But my flux sign convention can't be correct as it introduces an asymmetry in a symmetrical circuit, right?
Hold on to my answer video
Only from reading the first couple of comments i learned a lot totally new insights. Flux splitting and the summing of outputvoltages and so on. The drawing did me think of 3fase transformers from wich i never understand how the different fases can use the same core. I look so much out for the answervideolesson, in fact as always
Thanks for participating. Hold on to my answer video.
1) V1 = 2 V and V2 = 8 V The voltage of each secondary winding is related to the impedance reflected in the primary winding.
2) I1= 2.5 A The current of secondary 1 is equalised to the current delivered by secondary 2.
3) V2 = 10 V Since secondary 1 is short-circuited, the impedance seen from the primary is the load of secondary 2. Due to the 1:1 ratio, the voltage is the same as that of the primary.
👍👏Thanks for participating.
The more loaded a coil is, the more it will deflect the magnetic fields away from itself by creating an opposing field.
1 - 8V+2V. both coils will see half the magnetic flux of the middle one when open circuit. When loaded the more loaded one deflects the field away, so less voltage is induced.
2 - (10V/4Ohm)/2=1.75A. I1=Isupp/2, as it has to run just enough current to match the open circuit magnetic field magnitude.
3 - 10V. The magnetic path towards coil1(the right one) is effectively blocked, leaving out a 2 coil transformer.
2 is off. Thanks.
If R1 & R2 are infinite (I1=I2=0), then the answer is easy: the magnetic flux (Φ = B.A) generated by the centre winding splits equally between the other two core legs, and from Faradays law (V = NdΦ/dt) each of these windings sees half the voltage, ie: V1 = V2 = 5Vac.
However, this flux distribution is not maintained under the loaded condition, since I1 and I2 are not zero. In the reluctance diagram, the total MMF generated by centre leg sees two other MMFs in series. Since the nmbr of turns are the same, the currents must be the same, and the two voltages have to add to 10V. So we solve two simultaneous equations:
(1) V1 + V2 = 10.
(2) I1 = I2
(3) V1 / R1 = V2 / R2, where R1=1Ω, R2=4Ω.
Writing V2 in terms of V1 from (1), and subbing into (3) we get:
V1 / 1Ω = (10 - V1) / 4Ω --> V1= (2.5 - V1 /4Ω) --> 1.25x V1 = 2.5 ,
So V1 = 2, and V2 = 8.
And I1 = I2 = 2A.
Power: P1 = V1.I1 = 2V x 2A = 4W. P2 = V2.I2 = 8V x 2A = 16W.
Total power = 20W, so 2A in the centre winding (ignoring magnetising current).
If R1=0, then this situation **defines** the winding voltage to be zero , so V1=0, this means zero magnetic flux, so all the flux (and hence all of the primary voltage) is moved over to V2, so:
V2=10V, and I2 = 10V / 4Ω = 2.5A. Power = 10V x 2.5A = 25W.
Same for R2=0, V2 is forced to 0, so: V1=10V, and I1 = 10V / 1Ω = 10A. Power = 10V x 10A = 100W.
Another way to look at it: let's leave R1=1Ω, and let R2 vary from infinite to zero. For R2 open ckt (infinite), then all the voltage appears at V2, & V1=0. Total load power is zero. As R2 reduces, then V2 voltage starts to reduce, and V1 starts to increase, and load power increases. At R2=R1=1Ω, both the voltage and the currents will be equal, the voltages will sum to 10V, so 5V each, and 5A each, so 25W per winding, 50W total. As R2 reduces toward a 0, V2 reduces while V1 increases, until at R2=0 we get V2=0, V1=10V, I1=10A, and Power = 100W. So power increases quite rapidly as R2 reduces from 1Ω to 0Ω. We can think of this as a way of controlling how magnetic flux is split up from one path to another path.
Thanks for participating. Sorry incorrect, Hold on to my answer video.
Assuming unity coupling factor (otherwise it is incorrect) :
1. The load on V1 is 4 times greater, so 4*V1 = V2, as well as V = V1 + V2, yielding V1 = 2Vac, V2 = 8Vac.
2. If R1 = 0, V1 = 0, so V2 = 10Vac, therefore I2 = 2.5A. We must therefore have I1 = 2.5A.
3. V2 = 10Vac.
Good answer! Thanks for participating.
I would have assumed the flux divides evenly until considering short circuit cases, which tells me flux does not divide evenly. Lower resistance should get less flux.
1. V1=7.2 V2=2.8
(Thus P1=13W, P2=5.2W. Hmm.)
2. Any flux down this leg is resisted by the superconducting shorted turn. Seems undefined?
The current has to be nonzero to counteract the input flux. Power is zero.
I don’t know!
3. 10V, its a normal transformer on the left half since the right half is effectively disconnected.
Sorry incorrect. Thanks for response.
1. Amp turns must balance,half Mmf on each leg so 10v on both for conservation of energy.
2. As I 2.
3.10v .
Sorry incorrct. Hold on to my answer video. Thanks for participating .
1) 10V
2) 2.5A
3) 10V
The reluctance imposed by R1 is n^2.s/R1, where s=j.2.pi.f
The exact overall formula using non zero leg reluctances is tedious but straightforward.
Thanks for participating. Sorry , 1 is incorrect, Hold on to my answer video.
@@sambenyaakov1) V1=Vs*R1/(R1+R2)=2, V2=Vs*R2/(R1+R2)=8
@@sambenyaakovV1=Vs*R1/(R1+R2)=2V, V2=Vs*R2/(R1+R2)=8V, Is=2A
I do not know, but here is my 2 cents (PS: I see I'm completely wrong by other's answers... Therefore I look forward for you video)
-- If the core was ideal ( *no saturation* ):
1) V1 = V2 = 10/2 Vac (b/c the flux of the center portion splits 50/50 onto the other legs)
2) I1 → ∞
3) V2 would still be 10/2 Vac
-- With real ferrite core (i.e. *with saturation* ):
1) V1 = V2 = 10/2 Vac (if R1 and R2 are high enough for the core not to saturate)
2) As R1 → 0 the current in the "primary" increases. Eventually the core saturates and I think that then I1 will reach a maximum I1 = Isat and it will remain that way. V1 will drop to 0 as R1 → 0.
3) When the core saturates the magnetic flux will be Φsat and will stop increasing. V2 therefore will clamp also to V2_sat.
Bonus: when the core saturates I_primary → ∞ (i.e. will become a short and bad things will happen)
Thanks for participating. Sorry incorrect, Hold on to my answer video.
1. Since the ratios are 1:1:1, V1 & V2 are same as the driving source or 10V.
2. Assuming an ideal transformer and windings, i.e no other resistances than the values of R1 & R2, and that all the core limbs are equal x-section shorting out R1 will completely short out the flux and draw unlimited current.
3. Because of a short on R1 (all flux diverted to R1) and therefore same on the driving source there can't possibly be a voltage across R2, so V2 = 0V.
I'm now ready for the bad news as I've never considered this scenario before.
Completely wrong above.
Just did some actual measurements - did not have access to a suitable ferrite core but had a pair of steel laminated UNICOREs from AEM Cores
on which I wound up 3 coils as in the video, 23 turns each.
also the signal source I have was incapable of driving such low resistances shown in the video so I scaled the values to 1k for R1 and 4k for R2 keeping the same ratio of loads. in addition had to keep the drive signal to the middle coil at 1V rms to prevent overloading of the drive circuit when R1 shorted. The input frequency is 10kHz.
1. Measured V1 at 433.2 mV rms and V1 at 417.5 mV rms.
2. R1 shorted: I1 measured at 17.4 mA rms.
3. R1 shorted: V2 measured at 787mV rms.
As these are real values with real lossy magnetics etc, I'd be curious to see how they compare to the theoretical values using the resistors and input I used.
PS sorry for the James T. Kirk method of solving problems.😊
Cheers
Mark G
Sorry incorrect. Hold on to my answers video.
@@sambenyaakov Thanks for the response, look forward to the answers video.
Just out of curiosity, other than a possibility of errors in measurements is there a problem of scaling by using different values in my attempt?
EDIT:
Further investigation revealed some issues with my ac voltmeter, I also recalibrated my Rigol scope.
Using only the scope to do all the measurements I found that if the two secondary windings are either unloaded or loaded with equal resistances, the outputs end up being equal but only half the value of the applied voltage. this makes sense as the total applied flux is split in two equal values, one left and the other right of center limb - this is only true if the L & R magnetic path areas are identical - my cores seem to have a small difference between the two halves.
Unbalancing the two loads unbalances the voltages in the direction where the lower resistor develops a lower voltage and the opposite for the higher value resistor.
The sum of the two voltages add up to the value on the primary winding.
If either of the outside windings is shorted out the opposite winding ends up with almost full primary voltage across it, an ideal transformer without any losses I suspect would get a full primary voltage on the winding with resistive load whose current would be V2 / R2. so in case of R2 = 4R and Vmax across it = 10V, this would end up at 2.5A
Not sure what the current I1 would end up being, don't have a clamp meter to do the measurement, replacing R1 with a much lower resistor and measuring the voltage drop across it suggests that the current increases. Will be quite interesting to see the math behind it, hopefully it's not beyond my level of understanding.
Thanks for a nice little head scratcher leading up to a weekend.
Cheers
Mark G
@@markg1051 Thanks for your thoughts. Have a nice weekend
@@sambenyaakov you too.
V1=V/(1+R2/R1)=2V, V2=V/(1+R1/R2)=8V, 2.5A, 10V. Thank you for the video!
Because, V1+V2=V from flux split, and i+i2=0, i+i1=0, i1=i2, V1/R1=V2/R2 assuming mu=infinity, from mmf theorem.
Correct and elegant. Thanks for participating.
@@zaikindenis1775could ypu explain why does voltage divide this way (R1/R2) instead of R1/(R1+R2) like parallel currents through resistors would?
@@electrowizard2000 Actually, I have written it confusing. 1/(1+R1/R2) is the same as R2/(R1+R2).
👍🙏❤️
👍🙏🤞
Since the windings are on opposite sides of the middle core, I *think* it's easier to imagine this as two seperate transformers. .
Assuming stable power supply and no saturation, Initial voltage drop across each resistor Voltage drop across R1 should be 10Vac, with I1 10A & I2 2.5A. This already feels like I'm wrong. Shorting either resistor would immediately saturate the ferrite core on that side. But it would also saturate the primary core, dropping the impedance to the DC resistance (0 if ideal). Without a voltage drop on the primary you also can't have a secondary voltage, so.... Well, 0 makes me sure I'm even more wrong. By my calculations, the Professor has just invented the worse current-sense transformer ever conceived. I wouldn't normally post such an obviously wrong answer, but since there's only one other comment so far and I already told that guy that he was totally wrong, it only seems fair to admit that I don't have a clue either.
Drop the assumption that core will saturate. It will not.
@@sambenyaakov But that was the only idea I could come up with! 😁 At this point I either need to find a big ferrite core in my parts bin or wait for the answer video!
@@SkippiiKai Or, if you are familiar with, build a magnetic circuit diagram with reluctances etc. from which the problem and solution can be better envisioned. OR, wait to my video.
@@sambenyaakovIs it possible to see it as two normal transformers with instead of the middle leg the two primaries in series? Is that really an equivalent circuit?