
Originally Posted by
franko26
I copied and pasted the question at the top of the comment, your guess is as good as mine.
Originally Posted by
franko26
I copied and pasted the question at the top of the comment, your guess is as good as mine.
I've seen and worked on this voltage before on a temporary transformer with primary and secondary taps.
7200 if i remember is a rural voltage. (Line to neutral) PhPh will be 12,456v. Odd looking voltage I know but I'm sure this is what you will see on the nameplate.
So if the 7200 is Lineneutral than 1:20 is correct.
Unless for whatever strange reason they want an "indirect ratio or percent ratio" (a test equipment i have which i hate) Then the math is as follows.
360 / 7200 = 0.05x100% = 5 ; 1:5

Originally Posted by
MaurerAdam
I've seen and worked on this voltage before on a temporary transformer with primary and secondary taps.
7200 if i remember is a rural voltage. (Line to neutral) PhPh will be 12,456v. Odd looking voltage I know but I'm sure this is what you will see on the nameplate.
So if the 7200 is Lineneutral than 1:20 is correct.
Unless for whatever strange reason they want an "indirect ratio or percent ratio" (a test equipment i have which i hate) Then the math is as follows.
360 / 7200 = 0.05x100% = 5 ; 1:5
Based on my knowledge, I think for turn ratio here is the rule:
1 we need to just consider phph voltages not phground; so we don't care about delta or wye configuration.
2 turn ratio is always dividing primary over secondary not secondary over primary.
in this case, primary is 7200 and secondary is 360 so ratio is 20:1.
just we need to pay attention which one is primary and which one is secondary.

Correct!
Originally Posted by
iwanttotest
Wouldn't 1:20 be a stepup transformer? The transformer in question is a stepdown.
If the secondary winding has fewer turns than the primary, a lower voltage will be induced in the secondary. This type of transformer is called a stepdown transformer.
A secondary coil with twice as many turns as the primary will be cut twice as many times by the magnetic flux, and twice the applied primary voltage will be induced in the secondary. This transformer is known as a stepup transformer.
My answer would be 11.55:1 assuming 7200 is the line voltage.
You are 100% correct on this. 7,200V (Line) WYE = 4,160V (Phase). On the Delta side Vline=Vphase. You should always use the phase voltages to get a ratio, especially when dealing with a WYEDelta configuration. Using phase voltages your math should be 4,160V/360V = 11.55:1 Ratio not 1:11.55. Turns ration is stated with the Primary Turns First. So in this case the Primary voltage is higher so the ratio will begin with the higher number of turns.
Last edited by stevenl569; January 6, 2018 at 08:33 PM.
Reason: Typo

Originally Posted by
lester mcmanaway
My guess would be line voltage then. So 7200/1.732 = 4157/360 roughly 11.55:1 would have been what I was thinking.
I would have to agree with the 11.55:1 Ratio. I'm pretty sure in a WYE/DELTA you have to divide the primary by the square root of 3 first to get the correct ratio.

Originally Posted by
todgayle
I would have to agree with the 11.55:1 Ratio. I'm pretty sure in a WYE/DELTA you have to divide the primary by the square root of 3 first to get the correct ratio.
i agree.

Originally Posted by
iwanttotest
Wouldn't 1:20 be a stepup transformer? The transformer in question is a stepdown.
If the secondary winding has fewer turns than the primary, a lower voltage will be induced in the secondary. This type of transformer is called a stepdown transformer.
A secondary coil with twice as many turns as the primary will be cut twice as many times by the magnetic flux, and twice the applied primary voltage will be induced in the secondary. This transformer is known as a stepup transformer.
My answer would be 11.55:1 assuming 7200 is the line voltage.
If not stated its always assumed line voltage and what your suggesting is correct. I am not sure why they would do 20:1 as a correct answer in this problem or if that answer is confirmed or assumed correct. but who knows.

Originally Posted by
Lestracy31
If not stated its always assumed line voltage and what your suggesting is correct. I am not sure why they would do 20:1 as a correct answer in this problem or if that answer is confirmed or assumed correct. but who knows.
I just got the question and selected the 20:1 response and it was wrong.

Originally Posted by
franko26
Transformers
A wyedelta transformer has a primary voltage of 7200 V and a secondary voltage of 360 V. What is the turns ratio?
Your Answer: 1:20
The test considered this answer wrong. The primary to secondary voltage ratio is 1 to 20. I'm sure the correct answer is 20:1, but grammatically speaking, the 1:20 answer is more correct. The following is from the noted study material:
Since there is a ratio of 1 to 4 between the turns in the primary and secondary circuits, there must be a ratio of 1 to 4 between the primary and secondary voltage and a ratio of 4 to 1 between the primary and secondary current. Please advise.
This post took a very strange turn of events, as most people got stuck on the turns ratio vs how to actually solve the question.
The turns ratio is the ratio of turns of the primary to the number of turns to the secondary of a SINGLE PHASE TRANSFORMER. Since this is a Wye/Delta transformer, we know it is 3 phase, and will have to convert the voltages to phasetoground (single phase) voltages to perform the ratio.
The Mnemonic I use is WIDE, which stands for Wye transformers I is equal between Line and Phase, and Delta transformers E (electromagnetic force which is measured in volts) is equal between Line and Phase.
To find the opposite I (current) or E (voltage), you must divide by square root of 3 (or simplified to 1.732).
The high side is WYE at 7200V, thus we must divide by 1.732 to find Phase Voltage.
7200V/1.732=4157V
The low side is DELTA at 360V, thus Phase Voltage equals Line Voltage at 360V.
Turns ratio is derived from the equation, where A=turns ratio: A=N1/N2=V1/V2=I2/I1 thus A=V1/V2. (https://www.electricalclassroom.com/...ulatoronline/)
A=V1/V2
A=4157V/360
A=11.55
or stated as turns ratio is equal to 11.55:1 (the first number represents N1 or Primary winding turns, and the second number represents N2 or the secondary winding turns. A=N1/N2)

Yup
Originally Posted by
lester mcmanaway
My guess would be line voltage then. So 7200/1.732 = 4157/360 roughly 11.55:1 would have been what I was thinking.
WYEDELTA
H side would be 7200, you would divide /by 1.732 get you LL of 4157.044 divide this by your sec voltage of 360. 11:55:1. Good Stuff

Originally Posted by
Miller1085
WYEDELTA
H side would be 7200, you would divide /by 1.732 get you LL of 4157.044 divide this by your sec voltage of 360. 11:55:1. Good Stuff
I would disagree on one point. 7200V would be your V_{LL} and 4157.044V would be your V_{LG}. Anytime you read voltage on a WYE system, phase to phase (line to line) will be a larger number than your phase to ground (line to ground).
(V_{LL}/sqrt(3)) = V_{LG}
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