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A power system book asked me this question, but honestly I don't understand the circuit of question and the solution which author provided.

Question:

There are 3 primary currents, \$I_R,I_S\$,and \$I_T\$ with \$40\sqrt{3}A\$, of the three phase transformer, and the current ratio of the currentvtransformer (CT)is \$200/5\$. What is the current value of ammeter \$A\$?

enter image description here

The solution:

current value of Ammeter \$A =|\vec I_T-\vec I_R|=\sqrt{3}\vec I_R=\sqrt{3} \times 40\sqrt{3} \times \frac{5}{200}=3A\$

Sorry, I am not working on the power transmission and transformer, so I will have some basic problems, which the book did't mention.

  1. What does the symbol, which I circled in red, mean?
  2. Why does current value of ammeter \$A =|\vec I_T-\vec I_R|\$, not \$A =|\vec I_T+\vec I_R|\$, or \$A =|\vec I_T+\vec I_R+\vec I_S|\$
  3. Does \$I_R,I_S\$,and \$I_T\$ have some specific meaning, or I can just replace them with \$I_1, I_2\$, and \$I_3\$?
  4. Why does \$|\vec I_T-\vec I_R|=\sqrt{3}\vec I_R?\$
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  1. The symbol is of a current transformer, or more correctly, 2 wired in anti-parallel,

  2. As they are wired in opposite directions, if they had the same level of current they would cancel each other out, as such you end up with the difference in current between the 2 coils, IR - IT, There is no transformer measuring IS, so its not part of the math

  3. Without knowing the context, I would say your safe to substitute with I1,I2,I3, It appears to just be transformer phases

  4. Best refer you to here, the square root of 3 is a quirk of 3 phase systems when measuring between 2 phases, Square Root of 3

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As @Reroute says, these are current transformers (CTs). The primary winding is 1 turn and is not shown on schematics for simplicity. The overwhelming convention used is to draw the CTs with the implied understanding that the polarity dots are on the same side of the transformer (and we leave off the dots) for both windings. This is because instrument transformers (CTs and VTs) are subtractive polarity per standards (e.g. C57.12-1987 section 5.7.1).

snip

In the figure above the implied dots are on X1 and H1. If you have a reference current going into a dot on one side, the corresponding current on the other winding will be coming out of the dot (and vice versa)

Here is example with the dots and labels:

enter image description here

Here is same example except we drop the labels and dots because they are implied.

enter image description here

Ignoring affects of saturation or accuracy (which you must consider sometimes), the secondary current (in blue) has a magnitude that is the primary current divided by the transformer ratio. The phase angle is the same. So, in my last figure, if the primary (red) current is 100A @ 45 degrees (and the CT has a 20:1 ratio) then the secondary current (blue) is 5A @ 45 degrees.

I think from this you can easily solve you circuit. The \$\sqrt{3}\$ shows up in three-phase power when the quantities (voltages or currents) are perfectly balanced in magnitude and phase angle. Subtracting one from another produces a quantity that is \$\sqrt{3}\$ larger. e.g. 120@0 degrees minus 120@-120 degrees = 207.8@30 degrees. Magnitude is \$\sqrt{3}\$ larger than 120. This only occurs when they have same magnitude and are separated by 120 degrees. You are better off understanding this simple reason than trying to memorize when to use the \$\sqrt{3}\$.

R,S,T are just phase labels. In the U.S. we typically call them A, B, C.

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