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I understand that the secondary circuit in a current transformer cannot be left open, as it will result in a very potential difference being induced in the core of the transformer and permanently damage it. However, I'm having trouble understanding why the secondary winding is always shorted. Shouldn't it actually have a high resistance, since the voltage is high and current through it is low? Or is there perhaps a definition of 'shorted' which refers to something other than a low resistance path?

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    \$\begingroup\$ Why did you decide that "the voltage is high and current through it is low" in a secondary of a current transformer? \$\endgroup\$ – Dmitry Grigoryev Jan 18 at 12:36
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    \$\begingroup\$ According to my textbook, a current transformer is essentially a step up transformer (it steps up the voltage). It's used for the purpose of measuring high values of current using low range ammeters. \$\endgroup\$ – PhysicsStudent Jan 20 at 3:47
  • \$\begingroup\$ OK, fair enough. I would expect your textbook to also mention that shorting the secondary is exactly what makes the difference, if not that's indeed misleading. \$\endgroup\$ – Dmitry Grigoryev Jan 21 at 7:48
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Any resistance on the current tranformer secondary will be transformed into resistance on the primary by a factor of \$ \frac {1}{n^2} \$ where \$ n \$ is the turns ratio. To minimise the primary impedance you need to short the CT secondary.

You may find it helpful to consider CTs as complimentary to VTs.

  • VTs are happy with secondary terminals open-circuit. They are then unloaded.
  • CTs are happy with secondary terminals short-circuit. They are then unloaded!
  • VTs do not like short-circuits on the secondary.
  • CTs do not like open-circuits on the secondary.

Or is there perhaps a definition of 'shorted' which refers to something other than a low resistance path?

Nope. Use the best short-circuit you can make.

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  • \$\begingroup\$ Why would we want to minimise primary impedance? Isn't the entire point of the CT to step down primary current so as to measure it easily? \$\endgroup\$ – PhysicsStudent Jan 18 at 11:31
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    \$\begingroup\$ No, what I'm saying is that adding a CT and, for example, an ammeter on the secondary adds impedance (resistance) into the primary. The energy to power the meter has to come from the primary circuit. So, the idea is to keep the secondary resistance very low to minimise the effect on the primary circuit. \$\endgroup\$ – Transistor Jan 18 at 11:54
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For any transformer with AC we have

$$ \dfrac{N_p}{N_s} = \dfrac{V_p}{V_s} = \dfrac{I_s}{I_p} $$

When using a transformer for voltages we apply a voltage to the the primary and get one on the secondary. The current in the the primary is \$I_p = I_s \cdot\dfrac{N_s}{N_p}\$ plus a bit for the magnetising current. We can model the real transformer as above with an inductor in parallel with the primary winding to represent the magnetising inductance.

Now consider the action of a current transformer. We want the primary to look as much like a short circuit as possible so it does not change the current being measured. For this reason the load across the secondary wants to be low impedance. The secondary voltage reflected back to the primary is then very small and the magnetising inductance appears to be shorted out meaning primary current is not affected by adding the current transformer.

In practice the secondary usually has a small value resistor across it so you can sense the primary current as a voltage on the secondary. If this resistance is too big the primary voltage increases and the current transformer affects the primary current.

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