Why Current Transformer's secondary is short circuited ?

[Jeffrey Peter]

Consider a 1000/1 CT,
Secondary turns = 1000
Primary turns = 1

 

[Tahmid]

When operating, the secondary is not exactly shorted directly. A resistor is connected across the secondary. The resistor is used as a "current to voltage converter."

Let's say you have 10A current through the primary turn. So, current through secondary is 10/1000 A = 0.01A. By placing the resistor there, the current is "converted to voltage". V=IR. So, voltage across resistor (and thus across secondary) is 0.01*R. If we use a 200R resistor, the voltage will be 0.01*200V = 2V.

So, voltage drop across 200R resistor = 2V, when primary current = 10A.

However, remember that this is AC, and should be converted to DC. You may wish to use a precision rectifier.

Hope this helps.
Tahmid.

 

[FvM]

Two simple answers:
- by shorting the output, you also minimize the primary voltage drop across the CT
- minimizing the current transformer voltage reduces also the flux, transformer VA and required core size

 

[Jeffrey Peter]

In a current transformer,
if the current is stepped down then the voltage will be stepped up so if we have 230 V on primary then 1000*230 =230K on the secondary so I = 230K/R.... why we don't take in this manner ?
I'm a bit confused at this part ..

 

[FvM]

so if we have 230 V on primary

You don't have 230 V "on" the current transformer primary. I presume, you want 230 V at the load and about no voltage drop at the CT.

Your calculation can be understood however as a warning about the risk of operating the CT secondary open and connecting a heavy load to the primary circuit.