I define every circuit element of a primary coil of a transformer that is connected to a voltage source and calculate \$I_1\$ (current of the primary loop). I then connect it to a secondary coil with a known number (\$N\$) turns.
Regardless of the values of the secondary coil, I always can calculate \$I_2\$ (current in secondary coil) by using $$I_2=\frac{N_2}{N_1}I_1$$ Does it mean that an ideal transformer is an ideal current source?
Best Answer
An ideal transformer will have infinite magnetization inductance and, because of this, it will take zero primary current when there is zero secondary current. This doesn't make it an ideal current source.
Look at the equivalent circuit: -
An ideal transformer doesn't have any series elements so basically they become shorts. It doesn't have any parallel losses so Gc becomes open circuit and the only contentious thing left is the primary magnetization inductance shown with a blue square surrounding it. For an ideal transformer this is infinite in value and no current can flow into the primary if there is no load on the secondary. If there is load on the secondary then the primary impedance looking in will have an impedance of: -
\$P_Z = (\dfrac{N_1}{N_2})^2 \times S_Z\$
You could argue that a CT (current transformer) might be an ideal current source but it isn't. With a high primary current of (say) 100 amps it still has a primary magnetization inductance that will naturally limit the open circuit secondary output voltage despite rumours of thousands of volts being talked about on some websites.