How many loops is on the secondary coil of a set up transformer where the primary voltage is 120V and...?

Question

...the primary current is 0.6 A, the secondary voltage is 3600 V, and there are 350 turns on the primary coil? What current is flowing through the secondary coil?

Answer 1

20mA

The equation is (Vprim/Vsec)=(Isec/Iprim)

Substitute in your known, we get (120V/3600V) = (Isec/0.6A)

Cross multiply and find Isec. 120V * 0.6A = 3600V * X

If power in equals power out(negate losses), then you have 72W = 3600V * X

Divide both sides by 3600V, and you get 0.02A, or 20mA.

Answer 2

The transformer works in reality with AC (alternating modern-day) and if the open circuit transformer feels fairly warmth it has used a small volume of ability to warmth it. besides the undeniable fact that, as modern-day enters the established coil it builds up a magnetic field contained in the cord and iron middle yet even as modern-day reverses the field collapses giving again the potential used to verify it. even as the secondary coil is loaded and attracts ability the established coil can no longer recuperate each and each of the potential used to verify the magnetic field.

Answer 3

Transformer calculation:

Volts in = Primary turns
---------- -----------------
Volts out Secondary turns

Rearrange for variable solving and (assuming I did it right, I'm English, not Math) I get 10,500 turns on the secondary coil.

Transformers don't change power so watts in = watts out...watt is volt x amp. Rearranging, and assuming I did it right, I get 0.02 amps on secondary coil

Answer 4

N2 = N1 * (V2/V1) = 350 * (3600 / 120) = 10500 turns

I2 = I1 * (N1/N2) = 0.6 * (350 / 10500) = 0.02 A