Question

In an ideal transformer connected to a 240v A.C with number of turns in primary coil are 1000 and in secondary coil are 50 turns output connected to load of 10 ohm current passes through load?

Answer 1

Current, that passes through load

$$I=U2/R$$

where U2 - voltage on secondary coil;

R - resistance of load.

Voltage on secondary coil

$$U2=U1*(W2/W1)$$

where U1 - voltage on primary coil;

W1 - number of turns in primary coil;

W2 - number of turns in secondary coil.

So

$$I=U1*W2/W1*R = 240*50/1000*10 = 1.2 A$$