Answer:
The current in its secondary coil is 8 A.
Explanation:
Given;
number of turns in the primary coil, [tex]N_P[/tex] = 600 turns
number of turns in the secondary coil, [tex]N_S[/tex] = 150 turns
current in the primary coil, [tex]I_P[/tex] = 2 A
current in the secondary coil, [tex]I_S[/tex] = ?
For an Ideal transformer, the emf in the primary winding is equal to the emf in the secondary winding. [tex]N_PI_P = N_SI_S[/tex]
The current in the secondary coil, can be calculated by applying turn ratio;
[tex]\frac{N_P}{N_S} = \frac{I_S}{I_P} \\\\I_S = \frac{N_PI_P}{N_S} \\\\I_S = \frac{600*2}{150} = 8 A[/tex]
Therefore, the current in its secondary coil is 8 A.
