Respuesta :
Answer: 1A
Explanation: for an ideal transformer, the relationship between number of turns and current is given below as
Ip/ Is = Ns/ Np
Where Ip = current in primary loop = 3A
Is = current in secondary loop =?
Np = number of turns in primary loop = n
Ns = number of turns in secondary loop = 3n
By substituting the parameters, we have that
3/Is = 3n/n
3/Is = 3
3 = 3 × Is
Is = 3/3 = 1A
Current in the secondary loop is 1A
Answer:
the current in the secondary loop will be = 1.5A
Explanation:
In an ideal transformer, conservation of energy requires that power out to equal power in Thus, Ip x Vp = Vs x Is, or Ip = Vs x Is / Vp.
Since Vs = Ns x Vp / Np, Ip can be rewritten as Ip = ( Ns x Vp / Np ) x Is / Vp, or equivalently, Ip = Ns x Is / Np, or Ns x Is = Np x Ip
Thus, Np x Ip = Ns x Is
Where;
Ip = current in primary loop = 3A
Is = current in secondary loop
Np = number of turns in primary loop = n
Ns = number of turns in secondary loop = 3n
Now, we are told that the current in the primary loop is 3 A. So, Ip = 3A
We are also told that the secondary loop has twice as many turns as the primary loop. Thus, let's call the number of of turns in primary loop "y" Thus, Np = y and Ns = 2y
Let's plug these values into;
Np x Ip = Ns x Is
y x 3 = 2y x Is
Divide both sides by 2y;
3y/2y = 2y(Is)/2y
2y will cancel out, on the right and y will cancel out on the left,
so Is = 3/2A = 1.5A
