Question

Two loops of wire are arranged so that a changing current in one, the primary, will induce a current in the other, the secondary. The secondary loop has twice as many turns as the primary loop. As long as the current in the primary is steady at 3.0 A, the current in the secondary will be:

Answer 1

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 2

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