Solution

For two identical ideal transformers with a 1:3 turns ratio, the no‑load current is not zero due to the finite magnetizing inductance. Let$L_m$ be the magnetizing inductance referred to the primary of each transformer.

The input loop (100 V source) forces a current I through the series connection formed by the primary of the first transformer (P1) and the secondary of the second (S2).

In transformer 1, the secondary S1 is open‑circuited, so the ideal primary current is zero and the entire current I flows through the magnetizing inductance $L_m$:

$V_{P1} = \mathrm{j}\omega L_m I$

In transformer 2, the primary P2 is open; the secondary S2 carries I, and its magnetizing inductance referred to the secondary is $(3)^2L_m = 9L_m$so that

$V_{S2} = \mathrm{j}\omega (9L_m) I = 9\,V_{P1}$

The source imposes $V_{P1} + V_{S2} = 100, so 10\,V_{P1} = 100 \Rightarrow V_{P1} = 10\ \text{V}$

By the turns ratio of the first transformer, $V_{S1} = 3\,V_{P1} = 30\ \text{V}$
In the second transformer,$V_{P2} = V_{S2}/3 = 90/3 = 30\ \text{V}$
The output voltage between the free ends of S1 and P2 is the sum:

$V_{\text{out}} = V_{S1} + V_{P2} = 30 + 30 = 60\ \text{V}$

$\boxed{V_{\text{out}} = 60\ \text{V}}$