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| ### Statement |
| ### Statement |
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| $11.3.25.$ [Insert the problem statement] |
| $11.3.25.$ |
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| There are two identical ideal transformers with the same transformation co- |
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| efficient of 1 : 3. The primary winding of one of them is connected in series |
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| with the secondary of the second, and the free ends of these windings are con- |
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| nected to the AC network with a voltage of 100 V. The secondary winding of |
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| the first transformer is connected in series with the primary winding of the |
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| second. Determine the amplitude of the AC voltage between the other ends |
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| of the windings. |
| ### Solution |
| ### Solution |
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| 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. |
| 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. |
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| 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). | | 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). |
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| 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 $L_m$: | | 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 $L_m$: |
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| $V_{P1} = \mathrm{j}\omega L_m I$ | | $V_{P1} = \mathrm{j}\omega L_m I$ |
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| 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 | | 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 |
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| $V_{S2} = \mathrm{j}\omega (9L_m) I = 9\,V_{P1}$ | | $V_{S2} = \mathrm{j}\omega (9L_m) I = 9\,V_{P1}$ |
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| The source imposes $V_{P1} + V_{S2} = 100, so 10\,V_{P1} = 100 \Rightarrow V_{P1} = 10\ \text{V}$ | | The source imposes $V_{P1} + V_{S2} = 100, so 10\,V_{P1} = 100 \Rightarrow V_{P1} = 10\ \text{V}$ |
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| By the turns ratio of the first transformer, $V_{S1} = 3\,V_{P1} = 30\ \text{V}$ | | 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}$ | | 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: | | The output voltage between the free ends of S1 and P2 is the sum: |
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| $V_{\text{out}} = V_{S1} + V_{P2} = 30 + 30 = 60\ \text{V}$ | | $V_{\text{out}} = V_{S1} + V_{P2} = 30 + 30 = 60\ \text{V}$ |
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| $\boxed{V_{\text{out}} = 60\ \text{V}}$ | | $\boxed{V_{\text{out}} = 60\ \text{V}}$ |
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