The first question: the influence of the switch on the circuit. Answer: When the switches are inconsistent, there is only one resistor on the left and right sides of the transformer (the upper 2 ohm resistor), so there is no current on this resistor, because a loop cannot be formed. The resistance on it is like nothing. However, if the lower switch S is turned on, the upper resistor will generate current, so the two situations are different. The second question: The detailed steps to answer the first question are as follows: Remove the upper resistor, so that it is a 1: 2 transformer, and connect a 4 ohm resistor on the secondary side. What is the impedance seen in the primary? Are you familiar with the formula of impedance transformation? That is to say, the primary impedance ratio of the transformer is equal to the square of the turns ratio. Since the primary turns ratio is 1: 2 and the impedance ratio is of course 1: 4, now the secondary is connected with a 4 ohm resistor, which is of course 1 ohm from the primary point of view. The formula is derived as follows: if the turns ratio of the primary transformer is 1: n, the voltage u is connected to the primary and the secondary voltage is nu; If the secondary resistance r is connected, the current nu/r will be generated in the secondary; According to the law that the primary current of transformer is inversely proportional to the number of turns, the current N 2 * U/R will be generated in the primary coil, and the impedance z = u/(n from the primary point of view. Third, the second question of this question is more difficult, and the final answer is:-1 euro. Can you understand this? I remind you of two points: 1. Find the impedance between point A and point B, which coincides with Z=U/I, where U is the voltage between these two points, and there should be a point above U representing the vector, which can't be typed here; There should also be a point on I, indicating the current vector flowing from point A. Point 2: The negative sign of resistance here does not mean less than zero, just like the magnitude of force, -3 N, and the negative sign represents the direction of the vector. Here, the size of the resistor is 1ω, but when a sinusoidal AC voltage is applied to the resistor, the generated current is opposite to the applied voltage. The detailed answer to the second question is written in four steps. Please see:

The answer to the second question of your original question is too simple. The answer to the first step should be equal to -2, with a minus sign missing. So it's hard to understand. I would like to add again: an amendment to the answer to the second question. I've been thinking about this problem over and over again, and I'm not sure. I have carefully read what I have done for you, and I think there are flaws. The problem also lies in the phase of the current. What puzzled me at first was that the complex impedance is a negative real number, so the current generated by this circuit is opposite to the voltage at both ends, so the power consumed by the circuit should be negative, that is to say, the circuit not only does not consume active power, but also feeds back power to the power supply. This circuit is actually a power supply that is higher than the supply voltage, but this circuit is obviously not the case. I try other methods to solve this problem: Method 1: Draw the original circuit in another form, as shown below:

Because the switch at the bottom in the original drawing connects the primary side and the secondary side of the transformer together, it can be regarded as an autotransformer. Here, we only regard its function as impedance transformation. When the resistance of 2 Ω is switched to the primary side, it is still 2 Ω. The resistance of 4 Ω changes to the primary side and becomes1Ω. The parallel connection of 2 ω and 1 ω results in z = (2/3) ω. Method 2: Calculate the voltage on the resistors of 1 and 2 Ω with electric power, so the power is (u 2)/2; The voltage on the 4Ω resistor is 2U, so the power is (4U 2)/4, so the total power is the sum of them, which is equal to (U2)/2+U2 = (3U2)/2; 2. If the impedance is z from the input, there is p = u 2/z, and if the primary power of the ideal transformer is equal, there is u 2/z = (3u 2)/2, so Z=2/3. So, I think your answer is correct. Influenced by the answer in your question, I tried my best to get the same result as the original answer before I made an answer that was not my heart. Now I tend to think that the correct answer is (2/3) ω. But your problem-solving process is still insufficient. ) The above is for your reference.