Question 1
Assuming an ideal transformer, the Thevenin's equivalent voltage and impedance as seen from the terminals \(x\) and \(y\) for the circuit in figure are:

Solution
- The turn ratio is \(N_p:N_s = 1:2\).
- Thevenin Voltage (\(V_{Th}\)): \(V_{Th}\) is the open circuit voltage across the secondary. Since the primary current is zero, \(V_p\) equals \(V_S\). Equation\[V_{Th} = V_s \times \frac{N_s}{N_p} = \sin(\omega t) \times 2 = 2\sin(\omega t)\]
- Thevenin Impedance (\(Z_{Th}\)): The primary impedance \(Z_p\) is \(1\,\Omega\). Equation\[Z_{Th} = Z_p \times \left(\frac{N_s}{N_p}\right)^2 = 1\,\Omega \times (2)^2 = 4\,\Omega\]
Final Answer
Correct answer: (1) \(2\sin(\omega t),\,4\,\Omega\).