Let's break down the circuit and solve step by step: ### **Given:** - Voltage across \( 50 \Omega \) resistor, \( V_{50} = 100\,\text{V} \) (RMS) - Transformer ratio is **1:2** (primary:secondary) - The right circuit has \( j100\ \Omega \) in series with \( 50\ \Omega \), so the total secondary impedance is \( Z_2 = 50 + j100\ \Omega \) - The left circuit (primary) has \( 20\ \Omega \) and \( -j20\ \Omega \) in series with the source. --- ## **Step 1: Find Secondary Current \( I_2 \)** \[ I_2 = \frac{V_{50}}{50} = \frac{100}{50} = 2\,\text{A (RMS)} \] --- ## **Step 2: Reflect Secondary Impedance to Primary Side** The transformer's **turns ratio** is \( n = \frac{N_1}{N_2} = \frac{1}{2} \). The reflected impedance to the primary is: \[ Z_2' = Z_2 \times n^2 = (50 + j100) \times \left(\frac{1}{2}\right)^2 = (50 + j100) \times \frac{1}{4} = 12.5 + j25\ \Omega \] --- ## **Step 3: Find Primary Current \( I_1 \) (Source Current)** The current on the secondary side (\( I_2 \)) is related to the primary side (\( I_1 \)) by the turns ratio: \[ I_1 = I_2 \times n = 2 \times \frac{1}{2} = 1\,\text{A (RMS)} \] --- ## **Step 4: Find Total Primary Impedance** Total impedance seen by the source: \[ Z_{total} = 20 - j20 + Z_2' = 20 - j20 + 12.5 + j25 = (20 + 12.5) + (-20 + 25)j = 32.5 + j5 \] --- ## **Step 5: Find Source Voltage \( V \)** \[ V = I_1 \cdot Z_{total} = 1 \cdot (32.5 + j5) \] \[ |V| = \sqrt{32.5^2 + 5^2} = \sqrt{1056.25 + 25} = \sqrt{1081.25} \approx 32.89\,\text{V (RMS)} \] --- ## **Final Answers:** - **RMS value of the source voltage:** \[ \boxed{32.89~\text{V}} \] - **RMS value of the source current \( I_1 \):** \[ \boxed{1~\text{A}} \] --- **Let me know if you need the phasor angles or additional details!**