Let's break down the problem step by step. ### **Step 1: Understanding the Problem** - **RL circuit**: A resistor \( R \) and inductor \( L \) in series. - **90° leading phase shift**: Output voltage leads input by 90°. But in passive RL circuits, the output across the inductor **lags** the input; only in RC (capacitive) circuits do you get a leading output. - But let's proceed assuming you want a 90° phase shift (i.e., the output voltage is 90° out of phase with input, as in a pure inductor or capacitor). - **Input voltage**: \( V_{in} = 120 \) V (assume RMS). ### **Step 2: Series RL Circuit Phase Relationships** The total impedance of RL series is: \[ Z = R + j\omega L \] The phase angle between the input voltage and the current is: \[ \theta = \arctan\left(\frac{\omega L}{R}\right) \] - Current **lags** the input voltage by \( \theta \). - The voltage across the inductor **lags** the current by 90°, so overall \( V_L \) lags \( V_{in} \) by \( \theta \). - **Maximum phase shift** in RL circuit is 90°, when \( R = 0 \) (pure inductor). ### **Step 3: Condition for 90° Phase Shift** \[ \theta = \arctan\left(\frac{\omega L}{R}\right) = 90^\circ \] So, \[ \frac{\omega L}{R} \to \infty \implies R = 0 \] So, for 90° phase difference, you need a **pure inductor** (no resistor). ### **Step 4: Output Magnitude Calculation** In a pure inductor: - All the input voltage appears across the inductor. - The output (across L) is equal in magnitude to the input. - The phase shift is 90° (current lags voltage by 90°). So, \[ V_{out} = V_{in} = 120\,V \] ### **Step 5: Final Answers** #### **(a) Circuit Design** - Use a **pure inductance** (or make \( R \ll \omega L \) to approximate 90° phase shift). - Circuit: Connect **inductor L** only across the AC input. #### **(b) Output Magnitude** - The output voltage across L is **120 V** (same as input). --- ## **Summary Table** | Step | Calculation/Explanation | Result | |--------------|--------------------------------------------------------|------------------| | Phase shift | \(\theta = \arctan(\omega L / R) = 90^\circ\) | \(R = 0\) | | Output mag. | \(V_{out} = V_{in}\) | \(120\,V\) | --- ## **Final Answers** ### **1. RL Circuit Design:** - Use a pure inductor (or set \( R \ll \omega L \)) to achieve 90° phase shift. ### **2. Output Magnitude:** - The output across the inductor is **120 V**. --- **Note:** If you meant the output to be across the resistor, the phase shift would be 0°, and the magnitude would depend on the voltage divider. But for a 90° shift, only a pure inductor provides this in an RL circuit.