Let's break down the problem step by step: **Problem Recap:** Design an **RL circuit** to provide a **90° leading phase shift**. If the input is **120 V**, determine the **magnitude of the output**. --- ### Step 1: Understanding the Phase Shift in RL Circuits - An RL circuit (Resistor + Inductor) can be connected in two ways: **series** or **parallel**. - A **leading phase shift** means the output leads the input (e.g., output voltage across the inductor leads current or input voltage). - In a **series RL circuit**, the voltage across the inductor (L) leads the current by 90°, and the voltage across the resistor (R) is in phase with the current. #### **To provide a 90° leading phase shift:** - The output voltage across **L** (inductor) in a **series RL circuit** leads the input current by 90°. --- ### Step 2: Phasor Analysis of Series RL Circuit Let’s consider a series RL circuit with input voltage \( V_{in} \), resistor \( R \), and inductor \( L \): - Total impedance: \( Z = R + j\omega L \) - If we take output across **L**: \( V_{out} = V_L \) The phase angle between input and output is: \[ \text{Phase shift, } \theta = \arctan\left(\frac{\omega L}{R}\right) \] For **90° phase shift**: \[ \arctan\left(\frac{\omega L}{R}\right) = 90^\circ \implies \frac{\omega L}{R} \to \infty \implies R \to 0 \] So, for a **perfect 90° phase shift**, the resistor must be **zero**; i.e., a pure inductor. --- ### Step 3: Practical Circuit But in practice, we can't have exactly zero resistance, but let's proceed assuming **R = 0** for ideal case. - The circuit is just an inductor \( L \). - The output is taken across \( L \). **Input voltage:** \( V_{in} = 120 \) V (rms assumed). --- ### Step 4: Output Magnitude Calculation For a pure inductor: - Input voltage: \( V_{in} \) (applied across L) - Output voltage (across L): \( V_{out} = V_{in} \) - The **magnitude** is the same as the input, but it **leads the current by 90°**. So, \[ \boxed{V_{out} = 120~\text{V}~(\text{rms})} \] --- ### Step 5: Practical Note In real circuits, a small resistor is always present, so the phase shift will be just under 90°. To get close to 90°, select \( R \ll \omega L \). --- ## **Summary Table** | Step | Explanation | Result | |------|-------------------------------------------------------|----------------------------------------| | 1 | Series RL circuit, output across L | | | 2 | Phase shift formula: \(\arctan(\omega L / R)\) | For 90°, \(R=0\) | | 3 | Pure inductor, output leads current by 90° | | | 4 | Output magnitude equals input | \(V_{out} = 120~\text{V}\) | --- ## **Final Answer** ### **1. Circuit Design:** Use a **series RL circuit** with a negligible resistor (\(R \to 0\)), i.e., a pure inductor. Take the output across the inductor. This will provide a **90° leading phase shift** between the output (across L) and the input (current). ### **2. Output Magnitude:** If the input is **120 V**, the output magnitude (across L) is: \[ \boxed{120~\text{V}} \] **Note:** The output voltage leads the input current by 90°. In practice, use as small a resistor as possible to get close to 90° phase shift.