**Problem Statement:**Remove the capacitor from the circuit in the figure and set \( R = 151 \, \Omega \), \( L = 230 \, \text{mH} \), \( f_d = 60.0 \, \text{Hz} \), and \( \varepsilon_m = 42.0 \, \text{V} \). What are (a) \( Z \), (b) \( \phi \), and (c) \( I \)?**Circuit Diagram:**- The image shows a simple R-L series circuit.- It consists of an AC voltage source (\( \varepsilon_g \)), a resistor (\( R \)), and an inductor (\( L \)).- The capacitor (\( C \)) is labeled but needs to be removed as per instructions.- The current (\( i \)) flows through the circuit.**Given Values:**- \( R = 151 \, \Omega \)- \( L = 230 \, \text{mH} \)- \( f_d = 60.0 \, \text{Hz} \)- \( \varepsilon_m = 42.0 \, \text{V} \)**Required:**Determine the following:(a) \( Z \) - Impedance- Answer: \( 174 \, \Omega \)(b) \( \phi \) - Phase angle- Answer: \(-40.2 \, \text{degrees}\)(c) \( I \) - Current- Answer: \( 0.207 \, \text{A} \)---**Calculation Notes:**- To find the impedance \( Z \), use \( Z = \sqrt{R^2 + (X_L)^2} \), where \( X_L = 2\pi f_d L \).- The phase angle \( \phi \) is calculated using \( \phi = \tan^{-1}\left(\frac{X_L}{R}\right) \).- Current \( I \) is found using \( I = \frac{\varepsilon_m}{Z} \).This exercise demonstrates the analysis of an R-L circuit without a capacitor, focusing on calculating impedance, phase angle, and current flow.

Given:- R=151 Ω L=230 mH fd= 60.0 Hz εm=42.0 V Removing capacitor, Xc = 0

Answered: Remove the capacitor from the circuit…

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