Let's break down the problem step by step--- ## **Step 1: Understanding the and Given Data** **Explanation:- The circuit of an alternating source, a capacitor \( C \), and a resistor \( R \) in series. - The graph shows the reactance/capacitance ratio (\( X_C / R \)) on the y-axis vs. frequency (\( f \)) on the x-axis. - You are asked to find the value of the resistor \( R \). - \( C = 4.40 \, \mu F = 4.40 \times 10^{-6} \, F \). --- ## **Step 2: Using the Graph to Find R** **Explanation:** - The capacitive reactance is \( X_C = \frac{1}{2\pi f C} \). - At a chosen frequency, use the graph to read the value of \( X_C / R \). - Rearranging: \( \frac{X_C}{R} = \frac{1}{2\pi f C R} \implies R = \frac{1}{2\pi f C (X_C/R)} \). Suppose, from the graph, at \( f = 200 \) Hz, \( X_C / R = 1 \): - Plug values into the formula: \[ R = \frac{1}{2\pi (200)(4.40 \times 10^{-6})(1)} \] \[ R \approx \frac{1}{2\pi \times 200 \times 4.40 \times 10^{-6}} \] \[ R \approx \frac{1}{.00553} \] \[ R \approx 181 \, \Omega \] --- ## **Final Answer** \[ \boxed{R \approx 181 \, \Omega} \] **Note:** - The exact value may vary depending on the exact point you pick from the graph, but the method remains the same. - Always use the values directly read from your specific graph.