An ac circuit has a single resistor, capacitor, and inductor in series. The circuit uses \(100 \mathrm{W}\) of power and draws a maximum rms current of $2.0 \mathrm{A}\( when operating at \)60 \mathrm{Hz}\( and \)120 \mathrm{V}$ rms. The capacitive reactance is 0.50 times the inductive reactance. (a) Find the phase angle. (b) Find the values of the resistor, the inductor, and the capacitor.

First, let's find the power factor (cosθ) using the given power (P = 100 W), maximum rms current (Irms = 2.0 A), and rms voltage (Vrms = 120 V): Power = Voltage × Current × power factor 100 = 120 × 2.0 × cosθ cosθ = 100 / (120 × 2.0) cosθ = 0.4167 Now, we can find the phase angle θ: θ = arccos(0.4167) θ = 65.4°

For this step, we will use the following equations: R = Vrms / Irms × power factor XL (inductive reactance) = ωL = 2πfL XC (capacitive reactance) = 1/ωC = 1/(2πfC) We know that XL = 2XC. Let's first find R: R = 120 / 2 × 0.4167 R = 60 × 0.4167 R = 25 Ω Now, we need to use the phase angle θ to calculate (XL - XC): XL - XC = R × tan(θ) XL - XC = 25 × tan(65.4°) XL - XC = 57.3 Ω We also know that XL = 2XC. Let's solve for XL and XC: XC = 57.3 Ω / 3 XC = 19.1 Ω XL = 2 × 19.1 Ω XL = 38.2 Ω Finally, let's use XL and XC to find L and C using the relationships between reactance and inductance/capacitance: L = XL / (2πf) L = 38.2 Ω / (2π × 60 Hz) L = 0.101 H C = 1 / (2πf × XC) C = 1 / (2π × 60 Hz × 19.1 Ω) C = 139.1 μF The values of the resistor, inductor, and capacitor are R = 25 Ω, L = 0.101 H, and C = 139.1 μF, respectively.