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.

Short Answer

Expert verified
The phase angle, θ, for this AC circuit is 65.4°. (b) What are the values of the resistor (R), inductor (L), and capacitor (C)? The values of the resistor, inductor, and capacitor are: R = 25 Ω, L = 0.101 H, and C = 139.1 μF.

Step by step solution

01

(a) Find the phase angle θ

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°
02

(b) Find the values of the resistor R, inductor L, and capacitor C

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.

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