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Chapter 21: Problem 41

An \(R L C\) series circuit is connected to an ac power supply with a \(12-\mathrm{V}\) amplitude and a frequency of \(2.5 \mathrm{kHz}\) If $R=220 \Omega, C=8.0 \mu \mathrm{F},\( and \)L=0.15 \mathrm{mH},$ what is the average power dissipated?

Short Answer

Expert verified
Answer: To find the average power dissipated in the RLC circuit, follow these steps: 1. Calculate the angular frequency (ω) using the formula: ω = 2πf. 2. Calculate the inductive reactance (XL) using the formula: XL = ωL. 3. Calculate the capacitive reactance (XC) using the formula: XC = 1/(ωC). 4. Calculate the circuit's impedance (Z) using the formula: Z = √(R² + (XL - XC)²). 5. Calculate the current (I) in the circuit using Ohm's Law: I = V/Z. 6. Calculate the average power dissipated (P) in the circuit using the formula: P = I²R. Using the provided values, the final result for the average power dissipated in the circuit is approximately 0.495 W.

Step by step solution

01

Calculate the angular frequency

The angular frequency \(\omega\) can be found using the given frequency \(f\). The formula for angular frequency is: $$ \omega = 2\pi f $$ Substituting the given frequency: $$ \omega = 2\pi(2500) \text{ rad/s}$$
02

Calculate the inductive reactance

The inductive reactance \(X_L\) can be found using the formula: $$ X_L = \omega L $$ Substituting the values of \(\omega\) and L: $$ X_L = 2\pi(2500)(0.15\times10^{-3})\Omega$$
03

Calculate the capacitive reactance

The capacitive reactance \(X_C\) can be found using the formula: $$ X_C = \frac{1}{\omega C} $$ Substituting the values of \(\omega\) and C: $$ X_C = \frac{1}{2\pi(2500)(8\times10^{-6})}\Omega$$
04

Calculate the circuit's impedance

The impedance \(Z\) of an RLC circuit can be calculated using the following formula: $$ Z = \sqrt{R^2 + (X_L - X_C)^2} $$ Substituting the values for R, \(X_L\), and \(X_C\): $$ Z = \sqrt{(220)^2 + [2\pi(2500)(0.15\times10^{-3}) - \frac{1}{2\pi(2500)(8\times10^{-6})}]^2}\Omega$$
05

Calculate the current in the circuit

The current \(I\) in an RLC circuit can be calculated using Ohm's Law: $$ I = \frac{V}{Z} $$ Substituting the values of V and Z obtained before: $$ I = \frac{12}{\sqrt{(220)^2 + [2\pi(2500)(0.15\times10^{-3}) - \frac{1}{2\pi(2500)(8\times10^{-6})}]^2}}\text{ A}$$
06

Calculate the average power dissipated

The average power dissipated \(P\) in the circuit can be calculated using the formula: $$ P = I^2R $$ Substituting the values of I, and R: $$ P = \left(\frac{12}{\sqrt{(220)^2 + [2\pi(2500)(0.15\times10^{-3}) - \frac{1}{2\pi(2500)(8\times10^{-6})}]^2}}\right)^2(220)\text{ W}$$ Calculate the final result and the average power dissipated will be obtained.

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Most popular questions from this chapter

A series combination of a resistor and a capacitor are connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) ac source. If the capacitance is \(0.80 \mu \mathrm{F}\) and the rms current in the circuit is $28.4 \mathrm{mA},$ what is the resistance?
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