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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

An ac circuit contains a \(12.5-\Omega\) resistor, a \(5.00-\mu \mathrm{F}\) capacitor, and a \(3.60-\mathrm{mH}\) inductor connected in series to an ac generator with an output voltage of \(50.0 \mathrm{V}\) (peak) and frequency of \(1.59 \mathrm{kHz}\). Find the impedance,
A lightbulb is connected to a \(120-\mathrm{V}\) (rms), \(60-\mathrm{Hz}\) source. How many times per second does the current reverse direction?
Make a figure analogous to Fig. 21.4 for an ideal inductor in an ac circuit. Start by assuming that the voltage across an ideal inductor is \(v_{\mathrm{L}}(t)=V_{\mathrm{L}}\) sin \(\omega t .\) Make a graph showing one cycle of \(v_{\mathrm{L}}(t)\) and \(i(t)\) on the same axes. Then, at each of the times \(t=0, \frac{1}{8} T, \frac{2}{8} T, \ldots, T,\) indicate the direction of the current (or that it is zero), whether the current is increasing, decreasing, or (instantaneously) not changing, and the direction of the induced emf in the inductor (or that it is zero).
A television set draws an rms current of \(2.50 \mathrm{A}\) from a \(60-\mathrm{Hz}\) power line. Find (a) the average current, (b) the average of the square of the current, and (c) the amplitude of the current.
The voltage across an inductor and the current through the inductor are related by \(v_{\mathrm{L}}=L \Delta i / \Delta t .\) Suppose that $i(t)=I \sin \omega t .\( (a) Write an expression for \)v_{\mathrm{L}}(t) .$ [Hint: Use one of the relationships of Eq. \((20-7) .]\) (b) From your expression for \(v_{\mathrm{L}}(t),\) show that the reactance of the inductor is \(X_{\mathrm{L}}=\omega L_{\mathrm{r}}\) (c) Sketch graphs of \(i(t)\) and \(v_{\mathrm{L}}(t)\) on the same axes. What is the phase difference? Which one leads?
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