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Chapter 31: Problem 20

In an \(L-R-C\) series circuit, the components have the following values: \(L=20.0 \mathrm{mH}, C=140 \mathrm{nF},\) and \(R=350 \Omega .\) The generator has an rms voltage of 120 \(\mathrm{V}\) and a frequency of 1.25 \(\mathrm{kHz}\) . Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

Short Answer

Expert verified
The power supplied by the generator is approximately 4.65 W, and the power dissipated in the resistor is approximately 4.69 W.

Step by step solution

01

Calculate Angular Frequency

First, we need to find the angular frequency, \(\omega\), of the circuit using the formula \(\omega = 2\pi f\), where \(f\) is the frequency of the generator. Given \(f = 1.25 \text{ kHz}\), we have:\[\omega = 2\pi \times 1250 \text{ Hz} = 2500\pi \text{ rad/s}.\]
02

Calculate Inductive Reactance

Use the inductance \(L\) to find the inductive reactance, \(X_L\), using the formula \(X_L = \omega L\). With \(L = 20 \text{ mH} = 20 \times 10^{-3} \text{ H}\), we calculate:\[X_L = 2500\pi \times 0.020 = 50\pi \; \Omega.\]
03

Calculate Capacitive Reactance

Find the capacitive reactance, \(X_C\), using the formula \(X_C = \frac{1}{\omega C}\). Given \(C = 140 \text{ nF} = 140 \times 10^{-9} \text{ F}\), we have:\[X_C = \frac{1}{2500\pi \times 140 \times 10^{-9}} \approx 1.14 \times 10^3 \; \Omega.\]
04

Calculate Impedance

Determine the total impedance, \(Z\), using the formula \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). Substituting the known values:\[Z = \sqrt{350^2 + (50\pi - 1140)^2} \approx \sqrt{350^2 + (-982.15)^2} \approx 1035 \, \Omega.\]
05

Calculate RMS Current

The rms current, \(I_{rms}\), is calculated using Ohm's Law: \(I_{rms} = \frac{V_{rms}}{Z}\), where \(V_{rms} = 120\, \text{V}\). Thus:\[I_{rms} = \frac{120}{1035} \approx 0.116 \text{ A}.\]
06

Calculate Power Supplied by the Generator

The average power supplied by the generator is given by \(P = V_{rms} \times I_{rms} \times \cos\phi\), where \(\cos\phi = \frac{R}{Z}\). We calculate \(\cos\phi\) first:\[\cos\phi = \frac{350}{1035} \approx 0.338.\]Thus, the power is:\[P = 120 \times 0.116 \times 0.338 \approx 4.65 \text{ W}.\]
07

Calculate Power Dissipated in the Resistor

The power dissipated in the resistor, \(P_R\), is given by \(P_R = I_{rms}^2 \times R\). Thus:\[P_R = (0.116)^2 \times 350 \approx 4.69 \text{ W}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reactance Calculation
In an L-R-C circuit, reactance is a critical component when analyzing the behavior of inductors and capacitors in response to an alternating current (AC). Inductive reactance, denoted as \(X_L\), is determined by how an inductor resists changes in current and is calculated using the formula \(X_L = \omega L\), where \(\omega\) is the angular frequency and \(L\) is the inductance. For capacitors, the capacitive reactance, denoted as \(X_C\), reflects their ability to block changes in voltage, calculated by \(X_C = \frac{1}{\omega C}\), with \(C\) being the capacitance. Each type of reactance has an opposite effect on the circuit鈥檚 overall phase angle and impacts the impedance. These calculations are important for determining how energy is stored within the circuit's electromagnetic and electric fields.
Impedance in AC Circuits
Impedance, symbolized by \(Z\), is the total resistance to the flow of alternating current, combining both resistive (real) and reactive (imaginary) effects. It鈥檚 more comprehensive than simple resistance encountered in direct current circuits. The impedance in an L-R-C circuit is calculated using the formula \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), where \(R\) is the resistance, and \(X_L\) and \(X_C\) are the inductive and capacitive reactances respectively. This equation takes into account the phase angle differences between the voltage and current due to reactance. Understanding impedance is crucial for analyzing power flow and voltage/current relationships in AC powered systems. A complex impedance influences how efficiently power is transmitted and can affect the performance of electrical devices connected to the circuit.
Power in AC Circuits
Power in AC circuits differs from that in DC circuits due to the phase difference between current and voltage, which impacts how power is calculated and interpreted. The true or real power \(P\), given in watts, represents the power actually used in the circuit. It's computed as \(P = V_{rms} \times I_{rms} \times \cos\phi\), where \(V_{rms}\) is the root mean square voltage, \(I_{rms}\) is the root mean square current, and \(\cos\phi\) (power factor) reflects the cosine of the phase angle \(\phi\).
  • The power factor indicates the efficiency of power usage.
  • A power factor of 1 means all power is used for work; lower values indicate more power is wasted as reactive power.
In an L-R-C circuit, the power dissipated by the resistor, \(P_R = I_{rms}^2 \times R\), is always part of the real power since resistors consume energy. The remaining power, which oscillates back and forth in the circuit, is stored in the reactive components. Understanding these distinctions allows for improved design and operation of electrical systems.

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Most popular questions from this chapter

A \(100-\Omega\) resistor, a \(0.100-\mu \mathrm{F}\) capacitor, and a \(0.300-\mathrm{H}\) inductor are connected in parallel to a voltage source with amplitude 240 \(\mathrm{V}\) (a) What is the resonance angular frequency? (b) What is the maximum current through the source at the resonance frequency? (c) Find the maximum current in the resistor at resonance. (d) What is the maximum current in the inductor at resonance? (e) What is the maximum current in the branch containing the capacitor at resonance? (f) Find the maximum energy stored in the inductor and in the capacitor at resonance.

In an \(L-R-C\) series circuit the current is given by \(i=I \cos \omega t .\) The voltage amplitudes for the resistor, inductor, and capacitor are \(V_{R}, \quad V_{L},\) and \(V_{C}\) . (a) Show that the instantaneous power into the resistor is \(p_{R}=V_{R} I \cos ^{2} \omega t=\frac{1}{2} V_{R} I(1+\cos 2 \omega t)\) What does this expression give for the average power into the resistor? (b) Show that the instantaneous power into the inductor is \(p_{L}=-V_{L} I \sin \omega t \cos \omega t=-\frac{1}{2} V_{L} I \sin 2 \omega t .\) What does this expression give for the average power into the inductor? (c) Show that the instantaneous power into the capacitor is \(p_{C}=\) \(V_{C} I \sin \omega t \cos \omega t=\frac{1}{2} V_{C} I \sin 2 \omega t .\) What does this expression give for the average power into the capacitor? (d) The instantaneous power delivered by the source is shown in Section 31.4 to be \(p=V I \cos \omega t(\cos \phi \cos \omega t-\sin \phi \sin \omega t) .\) Show that \(p_{R}+p_{L}+\) \(p_{C}\) equals \(p\) at each instant of time.

In an \(L-R-C\) series circuit, \(R=300 \Omega, L=0.400 \mathrm{H},\) and \(C=6.00 \times 10^{-8} \mathrm{F} .\) When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 \(\mathrm{A}\) . (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

In an \(L-R-C\) series circuit, \(R=150 \Omega, L=0.750 \mathrm{H},\) and \(C=0.0180 \mu \mathrm{F} .\) The source has voltage amplitude \(V=150 \mathrm{V}\) and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with \(C=\) 0.0360\(\mu \mathrm{F}\) and the source frequency is adjusted to the new resonance value. Then what is the average power delivered by the source?

A resistance \(R,\) capacitance \(C,\) and inductance \(L\) are connected in series to a voltage source with amplitude \(V\) and variable angular frequency \(\omega .\) If \(\omega=\omega_{0},\) the resonance angular frequency, find (a) the maximum current in the resistor; (b) the maximum voltage across the capacitor; (c) the maximum voltage across the inductor; (d) the maximum energy stored in the capacitor; (e) the maximum energy stored in the inductor. Give your answers in terms of \(R, C, L,\) and \(V .\)
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