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A resistor \(R,\) inductor \(L,\) and capacitor \(C\) are connected in series to an AC source of rms voltage \(\Delta V\) and variable frequency. Find the energy that is delivered to the circuit during one period if the operating frequency is twice the resonance frequency.

Short Answer

Expert verified
The energy delivered to the circuit during one period at twice the resonance frequency is \( \frac{\Delta V^2}{2Rf_0} \).

Step by step solution

01

Identify the Resonance Frequency

The resonance frequency, also known as the natural frequency, for an LC circuit is given by the formula \( f_0 = \frac{1}{2\pi\sqrt{LC}} \). This is the frequency at which the inductive reactance and capacitive reactance cancel each other out.
02

Calculate the Operating Frequency

The operating frequency, \( f \), is given to be twice the resonance frequency. Therefore, \( f = 2f_0 \).
03

Determine the Reactance of the Inductor and Capacitor

At the operating frequency, the inductive reactance \( X_L \) is \( X_L = 2\pi f L \), and the capacitive reactance \( X_C \) is \( X_C = \frac{1}{2\pi f C} \). Since \( f = 2f_0 \), we have \( X_L = 4\pi f_0 L \) and \( X_C = \frac{1}{4\pi f_0 C} \) respectively.
04

Calculate the Impedance of the Circuit

At twice the resonance frequency, the impedance \( Z \) is purely resistive since \( X_L \) and \( X_C \) are still equal due to their dependence on \( f_0 \) and the given condition. Therefore, \( Z = R \).
05

Find the Average Power

The average power \( P_{avg} \) delivered to the circuit is \( P_{avg} = \frac{\Delta V^2}{Z} \) because the voltage and current are in phase at resonance, even though our frequency is twice the resonance frequency. Since the impedance is equal to \( R \), the average power becomes \( P_{avg} = \frac{\Delta V^2}{R} \).
06

Calculate the Energy Delivered per Period

The energy \( W \) delivered during one period \( T \) can be found using \( W = P_{avg}T \) where \( T = \frac{1}{f} \) is the period. Since \( f = 2f_0 \) and \( T_0 = \frac{1}{f_0} \) is the period at resonance frequency, \( W = P_{avg}\frac{1}{2f_0} = \frac{\Delta V^2}{2Rf_0} \).

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Key Concepts

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

Resonance Frequency
Resonance frequency holds a special place in the world of electricity and magnetism. Specifically, for an LC circuit—which is simply an inductor (L) and a capacitor (C) linked together—the resonance frequency, denoted as f0, marks the point at which the circuit naturally oscillates without external energy input. It's analogous to the perfect swing of a pendulum where gravity and inertia are in perfect balance.

The formula that describes this balancing act is f0 = 1 / (2π&sqrt;LC), connecting the inductance and capacitance by an inverse relationship. As either the inductance or capacitance increases, the resonance frequency decreases, and vice versa. This frequency signifies a condition where the reactance of the inductor perfectly cancels out the reactance of the capacitor, resulting in the circuit behaving as if it has zero impedance from the perspective of these reactive components. In educational practices, it might be compared to a dance where two partners move in such sync that they seem to be one entity.

Understanding resonance is crucial as it has practical applications, such as in tuning radios to certain frequencies or in wireless power transfer systems. However, deviating from this frequency can lead to less efficient operation, as seen when a circuit operates at a frequency different from its natural resonance.
Impedance in AC Circuits
Impedance, symbolized by Z, is a fundamental concept in AC (alternating current) circuits, acting as the AC equivalent of resistance in a DC (direct current) circuit. However, unlike resistance, impedance accounts for not only the resistive elements but also the circuit’s ability to store and release energy in the form of reactive components, such as inductors and capacitors.

In an AC circuit, the impedance is the combined effect of resistance (R), inductive reactance (XL), and capacitive reactance (XC). These components form a complex relationship, often expressed in a phasor diagram as a vector sum. The formula to calculate total impedance in a series circuit is Z = &sqrt;R2 + (XL - XC)2, which can greatly vary depending on the frequency of the AC supply.

At resonance frequency, something magical occurs—the inductive and capacitive reactances exactly cancel each other out, and the only opposition to current flow is due to the resistor, making Z = R. It would be like friction being the only force acting on a car coasting on a level road—the engine and brakes (inductive and capacitive effects) don't factor into how easily it rolls.
Reactance of Inductor and Capacitor
Reactance is what makes AC circuit analysis a bit more complex than DC circuits. It's present in components like inductors and capacitors, which store energy in magnetic fields and electric fields, respectively. The reactance of an inductor, denoted as XL, is directly proportional to the frequency of the AC current (f) and its inductance value, following XL = 2πfL. This means that as frequency increases, the opposition to the current flow provided by the inductor also increases.

For capacitors, the reactance, noted as XC, shows an inverse relationship: XC = 1 / (2πfC). Unlike an inductor, a capacitor provides more opposition at lower frequencies and less at higher frequencies. The reactance values are critical for understanding how a circuit behaves at various frequencies, particularly at resonance, where XL = XC and they effectively cancel each other out, allowing us to only consider resistive effects for our calculations.

In a classroom setting, you might illustrate this concept with analogies like a sprinter on a track (inductor) who speeds up with each lap (frequency increase), while a swimmer (capacitor) experiences more resistance as the water (frequency) becomes denser.

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Most popular questions from this chapter

At what frequency does the inductive reactance of a \(57.0-\mu \mathrm{H}\) inductor equal the capacitive reactance of a \(57.0-\mu \mathrm{F}\) capacitor?

A sinusoidal voltage \(\Delta v(t)=(40.0 \mathrm{V}) \sin (100 t)\) is applied to a series \(R L C\) circuit with \(L=160 \mathrm{mH}, C=99.0 \mu \mathrm{F},\) and \(R=68.0 \Omega .\) (a) What is the impedance of the circuit? (b) What is the maximum current? (c) Determine the numerical values for \(I_{\max }, \omega,\) and \(\phi\) in the equation \(i(t)=I_{\max } \sin (\omega t-\phi)\)

Impedance matching. Example 28.2 showed that maximum power is transferred when the internal resistance of a DC source is equal to the resistance of the load. A transformer may be used to provide maximum power transfer between two AC circuits that have different impedances Z1 and Z2, where 1 and 2 are subscripts and the Z’s are italic (as in the centered equation). (a) Show that the ratio of turns \(N_{1} / N_{2}\) needed to meet this condition is $$\frac{N_{1}}{N_{2}}=\sqrt{\frac{Z_{1}}{Z_{2}}}$$ (b) Suppose you want to use a transformer as an impedance-matching device between an audio amplifier that has an output impedance of 8.00 \(\mathrm{k} \Omega\) and a speaker that has an input impedance of 8.00\(\Omega\) . What should your \(N_{1} / N_{2}\) ratio be?

(a) For what frequencies does a \(22.0-\mu \mathrm{F}\) capacitor have a reactance below 175\(\Omega ?\) (b) What If? Over this same frequency range, what is the reactance of a \(44.0-\mu \mathrm{F}\) capacitor?

An RLC circuit consists of a \(150-\Omega\) resistor, a \(21.0-\mu \mathrm{F}\) capacitor, and a 460-mH inductor, connected in series with a 120-V, 60.0-Hz power supply. (a) What is the phase angle between the current and the applied voltage? (b) Which reaches its maximum earlier, the current or the voltage?

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