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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.

Short Answer

Expert verified
(a) \(5773\,\text{rad/s}\); (b) \(2.4\,\text{A}\); (c) \(2.4\,\text{A}\); (d) \(0.139\,\text{A}\); (e) \(0.139\,\text{A}\); (f) Both \(\approx 2.89\times 10^{-3} \,\text{J}\).

Step by step solution

01

Calculate Resonance Angular Frequency

At resonance, the angular frequency is given by the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \), where \( L \) is inductance and \( C \) is capacitance.Substituting the given values:\[ L = 0.300\,\text{H}, \quad C = 0.100\,\mu\text{F} = 0.100 \times 10^{-6}\,\text{F} \]\[ \omega_0 = \frac{1}{\sqrt{0.300 \times 0.100 \times 10^{-6}}} = \frac{1}{\sqrt{3 \times 10^{-8}}} = \frac{1}{\sqrt{30}} \times 10^{4} \approx 5773 \text{ rad/s} \]
02

Calculate Maximum Current Through the Source at Resonance

The impedance at resonance is the resistance \( R \), so the maximum current \( I_\text{max} \) through the source is given by:\[ I_\text{max} = \frac{V}{R} \]Given \( V = 240\,\text{V} \) and \( R = 100\,\Omega \):\[ I_\text{max} = \frac{240}{100} = 2.4\,\text{A} \]
03

Calculate Maximum Current in the Resistor at Resonance

The current through the resistor at resonance equals the source current since they are in parallel:\[ I_R = I_\text{max} = 2.4\,\text{A} \]
04

Calculate Maximum Current in the Inductor at Resonance

The maximum current in the inductor at resonance will be:\[ I_L = \frac{V}{\omega_0 L} \]\[ I_L = \frac{240}{5773 \times 0.300} \approx 0.139\,\text{A} \]
05

Calculate Maximum Current in the Capacitor Branch at Resonance

The maximum current in the capacitor at resonance is given by:\[ I_C = \omega_0 C V \]\[ I_C = 5773 \times 0.100 \times 10^{-6} \times 240 \approx 0.139\,\text{A} \]
06

Calculate Maximum Energy Stored in the Inductor

The maximum energy stored in the inductor is given by \( \frac{1}{2} L I_L^2 \):\[ E_L = \frac{1}{2} \times 0.300 \times (0.139)^2 \approx 2.89 \times 10^{-3}\,\text{J} \]
07

Calculate Maximum Energy Stored in the Capacitor

The maximum energy stored in the capacitor is calculated as \( \frac{1}{2} C V^2 \):\[ E_C = \frac{1}{2} \times 0.100 \times 10^{-6} \times 240^2 \approx 2.89 \times 10^{-3}\,\text{J} \]

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Key Concepts

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

Angular Frequency
In resonant circuits, angular frequency plays a crucial role in determining the rate at which energy is stored and released by the system. Angular frequency, symbolized by \( \omega \), measures how quickly the oscillations occur in a resonant circuit. It is related to the frequency \( f \) of the AC signal by \( \omega = 2\pi f \). Particularly important is the resonance angular frequency \( \omega_0 \), calculated using the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance.
This frequency is special because it allows the circuit to oscillate at its natural frequency, maximizing energy flow. Without the right angular frequency, the circuit wouldn't reach its peak performance. Knowing how to calculate and understand angular frequency helps you predict how the circuit behaves at resonance.
Inductor
An inductor is a passive electronic component that stores energy in a magnetic field when electrical current flows through it. In our exercise, the inductor has a value of \( L = 0.300 \text{ H} \). It's crucial in resonant circuits because it works alongside the capacitor to determine the circuit's resonant frequency. When at resonance, the inductor's impedance cancels out the capacitor's impedance.
The current through the inductor at resonance can be calculated using the formula: \[ I_L = \frac{V}{\omega_0 L} \]Where \( V \) is the voltage supply. At resonance, the inductor effectively balances out energy flows with the capacitor, allowing for pure resistance, which means the peak voltages and currents won't be affected by inductive or capacitive reactance.
Capacitor
Capacitors play an essential role in resonant circuits by storing energy in an electric field. In our scenario, we have a capacitor with a capacitance of \( C = 0.100 \mu\text{F} \). Capacitors in a circuit like this act in tandem with inductors to achieve resonance.
The maximum current in the capacitor at the resonance frequency is determined by:\[ I_C = \omega_0 C V \]Where \( V \) is the voltage amplitude. At resonance, the reactance of a capacitor is exactly offset by the inductive reactance, allowing the alternating current (AC) circuit to efficiently transfer energy. Capacitance and resonance frequency also determine how long the system can naturally oscillate after external energy is removed.
Resistor
A resistor is a component that opposes the flow of electrical current, creating resistance. In this exercise, the resistor value is \( R = 100 \text{ } \Omega \).
A resistor's role in a resonant circuit is substantial since it affects the damping of oscillations. At resonance, in the given circuit, the effect of reactance cancels out, and the circuit's impedance is purely resistive. Therefore, the current through the resistor becomes:\[ I_R = \frac{V}{R} \]This also equals the total current through the source when in parallel resonance. Resistors set the allowable current in a circuit, preventing damages due to excessive current, and thus play a vital role in maintaining circuit stability.
Energy Storage
In resonant circuits, energy storage is primarily managed by inductors and capacitors. These components are key to short-term energy storage, storing energy temporarily and releasing it either in the form of a magnetic or electric field.
The energy stored in an inductor is given by:\[ E_L = \frac{1}{2} L I_L^2 \]And for a capacitor, it is:\[ E_C = \frac{1}{2} C V^2 \]Both expression signify that the energy is proportional to the square of the current or voltage. As such, these components dictate how current and voltage are handled in the circuit. At resonance, the energy continually swaps between the inductor and capacitor, leading to a highly efficient energy exchange system, optimized for maximum energy flow.

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Most popular questions from this chapter

A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?

An \(L-R-C\) series circuit is connected to an ac source of constant voltage amplitude \(V\) and variable angular frequency \(\omega\) (a) Show that the current amplitude, as a function of \(\omega,\) is $$I=\frac{V}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}}$$ (b) Show that the average power dissipated in the resistor is $$P=\frac{V^{2} R / 2}{R^{2}+(\omega L-1 / \omega C)^{2}}$$ (c) Show that \(I\) and \(P\) are both maximum when \(\omega=1 / \sqrt{L C},\) the resonance frequency of the circuit. (d) Graph \(P\) as a function of \(\omega\) for \(V=100 \mathrm{V}, R=200 \Omega, L=2.0 \mathrm{H},\) and \(C=0.50 \mu \mathrm{F}\) . Compare to the light purple curve in Fig. \(31.19 .\) Discuss the behavior of \(I\) and \(P\) in the limits \(\omega=0\) and \(\omega \rightarrow \infty\) .

A large electromagnetic coil is connected to a 120 -Hz ac source. The coil has resistance \(400 \Omega,\) and at this source frequency the coil has inductive reactance 250\(\Omega\) (a) What is the inductance of the coil? (b) What must the rms voltage of the source be if the coil is to consume an average electrical power of 800 \(\mathrm{w} ?\)

You want to double the resonance angular frequency of an L-R-C series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

An \(L \cdot R-C\) series circuit has \(R=500 \Omega, L=2.00 \mathrm{H}\) \(C=0.500 \mu \mathrm{F},\) and \(V=100 \mathrm{V} .\) (a) For \(\omega=800 \mathrm{rad} / \mathrm{s},\) calculate \(V_{R}, V_{L}, V_{C},\) and \(\phi .\) Using a single set of axes, graph \(v, v_{R}, v_{L},\) and \(v_{C}\) as functions of time. Include two cycles of \(v\) on your graph. (b) Repeat part (a) for \(\omega=1000 \mathrm{rad} / \mathrm{s} .\) (c) Repeat part (a) for \(\omega=1250 \mathrm{rad} / \mathrm{s}\)

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