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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 \(V\)

Short Answer

Expert verified
The maximum current in the resistor is \(\frac{V}{R}\), the maximum voltage across the capacitor is \(\frac{V}{\omega_{0}C \cdot R}\), and the maximum voltage across the inductor is \(\frac{V \cdot \omega_{0}L}{R}\). The maximum energy stored in the capacitor and the inductor is \(\frac{V^{2}}{2\omega_{0}^{2}C^{2}R^{2}}\) and \(\frac{V^{2}\omega_{0}^{2}L^{2}}{2R^{2}}\) , respectively.

Step by step solution

01

Find the maximum current in the resistor

At resonance, the capacitive reactance (\(X_{C}\)) and inductive reactance (\(X_{L}\)) cancel out each other. The current in the circuit is given by Ohm's law \(V = I \cdot R\). So at resonance, the current \(I\) can be written as \(I_{max} = \frac{V}{R}\).
02

Find the maximum voltage across the capacitor

The voltage across the capacitor at resonance is obtained by \(V_{C} = I_{max} \cdot X_{C}\). Since \(X_{C} = \frac{1}{\omega_{0}C}\), the maximum voltage across the capacitor is \(V_{C_{max}} = \frac{V}{\omega_{0}C \cdot R}\).
03

Find the maximum voltage across the inductor

The voltage across the inductor at resonance is calculated by \(V_{L} = I_{max} \cdot X_{L}\). Since \(X_{L} = \omega_{0}L\), the maximum voltage across the inductor is \(V_{L_{max}} = \frac{V \cdot \omega_{0}L}{R}\).
04

Find the maximum energy stored in the capacitor

The energy stored in the capacitor at resonance is given by \(U_{C} = \frac{1}{2}CV^{2}\). The maximum energy in the capacitor can be calculated as \(U_{C_{max}} = \frac{1}{2}C(V_{C_{max}})^{2} = \frac{V^{2}}{2\omega_{0}^{2}C^{2}R^{2}}\).
05

Find the maximum energy stored in the inductor

The energy stored in the inductor at the resonance is given by \(U_{L} = \frac{1}{2}LI^{2}\). The maximum energy in the inductor can be calculated as \(U_{L_{max}} = \frac{1}{2}L(I_{max})^{2} = \frac{V^{2}\omega_{0}^{2}L^{2}}{2R^{2}}\).

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Key Concepts

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

Series RLC Circuit
Imagine three amigos—Resistance (R), Inductance (L), and Capacitance (C)—taking a walk on a single path, one behind the other; this is much like a series RLC circuit in the world of electronics. When these components are connected in series with a voltage source, they have a fascinating way of interacting with one another. As the voltage source starts to oscillate with a frequency, it sends waves of electric current down the line.

In such a setup, Resistance opposes the current uniformly at all times, while Capacitance stores energy in an electric field, and Inductance in a magnetic field, both releasing it back into the circuit at different phases. This interplay brings us to a point called electrical resonance, where Capacitance and Inductance perfectly balance each other out, so the current only has to overcome the Resistance. The series RLC circuit then behaves as if it's a simple resistor circuit, which is crucial for understanding resonance behaviors in practical applications like radio tuners.
Resonance Angular Frequency
Now, let’s turn up the frequency and hit that sweet spot—resonance angular frequency (\r\(\r\omega_0\r\)\r)). This is the frequency at which our series RLC circuit decides to sing in harmony. It's determined mathematically by the formula \r\(\r\omega_0 = 1 / \sqrt{LC}\r\)\r, representing the natural frequency for the RLC system to resonate.

At this particular frequency, the circuit allows the maximum current to flow through because the reactive opposition from both the capacitor and inductor cancel each other out. It's as if they have a truce for the moment. This concept is super useful because it lets us pinpoint the frequency at which the circuit can operate most efficiently, making it a critical parameter in designing electronics that filter or select specific frequencies—like picking your favorite radio station!
Reactance
Reactance is a concept that may sound intimidating, but really, it just refers to how the inductor and the capacitor in our circuit say 'Not so fast!' to the alternating current.

Capacitive reactance (\r\(\rX_C\r\)\r)) is the resistance to change in voltage, and inductive reactance (\r\(\rX_L\r\)\r)) is the resistance to change in current. They are calculated by the formulas \r\(\rX_C = 1 / (\omega C)\r\)\r and \r\(\rX_L = \omega L\r\)\r, respectively. When the circuit reaches that special resonance angular frequency, they're equal and opposite, neutralizing each other.

This idea is essential because understanding how reactance changes with frequency can help us manipulate circuits to do what we want, like filtering noise or matching impedances in antenna systems.
Ohm's Law
Lastly, let’s not forget about Ohm’s law, the backbone of understanding electrical circuits. Ohm’s law is like the golden rule that current, voltage, and resistance follow: \r\(\rV = I \times R\r\)\r. It says that the voltage (\r\(\rV\r\)\r)) across a resistor is equal to the current (\r\(\rI\r\)\r)) flowing through it multiplied by the resistance (\r\(\rR\r\)\r)).

In the context of our resonance scenario in the series RLC circuit, once the capacitive and inductive reactances cancel each other out, Ohm's law directly applies to find the maximum current. We can easily figure out how a series RLC circuit will behave at any given moment, provided we know these three simple values. Ohm’s law is not just a principle; it's a fundamental tool used everywhere electricity flows, from enormous power stations to the tiny circuits in our smartphones.

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Most popular questions from this chapter

Cell phones that use \(4 \mathrm{G}\) technology receive signals broadcast between \(2 \mathrm{GHz}\) and \(8 \mathrm{GHz}\) (a) If you want to create a simple \(L-R-C\) series circuit to detect a \(4.0 \mathrm{GHz}\) cell phone signal, what is the relevant value of the product \(L C,\) where \(L\) is the inductance and \(C\) is the capacitance? (b) If you choose a capacitor that has \(C=1.0 \times 10^{-15} \mathrm{~F}\), what inductance do you need? (c) Suppose you want to wind your own toroidal inductor and fit it inside a box as thin as your cell phone. Based on the size of your phone, estimate the largest cross- sectional area possible for this. (d) Assume the largest allowable radius of the toroid is \(1.0 \mathrm{~cm}\) and estimate the lowest number of windings needed to create your inductor, assuming the material inside has a relative permeability of 1 .

The power of a certain CD player operating at \(120 \mathrm{~V} \mathrm{rms}\) is \(20.0 \mathrm{~W}\). Assuming that the CD player behaves like a pure resistor, find (a) the maximum instantaneous power; (b) the rms current; (c) the resistance of this player.

A series circuit has an impedance of \(60.0 \Omega\) and a power factor of 0.720 at \(50.0 \mathrm{~Hz}\). The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

In an \(L-R-C\) series circuit, \(L=0.280 \mathrm{H}\) and \(C=4.00 \mu \mathrm{F}\). The voltage amplitude of the source is \(120 \mathrm{~V}\). (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 A. What is the resistance \(R\) of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

A circuit has an ac voltage source and a resistor and capacitor connected in series. There is no inductor. The ac voltage source has voltage amplitude \(900 \mathrm{~V}\) and angular frequency \(\omega=20.0 \mathrm{rad} / \mathrm{s}\). The voltage amplitude across the capacitor is \(500 \mathrm{~V}\). The resistor has resistance \(R=300 \Omega\). (a) What is the voltage amplitude across the resistor? (b) What is the capacitance \(C\) of the capacitor? (c) Does the source voltage lag or lead the current? (d) What is the average rate at which the ac source supplies electrical energy to the circuit?

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