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A series circuit contains a \(100.0-\Omega\) resistor, a \(0.500-\mathrm{H}\) inductor, a 0.400 - \(\mu \mathrm{F}\) capacitor, and a source of time-varying emf providing \(40.0 \mathrm{~V}\) a) What is the resonant angular frequency of the circuit? b) What current will flow through the circuit at the resonant frequency?

Short Answer

Expert verified
The resonant angular frequency of the circuit is \(\omega_0 = 2236.07 \, \mathrm{rad/s}\). b) What is the rms current through the circuit at the resonant frequency? The rms current flowing through the circuit at the resonant frequency is \(I = 0.4 \, \mathrm{A}\).

Step by step solution

01

Find the resonant angular frequency of the circuit

To find the resonant angular frequency of the RLC circuit, we can use the formula \(\omega_0 = \frac{1}{\sqrt{LC}}\), where \(\omega_0\) is the resonant angular frequency, \(L\) is the inductance of the inductor, and \(C\) is the capacitance of the capacitor. Here, we are given \(L=0.500 H\) and \(C= 0.400 \times 10^{-6} F\). Plugging the values into the formula, we get: \(\omega_0 = \frac{1}{\sqrt{(0.500 H)(0.400\times10^{-6} F)}}\) Calculate the value of the resonant angular frequency: \(\omega_0 = \frac{1}{\sqrt{2\times10^{-7}}} = 2236.07 \, \mathrm{rad/s}\)
02

Calculate the impedance of the circuit at the resonant frequency

At the resonant frequency, the reactive components of the impedance (inductor and capacitor) will cancel each other out, resulting in only the resistive component being present. This means that the impedance of the circuit at resonance, \(Z_0\), is equal to the resistance \(R\): \(Z_0=R = 100.0 \, \Omega\)
03

Calculate the rms current through the circuit at the resonant frequency

Now that we have the impedance of the circuit at the resonant frequency, we can use Ohm's Law to find the rms current flowing through the circuit. The formula for Ohm's Law is \(I = \frac{V}{Z}\), where \(I\) is the rms current, \(V\) is the rms voltage, and \(Z\) is the impedance of the circuit. We are given an rms voltage of \(40.0 V\), and we calculated the impedance at resonance to be \(100.0 \, \Omega\). Plugging these values into the formula, we get: \(I = \frac{40.0 V}{100.0 \, \Omega} = 0.4 \, \mathrm{A}\) The rms current flowing through the circuit at the resonant frequency is \(0.4 A\).

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Key Concepts

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

Resonant Angular Frequency
The resonant angular frequency is a fundamental concept in RLC series circuits. It refers to the frequency at which the inductive reactance and capacitive reactance are equal in magnitude but opposite in phase, causing them to cancel each other out. As a result, the circuit exhibits purely resistive behavior with minimal impedance.

According to the formula \(\omega_0 = \dfrac{1}{\sqrt{LC}}\), where \(L\) represents inductance and \(C\) capacitance, we can determine the point of resonance. For example, in a circuit with an inductance of \(0.500 \, \mathrm{H}\) and a capacitance of \(0.400 \times 10^{-6} \mathrm{F}\), the resonant angular frequency is calculated to be approximately \(2236.07 \, \mathrm{rad/s}\).

This frequency is crucial as it is the point where the circuit will respond most strongly to an external signal, often used in radio tuning and other applications requiring frequency selection.
Impedance in RLC Circuit
In the context of an RLC series circuit, impedance is the measure of opposition that the circuit offers to the flow of alternating current. It is a combination of resistance, inductive reactance, and capacitive reactance. However, at the resonant frequency, the inductive and capacitive reactances cancel out, simply leaving the resistive component.

For practical calculations, at resonance, Impedance \(Z_0\) is equivalent to the resistance of the circuit \(R\), which ignores the effects of induction and capacitance. So in a circuit with a resistor of \(100.0 \, \Omega\), the impedance at resonance would be \(100.0 \, \Omega\). The understanding of impedance is key to predicting how the circuit will perform under different frequencies and is an essential parameter for designing electronic circuits.
Ohm's Law
  • Ohm's Law is a crucial principle in electrical circuits, defining the relationship between voltage \(V\), current \(I\), and resistance \(R\).
  • It is commonly stated as \(I = \dfrac{V}{R}\), expressing that the current through a conductor between two points is directly proportional to the voltage across the two points.
  • In the setting of an RLC series circuit, Ohm’s Law helps us determine the root mean square (RMS) current, given the RMS voltage and the impedance.
  • At resonance, where the circuit acts purely resistive \(Z = R\), using Ohm's Law simplifies the process of finding the current running through the circuit.
For instance, with a known resistor of \(100.0\, \Omega\) and an applied RMS voltage of \(40.0\, V\), Ohm's Law calculates the RMS current as \(0.4\, A\).
Root Mean Square (RMS) Current
The root mean square (RMS) current is a significant term in AC circuits, and it represents the effective current that would produce the same amount of heat in a resistor as a direct current of the same value. This measurement is crucial because it indicates the real power consumption in a circuit.

The RMS value is derived from the peak current value but accounts for the fact that AC current varies with time. Using the Ohm’s Law formula \(I = \frac{V}{Z}\) at the resonant frequency where impedance is purely resistive, we can calculate the RMS current that flows through the circuit. For an RMS voltage of \(40.0 V\) and an impedance of \(100.0 \Omega\) at resonance, the circuit will have an RMS current of \(0.4 A\). This value allows us to accurately describe the power conditions in the circuit without referring to the constantly changing instantaneous values of current and voltage.

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Most popular questions from this chapter

A 75,000 -W light bulb (yes, there are such things!) operates at \(I_{\mathrm{rms}}=200 . \mathrm{A}\) and \(V_{\mathrm{rms}}=440 . \mathrm{V}\) in a \(60.0-\mathrm{Hz} \mathrm{AC}\) circuit. Find the resistance, \(R,\) and self- inductance, \(L,\) of this bulb. Its capacitive reactance is negligible.

An inductor with inductance \(L\) is connected to an \(A C\) power source that supplies \(V_{\mathrm{emf}}=20.7 \mathrm{~V}\) at \(f=733 \mathrm{~Hz}\). If the reactance of the inductor is to be \(81.52 \Omega\), what should the value of \(L\) be?

A transformer contains a primary coil with 200 turns and a secondary coil with 120 turns. The secondary coil drives a current \(I\) through a \(1.00-\mathrm{k} \Omega\) resistor. If an input voltage of \(V_{\mathrm{rms}}=75.0 \mathrm{~V}\) is applied across the primary coil, what is the power dissipated in the resistor?

The figure shows a simple FM antenna circuit in which \(L=8.22 \mu \mathrm{H}\) and \(C\) is variable (the capacitor can be tuned to receive a specific station). The radio signal from your favorite FM station produces a sinusoidal time-varying emf with an amplitude of \(12.9 \mu \mathrm{V}\) and a frequency of \(88.7 \mathrm{MHz}\) in the antenna. a) To what value, \(C_{0}\), should you tune the capacitor in order to best receive this station? b) Another radio station's signal produces a sinusoidal time-varying emf with the same amplitude, \(12.9 \mu \mathrm{V}\), but with a frequency of \(88.5 \mathrm{MHz}\) in the antenna. With the circuit tuned to optimize reception at \(88.7 \mathrm{MHz},\) what should the value, \(R_{0},\) of the resistance be in order to reduce by a factor of 2 (compared to the current if the circuit were optimized for \(88.5 \mathrm{MHz}\) ) the current produced by the signal from this station?

Laboratory experiments with series RLC circuits require some care, as these circuits can produce large voltages at resonance. Suppose you have a \(1.00-\mathrm{H}\) inductor (not difficult to obtain) and a variety of resistors and capacitors. Design a series RLC circuit that will resonate at a frequency (not an angular frequency) of \(60.0 \mathrm{~Hz}\) and will produce at resonance a magnification of the voltage across the capacitor or the inductor by a factor of 20.0 times the input voltage or the voltage across the resistor.

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