/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A series \(R L C\) circuit has c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A series \(R L C\) circuit has components with following values: \(L=20.0 \mathrm{mH}, C=100 \mathrm{nF}, R=20.0 \Omega,\) and \(\Delta V_{\max }=100 \mathrm{V}\) with \(\Delta v=\Delta V_{\ln \ln x} \sin \omega t .\) Find \((a)\) the resonant frequency, (b) the amplitude of the current at the resonant frequency, (c) the \(Q\) of the circuit, and (d) the amplitude of the voltage across the inductor at resonance.

Short Answer

Expert verified
a) The resonant frequency is \(f = \frac{1}{2\pi\sqrt{LC}} \) Hz, b) The amplitude of the current at the resonant frequency is \(I = 5 A\), c) The quality factor of the circuit is \(Q = \frac{1}{R}\sqrt{\frac{L}{C}}\), and d) The amplitude of the voltage across the inductor at resonance is \(V_L = Q \times 100 V\).

Step by step solution

01

Find the Resonant Frequency

The resonant frequency can be calculated by the formula \(f = \frac{1}{2\pi\sqrt{LC}} \). Substitute the given values of inductance \(L = 20.0 mH = 20 \times 10^{-3} H \) and capacitance \(C = 100 nF = 100 \times 10^{-9} F\) into this formula. The resonant frequency is calculated as \(f = \frac{1}{2\pi\sqrt{20 \times 10^{-3} H \times 100 \times 10^{-9} F}} \).
02

Find the Amplitude of the Current

The amplitude of the current can be obtained by the formula \(I = \frac{\Delta V_{max}}{R}\). Substitute the given values into this formula: \(I = \frac{100}{20}\). The amplitude of the current at the resonant frequency is then equal to \(I = 5 A\).
03

Find the Q of the Circuit

The quality factor Q can be calculated by the formula \(Q = \frac{1}{R}\sqrt{\frac{L}{C}}\). Replace \( L = 20 \times 10^{-3} H, C = 100 \times 10^{-9} F, R = 20 \Omega \) into the formula: The Q of the circuit is \(Q = \frac{1}{20}\sqrt{\frac{20 \times 10^{-3}}{100 \times 10^{-9}}}\).
04

Find the Amplitude of the Voltage

The amplitude of the voltage across the inductor at resonance can be calculated by the formula \(V_L = Q \times \Delta V_{max}\). Substitute the previously calculated Q and the given maximum value of \( \Delta V_{max} = 100 V \) into the formula: \(V_L = Q \times 100\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Resonant Frequency Formula
Understanding the resonant frequency of an RLC circuit is crucial for grasping how these systems maximize the transfer of power at a particular frequency. The resonant frequency is where the impedance of the circuit is at its minimum and the circuit behaves like a purely resistive load. To find the resonant frequency, the formula you'll need is
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
where \(L\) is the inductance, \(C\) is the capacitance, and \(f\) is the frequency. Resonance occurs due to the balance between the reactance of the inductor and the capacitor, which at this specific frequency cancel each other out. This concept is not just theoretical; it has practical applications in radio transmission, audio systems, and wireless power transfer, where tuning to the resonant frequency allows the systems to operate efficiently.
Amplitude of Current
The amplitude of the current in an RLC circuit is a measurement of how much charge flows through the circuit per time at its peak. At resonance, this is particularly important because it is when the circuit allows the highest current to flow for a given voltage.
The formula to find the amplitude of current at resonant frequency is quite straightforward: \[ I = \frac{\Delta V_{max}}{R} \]
where \(I\) is the current amplitude, \(\Delta V_{max}\) is the maximum voltage, and \(R\) is resistance. This value is maximized at resonance because the inductive and capacitive reactances cancel out, leaving the resistance as the only opposition to current flow.
For applications like broadcasting, the current amplitude at resonant frequency is essential as it relates to the signal's strength transmitted.
Quality Factor (Q)
The Quality Factor, often denoted by \(Q\), provides a measure of the sharpness of the resonance of the circuit. A higher \(Q\) translates to a narrower, more 'selective' peak of resonance, which means that the RLC circuit is more selective in its resonant frequency. The formula to compute the quality factor is: \[ Q = \frac{1}{R}\sqrt{\frac{L}{C}} \]
where \(R\) is the resistance, \(L\) is the inductance, and \(C\) is the capacitance. Quality factor is critical in determining the bandwidth of the circuit and is particularly important for filters in communication systems, where distinguishing between closely spaced frequencies is vital.
Voltage Across Inductor
At resonance, the voltage across the inductor can be quite a surprising phenomenon to consider. Because the reactive impedance of the inductor and capacitor are equal and opposite, it's possible for the voltage across one component to be significantly larger than the voltage across the series combination of components. The formula used for finding the amplitude of voltage across the inductor at resonance is: \[ V_L = Q \times \Delta V_{max} \]
This relationship showcases that the voltage across the inductor can be a multiple of the applied voltage, depending on the quality factor.
The concept of voltage magnification is important in the workings of devices like metal detectors and resonant transformers, where the voltage across the inductor is used to detect metal or to step-up voltage efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An AC voltage of the form \(\Delta v=(100 \mathrm{V}) \sin (1000 t)\) is applied to a series \(R L C\) circuit. Assume the resistance is \(400 \Omega,\) the capacitance is \(5.00 \mu \mathrm{F},\) and the inductance is \(0.500 \mathrm{H} .\) Find the average power delivered to the circuit.

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.

Suppose you manage a factory that uses many electric motors. The motors create a large inductive load to the electric power line, as well as a resistive load. The electric company builds an extra-heavy distribution line to supply you with a component of current that is \(90^{\circ}\) out of phase with the voltage, as well as with current in phase with the voltage. The electric company charges you an extra fee for "reactive volt-amps," in addition to the amount you pay for the energy you use. You can avoid the extra fee by installing a capacitor between the power line and your factory. The following problem models this solution. In an \(R L\) circuit, a \(120-\mathrm{V}\) (rms), \(60.0-\mathrm{Hz}\) source is in series with a 25.0 -mH inductor and a \(20.0-\Omega\) resistor. What are (a) the rms current and (b) the power factor? (c) What capacitor must be added in series to make the power factor \(1 ?\) (d) To what value can the supply voltage be reduced, if the power supplied is to be the same as before the capacitor was installed?

An \(80.0-\Omega\) resistor and a \(200-\mathrm{mH}\) inductor are connected in parallel across a \(100-\mathrm{V}\) (rms), \(60.0-\mathrm{Hz}\) source. (a) What is the rms current in the resistor? (b) By what angle does the total current lead or lag behind the voltage?

A series \(R L C\) circuit has a resistance of \(45.0 \Omega\) and an impedance of \(75.0 \Omega .\) What average power is delivered to this circuit when \(\Delta V_{\mathrm{rms}}=210 \mathrm{V} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.