/*! 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 8 Find the resonant frequency of a... [FREE SOLUTION] | 91影视

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Find the resonant frequency of a circuit containing a \(10.0-\mu \mathrm{F}\) capacitor in series with a \(37.5-\mu \mathrm{H}\) inductor.

Short Answer

Expert verified
The resonant frequency is approximately 8220.78 Hz.

Step by step solution

01

Recall the Formula for Resonant Frequency

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02

Convert Unit Measurements

To start with, convert the capacitance value of the capacitor and the inductance value of the inductor into standard units. The capacitance value is given as 10.0 碌F, which is equivalent to \(10.0 \times 10^{-6}\) F. The inductance value is given as 37.5 碌H, which is equivalent to \(37.5 \times 10^{-6}\) H.
03

Substitute the Values into the Formula

Insert the values of the inductance and capacitance into the resonant frequency formula:\[ f = \frac{1}{2\pi \sqrt{LC} } \]Given that \(L = 37.5 \times 10^{-6}\) H and \(C = 10.0 \times 10^{-6}\) F, substitute these values into the formula to find the resonant frequency.
04

Calculate the Expression Under the Square Root

Calculate the product of the inductance \(L\) and capacitance \(C\): \[ LC = (37.5 \times 10^{-6}) \times (10.0 \times 10^{-6}) = 375.0 \times 10^{-12} \] This value (\(LC\)) will be used to find the square root in the next step.
05

Compute the Square Root and Resulting Frequency

Find the square root of the product calculated in Step 4: \[ \sqrt{LC} = \sqrt{375.0 \times 10^{-12}} = 19.36 \times 10^{-6} \] Substitute back into the frequency equation:\[ f = \frac{1}{2\pi (19.36 \times 10^{-6})} \approx \frac{1}{121.71 \times 10^{-6}} \approx 8220.78 \text{ Hz} \] The resonant frequency is approximately 8220.78 Hz.

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Key Concepts

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

Circuit Analysis
Analyzing circuits is a fundamental skill in electronics and physics, involving the study of how electrical components are arranged and interact. It requires understanding different components like resistors, capacitors, and inductors, all of which play a unique role in current flow and energy storage. In this instance, we are dealing with a series circuit, which means the components are arranged in a single path for the current. This setup is quite common in physics problems involving alternating current (AC) circuits. With every component having specific characteristics, circuit analysis allows us to predict how the circuit will behave when powered. It involves using mathematical equations to describe the relationships between voltage, current, and resistance.
Capacitance
Capacitance is the ability of a component to store and release electrical energy in the form of an electric charge. In an AC circuit, capacitors temporarily store energy in an electric field, which can later be released back into the circuit. Capacitance is measured in Farads (F), but due to the usually small amount of capacitance required, we often see microfarads (碌F) used, as in the 10.0 碌F capacitor of this exercise.
Capacitors in AC circuits help to maintain voltage levels and manage power supply, crucial for devices that require a steady voltage to function correctly. The capacitance of a capacitor influences the resonant frequency of a circuit, determining how it will vibrate at certain frequencies.
Inductance
Inductance is a property of electrical conductors, such as coils, which enables them to resist changes in current flow. This resistance is due to the magnetic field generated around the conductor when electrical current passes through it. Inductors, like capacitors, store energy, but in a magnetic field rather than an electric one. Inductance is measured in Henrys (H), and in the case of this exercise, we handle a relatively small inductance value of 37.5 碌H.
Inductors are essential in AC circuits as they can filter signals, manage power, and influence the circuit's resonant frequency, acting as a complement to capacitors. The interaction between inductance and capacitance is a pivotal factor in defining the circuit's behavior, especially concerning resonance.
Physics Problem Solving
Solving physics problems often involves translating words into mathematical equations, a skill that is developed through practice. This conversion requires a firm understanding of the units and relationships between different physical quantities. In our exercise, understanding the resonant frequency involves using a specific formula involving capacitance and inductance.
  • Identify the given values and their units.
  • Convert units where necessary to ensure consistency.
  • Substitute the values into the relevant formula.
  • Compute the results carefully, step by step.
Staying organized and systematic is key, as is double-checking calculations for accuracy.
AC Circuits
Alternating Current (AC) circuits are those where the current changes direction periodically. This is different from Direct Current (DC) circuits, where the flow of electricity is in a constant direction. In AC circuits, elements such as capacitors and inductors play key roles in how the circuit responds to different frequencies.
AC circuits are critical in residential and industrial settings as they can efficiently transmit power over long distances. Understanding how AC circuits work, including the concepts of impedance, reactance, and resonance, gives insight into how to maximize efficiency and safely operate electrical systems. In particular, the resonant frequency, calculated in the exercise, is significant because it represents the frequency at which the circuit naturally oscillates with the greatest amplitude due to the resonance between inductance and capacitance.

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