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(a) For what frequencies does a \(22.0-\mu \mathrm{F}\) capacitor have a reactance below 175\(\Omega ?\) (b) What If? Over this same frequency range, what is the reactance of a \(44.0-\mu \mathrm{F}\) capacitor?

Short Answer

Expert verified
The frequency must be greater than approximately 51.3 Hz for a 22.0-\textmu F capacitor to have a reactance below 175\textOmega. For the 44.0-\textmu F capacitor, over this same frequency range, the reactance will be half of that for the 22.0-\textmu F capacitor.

Step by step solution

01

Understand the concept of capacitive reactance

The capacitive reactance (Xc) for a capacitor is given by the formula \( X_c = \frac{1}{2 \pi f C} \), where \( f \) is the frequency in hertz, \( C \) is the capacitance in farads, and \( \pi \) is pi, approximately 3.14159. The reactance of a capacitor decreases with increasing frequency. We are given a capacitance of \( 22.0-\mu\mathrm{F} \), and we need to find the frequencies that result in a reactance below \( 175\Omega \).
02

Convert capacitance to farads

First, convert the capacitance from microfarads to farads. \( 22.0-\mu\mathrm{F} = 22.0 \times 10^{-6} \mathrm{F} \).
03

Set up the inequality to solve for frequency

Since we need to find the frequency where the reactance is below \( 175\Omega \), we set up the inequality as \( \frac{1}{2 \pi f C} < 175 \). Substitute the capacitance value in farads: \( \frac{1}{2 \pi f \times 22.0 \times 10^{-6}} < 175 \).
04

Solve for the frequency

Rearrange the inequality to solve for \( f \). Multiply both sides by \( f \) and then by \( 2 \pi C \), and then divide both sides by 175 to get: \( f > \frac{1}{2 \pi \times 175 \times 22.0 \times 10^{-6}} \). Compute the right-hand side to find the minimum frequency. Using a calculator, \( f > \frac{1}{2 \pi \times 175 \times 22.0 \times 10^{-6}} \approx 51.3 \) Hz.
05

Double the capacitance to find new reactance

For the second part of the problem, we now double the capacitance. Since reactance is inversely proportional to capacitance, the new reactance will be half of the original for a given frequency. So if the original reactance was \( X_{c1} \), the new reactance \( X_{c2} \) for the \( 44.0-\mu\mathrm{F} \) capacitor will be \( X_{c2} = \frac{1}{2} X_{c1} \). In terms of frequency: \( X_{c2} = \frac{1}{2 \pi f \times 44.0 \times 10^{-6}} \). Since the frequency range is the same as the one found in the previous step, the reactance of the \( 44.0-\mu\mathrm{F} \) capacitor will also be below \( 175\Omega \) in this range.

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Key Concepts

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

Frequency and Reactance Relation
Reactance is a fundamental concept in electrical circuits, particularly when dealing with capacitors. The key takeaway here is that capacitive reactance, denoted as Xc, is inversely proportional to the frequency of the electrical signal passing through the capacitor. In other words, as the frequency increases, the opposition to current flow posed by the capacitor's reactance decreases.

This relationship can be mathematically represented by the formula:
\[\begin{equation}X_c = \frac{1}{2 \pi f C}\end{equation}\]where f stands for the frequency in hertz and C represents the capacitance in farads. It's essential to consider this relationship when designing circuits, as it influences how a signal will be affected by a capacitor at different frequencies. For instance, a capacitor may act as a near short-circuit at high frequencies, while significantly impeding the flow of current at low frequencies. This principle is used in various applications such as filters and tuning circuits.
Electrical Impedance
Electrical impedance is the measure of the total opposition that a circuit presents to the flow of alternating current. It is often symbolized by Z and combines both resistive and reactive components of a circuit into one complex number. Reactance, which includes capacitive reactance Xc and inductive reactance XL, is a part of this overall impedance.

Impedance is not just about resistance; it also entails how the circuit behaves in response to different frequencies due to these reactive elements. Capacitive reactance specifically shows that capacitors present less impedance as the frequency increases, illustrating their frequency-dependent nature. An interesting note is that impedance is at its minimum at a frequency called the resonant frequency, where capacitive and inductive reactances cancel each other out. Overall, understanding impedance is crucial for analyzing AC circuits and optimizing their performance.
Capacitance in Farads
Capacitance is a measure of a capacitor’s ability to store charge per unit voltage. It is quantified in farads (F), a unit named after English physicist Michael Faraday. In real-world scenarios, capacitance values are often much smaller than a farad, commonly expressed in microfarads (µF), which are one millionth of a farad.

When calculating the capacitive reactance for a capacitor in an AC circuit, it's vital to convert the capacitance from microfarads to farads to ensure proper units are used in the formula:
\[\begin{equation} X_c = \frac{1}{2 \pi f C} \end{equation}\]Precision in these conversions is important for accurate calculations. The capacitance of a capacitor is determined by factors such as the area of the plates, the distance between them, and the dielectric material used. This property doesn't change with the applied voltage or current; however, it's a critical factor in determining how much reactance the capacitor will have to an AC signal at a given frequency.

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Most popular questions from this chapter

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance R is equal to the inductive reactance. If the plate separation of the capacitor is reduced to half of its original value, the current in the circuit doubles. Find the initial capacitive reactance in terms of R.

A step-down transformer is used for recharging the batteries of portable devices such as tape players. The turns ratio inside the transformer is 13:1 and it is used with 120-V (rms) household service. If a particular ideal transformer draws 0.350 A from the house outlet, what are (a) the voltage and (b) the current supplied to a tape player from the transformer? (c) How much power is delivered?

An inductor \((L=400 \mathrm{mH}),\) a capacitor \((C=4.43 \mu \mathrm{F}),\) and a resistor \((R=500 \Omega)\) are connected in series. A \(50.0-\mathrm{Hz}\) AC source produces a peak current of 250 \(\mathrm{mA}\) in the circuit. (a) Calculate the required peak voltage \(\Delta V_{\text { max }}\) (b) Determine the phase angle by which the current leads or lags the applied 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 \(\Delta V_{\mathrm{rms}}=210 \mathrm{V} ?\)

A series \(R L C\) circuit consists of an \(8.00-\Omega\) resistor, a \(5.00-\mu \mathrm{F}\) capacitor, and a 50.0 \(\mathrm{-mH}\) inductor. A variable frequency source applies an emf of 400 V (rms) across the combination. Determine the power delivered to the circuit when the frequency is equal to half the resonance frequency.

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