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Chapter 33: Problem 14

(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 22.0-碌F capacitor has a reactance below 175 惟 for frequencies above 412.53 Hz. The reactance of a 44.0-碌F capacitor over the same frequency range is 87.5 惟.

Step by step solution

01

Find the frequency range for the 22.0-碌F capacitor

Given that \(X_C = 175 惟\) and \(C = 22.0 碌F\), we can use the formula for capacitive reactance to solve for the frequency \(f\), by rearranging the formula to \(f = \frac{1}{2蟺*X_C*C}\). This gives \(f = \frac{1}{2 * 蟺 * 175 惟 * 22.0 * 10^{-6} F} = 412.53 Hz\). Therefore, the 22.0-碌F capacitor has a reactance below 175 惟 for frequencies above 412.53 Hz.
02

Calculate reactance for the 44.0-碌F capacitor

Using the same frequency range (since the frequency does not change), we can now find the reactance of the 44.0-碌F capacitor using the same formula. Here, \(C = 44.0 碌F\) and \(f = 412.53 Hz\). Substituting these into the formula gives \(X_C = \frac{1}{2 * 蟺 * 412.53 Hz * 44.0 * 10^{-6} F} = 87.5 惟\). So, the reactance of a 44.0-碌F capacitor over the same frequency range is 87.5 惟.

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Key Concepts

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

Frequency
Frequency is an essential concept when dealing with capacitive reactance. It refers to how often an event occurs per unit of time. In the context of electrical circuits, it measures how many cycles a waveform completes in one second. This is measured in Hertz (Hz).

When considering capacitive reactance, the frequency of the circuit determines how quickly the capacitor can charge and discharge. A higher frequency means a quicker charge and discharge cycle.

In formula terms, frequency is part of the calculation for reactance: \( f = \frac{1}{2\pi X_C C} \). Understanding frequency helps in predicting how components like capacitors behave at different speeds of operation in a circuit.
Capacitance
Capacitance is the ability of a component to store charge. It's essentially the capacity of a capacitor. Measured in Farads (F), capacitance determines how much charge a capacitor can hold at a given voltage.

There are different types of capacitors, each with unique characteristics. However, they all serve the fundamental purpose of holding and releasing electrical energy in circuits. The greater the capacitance, the more charge a capacitor can store.

In our calculations, capacitance is pivotal in determining the capacitive reactance: the opposition a capacitor offers to changes in voltage over time. It's part of the formula \( X_C = \frac{1}{2\pi f C} \), which helps calculate reactance given a specific frequency and capacitance.
Reactance Formula
The reactance formula is a mathematical representation of how capacitors react to alternating current (AC) circuits. Reactance is measured in Ohms (惟) and specifically indicates how much a capacitor resists the flow of AC.

The formula for capacitive reactance is \( X_C = \frac{1}{2 \pi f C} \), where:
  • \(X_C\) is the capacitive reactance,
  • \(f\) is the frequency in Hertz,
  • \(C\) is the capacitance in Farads.
The reactance decreases with an increase in frequency or capacitance.

Using this formula is crucial when designing and analyzing circuits as it determines how capacitors will influence the flow of current. Understanding the reactance allows engineers to adjust either frequency or capacitance to meet the desired specifications of the circuit.

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Most popular questions from this chapter

Consider a series \(R L C\) circuit having the following circuit parameters: \(R=200 \Omega, L=663 \mathrm{mH},\) and \(C=26.5 \mu \mathrm{F}\) The applied voltage has an amplitude of \(50.0 \mathrm{V}\) and a frequency of \(60.0 \mathrm{Hz}\). Find the following amplitudes: (a) The current \(I_{\max },\) including its phase constant \(\phi\) relative to the applied voltage \(\Delta v,\) (b) the voltage \(\Delta V_{R}\) across the resistor and its phase relative to the current, (c) the voltage \(\Delta V_{C}\) across the capacitor and its phase relative to the current, and (d) the voltage \(\Delta V_{L}\) across the inductor and its phase relative to the current.

The tuning circuit of an AM radio contains an \(L C\) combination. The inductance is \(0.200 \mathrm{mH},\) and the capacitor is variable, so that the circuit can resonate at any frequency between \(550 \mathrm{kHz}\) and \(1650 \mathrm{kHz}\). Find the range of values required for \(C.\)

A coil of resistance \(35.0 \Omega\) and inductance \(20.5 \mathrm{H}\) is in series with a capacitor and a \(200-\mathrm{V}\) (rms), \(100-\mathrm{Hz}\) source. The rms current in the circuit is \(4.00 \mathrm{A}\). (a) Calculate the capacitance in the circuit. (b) What is \(\Delta V_{\mathrm{rms}}\) across the coil?

A transmission line that has a resistance per unit length of \(4.50 \times 10^{-4} \Omega / \mathrm{m}\) is to be used to transmit \(5.00 \mathrm{MW}\) over 400 miles \(\left(6.44 \times 10^{5} \mathrm{m}\right) .\) The output voltage of the generator is \(4.50 \mathrm{kV} .\) (a) What is the line loss if a transformer is used to step up the voltage to \(500 \mathrm{kV} ?\) (b) What fraction of the input power is lost to the line under these circumstances? (c) What If? What difficulties would be encountered in attempting to transmit the \(5.00 \mathrm{MW}\) at the generator voltage of \(4.50 \mathrm{kV} ?\)

The secondary voltage of an ignition transformer in a furnace is \(10.0 \mathrm{kV} .\) When the primary operates at an rms voltage of \(120 \mathrm{V}\), the primary impedance is \(24.0 \Omega\) and the transformer is \(90.0 \%\) efficient. (a) What turns ratio is required? What are (b) the current in the secondary and (c) the impedance in the secondary?
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