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Chapter 11: Problem 38

A series \(\mathrm{RC}\) circuit with \(R=10 \Omega\) has an impendance with phase angle of \(45^{\circ}\) at frequency \(f=500 \mathrm{~Hz}\). Find the frequency for which the magnitude of the impedance is (a) twice that off, (b) one-half that at \(f\).

Short Answer

Expert verified
For twice the impedance, frequency is reduced. For half, frequency increases.

Step by step solution

01

Understanding the Resistor and Capacitor Impedance

In a series RC circuit, impedance \( Z \) is given by \( Z = \sqrt{R^2 + (X_C)^2} \), where \( X_C \) is the capacitive reactance. The phase angle \( \theta \) is related to these components by \( \tan \theta = \frac{X_C}{R} \). Given \( \theta = 45^{\circ} \), \( \tan 45^{\circ} = 1 \), thus \( X_C = R = 10 \Omega \).
02

Calculating Capacitive Reactance

At \( f = 500 \) Hz, the capacitive reactance \( X_C \) can be calculated using \( X_C = \frac{1}{2\pi fC} \). Substituting \( X_C = 10 \Omega \) and \( f = 500 \) Hz we get \( C = \frac{1}{2\pi \times 500 \times 10} \).
03

Solving for Twice the Impedance Magnitude

The new impedance for twice the magnitude, \( 2Z \), is \( 2 \times \sqrt{2}R \), since \( Z = \sqrt{2}R \) at \( 45^{\circ} \). Solve \( \sqrt{R^2 + (X_C')^2} = 2\sqrt{2}R \) with \( X_C' = \frac{1}{2\pi f'C} \) and find \( f' \).
04

Solving for Half the Impedance Magnitude

For half the impedance, \( Z' = \frac{1}{2} \times \sqrt{2}R \), similarly resolve using \(\sqrt{R^2 + (X_C'')^2} = \frac{1}{2}\sqrt{2}R\). Calculate \( f'' \) using \( X_C'' = \frac{1}{2\pi f''C} \).
05

Calculation and Reasoning Verification

Substitute \( C \) into equations from steps 3 and 4, solve for \( f' \) and \( f'' \), verifying logical consistency and constraint satisfaction given circuit properties.

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Key Concepts

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

Impedance Calculation
Impedance in an RC circuit refers to the combination of resistive and capacitive effects that oppose the flow of alternating current. It is represented by the symbol \( Z \) and can be thought of as the total opposition to the current flow. Impedance is a complex quantity, typically involving both magnitude and phase components. To calculate the impedance in an RC circuit, the following formula is used:
  • \( Z = \sqrt{R^2 + X_C^2} \), where \( R \) is the resistance and \( X_C \) is the capacitive reactance.
The magnitude of the impedance is essentially how much the circuit "resists" the current. It's important to note that impedance is frequency-dependent. An increase or decrease in frequency can alter the capacitive reactance, \( X_C \), and thus change the overall impedance. This equation allows for calculating the impedance when the resistance \( R \) and the frequency-dependent reactance \( X_C \) are known.
Phase Angle in RC Circuits
In an RC circuit, the phase angle \( \theta \) represents the phase difference between the voltage and the current. It quantifies how much the current waveform is "ahead" or "behind" the voltage waveform. This phase difference occurs due to the property of the capacitor which causes the current to lead the voltage. The phase angle can be calculated using the relationship:
  • \( \tan \theta = \frac{X_C}{R} \), where \( X_C \) is the capacitive reactance and \( R \) is the resistance of the circuit.
For instance, a phase angle of \( 45^{\circ} \) indicates a specific scenario where the capacitive reactance is equal to the resistance, leading to a condition known as "equal phase shift." This means the circuit presents a balanced characteristic to both resistive and capacitive influences. Understanding the phase angle in RC circuits is crucial because it affects how power is stored and released by the circuit, impacting its efficiency.
Capacitive Reactance
Capacitive reactance, \( X_C \), is a measure of a capacitor's opposition to the change of voltage across it in an AC circuit. It is a crucial element of determining the impedance in an RC circuit. The formula for calculating capacitive reactance is:
  • \( X_C = \frac{1}{2\pi f C} \), where \( f \) is the frequency of the AC signal and \( C \) is the capacitance.
This equation highlights the inverse relationship between capacitive reactance and frequency. As frequency increases, capacitive reactance decreases, making the capacitor "less resistive" to the flow of the AC signal. Conversely, a lower frequency means higher capacitive reactance, offering more "resistance" to current changes.Understanding capacitive reactance is key because it dictates how a capacitor will behave at different frequencies and affects the total impedance of the circuit.

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Most popular questions from this chapter

An \(L-R-C\) series circuit consists of a source with voltage amplitude \(120 \mathrm{~V}\) and angular frequency \(50.0 \mathrm{rad} / \mathrm{s}\), a resistor with \(R=400 \Omega\), an inductor with \(L=9.00 \mathrm{H}\), and a capacitor with capacitance \(C\). (a) For what value of \(C\) will the current amplitude in the circuit be a maximum? (b) When \(C\) has the value calculated in part (a), what is the amplitude of the voltage across the inductor?

An inductor with \(L=10 \mathrm{mH}\) is connected across an ac source that has voltage amplitude \(44 \mathrm{~V}\). (a) What is the phase angle \(\phi\) for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What value for the frequency of the source results in a current amplitude of \(4 \mathrm{~A}\) ?

You want to double the resonance angular frequency of an \(L-R-C\) series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply \(12.0 \mathrm{~V}\) (rms) to a portable electronic device. The load resistance in the secondary is \(5.00 \Omega\). (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply?

Off to Europe! You plan to take your hair dryer to Europe, where the electrical outlets put out \(240 \mathrm{~V}\) instead of the \(120 \mathrm{~V}\) seen in the United States. The dryer puts out \(1600 \mathrm{~W}\) at \(120 \mathrm{~V}\). (a) What could you do to operate your dryer via the \(240-\mathrm{V}\) line in Europe? (b) What current will your dryer draw from a European outlet? (c) What resistance will your dryer appear to have when operated at \(240 \mathrm{~V}\) ?
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