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Ch. 30 - Inductance, Electromagnetic Oscillations, and AC Circuits
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 30 - Inductance, Electromagnetic Oscillations, and AC CircuitsProblem 47
Chapter 29, Problem 47

A 1.50-k Ω resistor in series with a 370-mH inductor is driven by an ac power supply. At what frequency is the impedance double that of the impedance at 60.0 Hz?

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Start by recalling the formula for the impedance of an R-L circuit: \( Z = \sqrt{R^2 + (\omega L)^2} \), where \( R \) is the resistance, \( \omega \) is the angular frequency \( \omega = 2\pi f \), and \( L \) is the inductance.
The problem states that the impedance \( Z \) at the unknown frequency \( f \) is double the impedance at \( f = 60.0 \ \text{Hz} \). Write this relationship as \( Z_{\text{new}} = 2Z_{60} \).
Substitute the impedance formula into the relationship: \( \sqrt{R^2 + (\omega_{\text{new}} L)^2} = 2 \sqrt{R^2 + (\omega_{60} L)^2} \), where \( \omega_{\text{new}} = 2\pi f_{\text{new}} \) and \( \omega_{60} = 2\pi \times 60.0 \).
Square both sides of the equation to eliminate the square root: \( R^2 + (\omega_{\text{new}} L)^2 = 4[R^2 + (\omega_{60} L)^2] \). Expand and rearrange to isolate \( \omega_{\text{new}} \): \( (\omega_{\text{new}} L)^2 = 4R^2 + 4(\omega_{60} L)^2 - R^2 \).
Solve for \( \omega_{\text{new}} \) by taking the square root: \( \omega_{\text{new}} = \sqrt{\frac{4R^2 + 4(\omega_{60} L)^2 - R^2}{L^2}} \). Finally, convert \( \omega_{\text{new}} \) back to frequency using \( f_{\text{new}} = \frac{\omega_{\text{new}}}{2\pi} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Impedance in AC Circuits

Impedance is the total opposition that a circuit offers to the flow of alternating current (AC) and is represented as a complex number. It combines resistance (R) and reactance (X), where reactance arises from inductors and capacitors. The impedance of a series circuit with a resistor and inductor can be calculated using the formula Z = √(R² + (ωL)²), where ω is the angular frequency and L is the inductance.
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Inductive Reactance

Inductive reactance (X_L) is the opposition to the change in current flow through an inductor in an AC circuit, given by the formula X_L = ωL, where ω is the angular frequency (ω = 2πf) and L is the inductance. As frequency increases, the inductive reactance increases, which affects the overall impedance of the circuit. This concept is crucial for understanding how the impedance changes with frequency.
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Frequency and Impedance Relationship

The relationship between frequency and impedance in an R-L circuit is significant because the impedance varies with frequency due to the inductive reactance. To find the frequency at which the impedance is double that at a given frequency (e.g., 60 Hz), one must analyze how the reactance changes with frequency and apply the impedance formula accordingly. This relationship is essential for solving problems involving AC circuits.
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Related Practice
Textbook Question

The frequency of the ac voltage source (peak voltage Vo) in an LRC circuit is tuned to the circuit’s resonant frequency f₀ = 1 / (2π√LC). (a) Show that the peak voltage across the capacitor is Vco = VoTo/ (2πτ), where To ( =1/fo) is the period of the resonant frequency and τ = RC is the time constant for charging the capacitor C through a resistor R. (b) Define β = To/ (2πτ) so that Vco = βVo. Then β is the “amplification” of the source voltage across the capacitor. If a particular LRC circuit contains a 2.0-nF capacitor and has a resonant frequency of 5.0 kHz, what value of R will yield β = 125?

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Textbook Question

An ac voltage source is connected in series with a 2.0-μF capacitor and a 750-Ω resistor. Using a digital ac voltmeter, the voltage source is measured to be 4.0 V rms, and the voltages across the resistor and across the capacitor are found to be 3.0 V rms and 2.7 V rms, respectively. Determine the frequency of the ac voltage source. Why is the voltage measured across the voltage source not equal to the sum of the voltages measured across the resistor and across the capacitor?

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Textbook Question

A 10.0-k Ω resistor is in series with a 34.0-mH inductor and an ac source. Calculate the impedance of the circuit if the source frequency is (a) 55.0 Hz; (b) 55.0 kHz.

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Textbook Question

(II) A 25-mH coil whose resistance is 0.80 Ω is connected to a capacitor C and a 420-Hz source voltage. If the current and voltage are to be in phase, what value must C have?

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Textbook Question

An average power output of 150 W is sent into a 4-Ω loudspeaker (see Fig. 25–14). What are the rms voltage and the rms current fed to the speaker at 1.0 W when the volume is turned down?

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Textbook Question

(II) A capacitor is placed in parallel with some device, B, as in Fig. 30–18b, to filter out stray high-frequency signals, but to allow ordinary 60.0-Hz ac to pass through with little loss. Suppose that circuit B in Fig. 30–18b is a resistance R = 530 Ω connected to ground, and that C = 0.35 μF. Calculate the ratio of the capacitor’s current amplitude to the incoming current’s amplitude if the incoming current has a frequency of (a) 60.0 Hz; (b) 60.0 kHz.