Ch 31: Alternating Current

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 31: Alternating Current

Problem 19a

Chapter 31, Problem 19a

The power of a certain CD player operating at 120 V rms is 20.0 W. Assuming that the CD player behaves like a pure resistor, find the maximum instantaneous power.

Verified step by step guidance

Understand that the CD player is modeled as a pure resistor, which means it follows Ohm's Law. The power formula for a resistor is P = V^2 / R, where P is the power, V is the voltage, and R is the resistance.

Given that the CD player operates at 120 V rms and has a power of 20.0 W, use the formula P = V_rms^2 / R to find the resistance R. Rearrange the formula to R = V_rms^2 / P.

Calculate the resistance using the given values: R = (120 V)^2 / 20.0 W.

To find the maximum instantaneous power, recall that the maximum power in an AC circuit occurs at the peak voltage, which is V_peak = V_rms * sqrt(2).

Use the formula for instantaneous power P_max = V_peak^2 / R to find the maximum power. Substitute V_peak and the resistance R calculated earlier into this formula.

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Key Concepts

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

RMS Voltage

RMS (Root Mean Square) voltage is a measure of the effective value of an alternating current (AC) voltage, equivalent to a DC voltage that delivers the same power. It is calculated as the square root of the average of the squares of the instantaneous values over one cycle. For a sinusoidal waveform, RMS voltage is 0.707 times the peak voltage.

Power in Resistors

Power in resistors is calculated using the formula P = V^2/R, where P is the power, V is the voltage across the resistor, and R is the resistance. In AC circuits, the power can also be expressed using RMS values, as P = Vrms^2/R, which gives the average power consumed by the resistor over time.

Instantaneous Power

Instantaneous power in an AC circuit is the power at any given moment, calculated as p(t) = v(t) * i(t), where v(t) is the instantaneous voltage and i(t) is the instantaneous current. For a pure resistor, the maximum instantaneous power occurs at the peak voltage and current, which is twice the average power calculated using RMS values.

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Textbook Question

In an L-R-C series circuit, the components have the following values: L = 20.0 mH, C = 140 nF, and R = 350 Ω.The generator has an rms voltage of 120 V and a frequency of 1.25 kHz. Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

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Textbook Question

You have a 200-Ω resistor, a 0.400-H inductor, and a 6.00-μF capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude 30.0 V and an angular frequency of 250 rad/s. What is the impedance of the circuit?

Textbook Question

An L-R-C series circuit with L = 0.120 H, R = 240 Ω, and C = 7.30 μF carries an rms current of 0.450 A with a frequency of 400 Hz. What are the phase angle and power factor for this circuit?

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Textbook Question

An L-R-C series circuit is connected to a 120-Hz ac source that has V_rms = 80.0 V. The circuit has a resistance of 75.0 Ω and an impedance at this frequency of 105 Ω. What average power is delivered to the circuit by the source?

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Textbook Question

A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 286 W. What is (a) the impedance Z of the circuit; (b) the amplitude of the voltage across the inductor; (c) the power factor?

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Textbook Question

You have a 200-Ω resistor, a 0.400-H inductor, and a 6.00-μF capacitor. They are connected to form an L-R-C series circuit. What is the current amplitude?