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Chapter 20: Problem 37

The average power used by a stereo speaker is 55 W. Assuming that the speaker can be treated as a \(4.0-\Omega\) resistance, find the peak value of the ac voltage applied to the speaker.

Short Answer

Expert verified
The peak voltage applied to the speaker is approximately 20.96 V.

Step by step solution

01

Understand the Given Data and Formula

The given data includes the average power of the speaker, which is 55 W, and the resistance of the speaker, which is 4.0 惟. We are asked to find the peak voltage of the AC applied. The formula connecting these variables is \[P_{avg} = \frac{V_{rms}^2}{R},\]where \( P_{avg} \) is the average power, \( V_{rms} \) is the root mean square voltage, and \( R \) is the resistance. We aim to find the peak voltage \( V_{peak} \).
02

Calculate the RMS Voltage

Using the formula \( P_{avg} = \frac{V_{rms}^2}{R} \), we can rearrange it to find the RMS voltage:\[V_{rms} = \sqrt{P_{avg} \times R}.\]Substitute the given values into the formula:\[V_{rms} = \sqrt{55 \times 4.0} = \sqrt{220} = 14.83 \text{ V}.\]Thus, the RMS voltage is approximately 14.83 V.
03

Find the Peak Voltage from RMS Voltage

The relationship between peak voltage \( V_{peak} \) and RMS voltage \( V_{rms} \) for an AC circuit is given by:\[V_{peak} = \sqrt{2} \times V_{rms}.\]Substitute the RMS voltage into the formula to find the peak voltage:\[V_{peak} = \sqrt{2} \times 14.83 \approx 1.414 \times 14.83 = 20.96 \text{ V}.\]Therefore, the peak voltage is approximately 20.96 V.

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Key Concepts

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

Understanding Average Power
Average power is the mean power consumed over a period of time by an electrical device. It provides a measure of the overall energy used by a circuit. In AC circuits, power varies periodically, so the average power helps us understand the effective power use over a single cycle.
  • In formula terms, for AC circuits with resistance, the average power \[P_{avg} = \frac{V_{rms}^2}{R},\]where \(P_{avg}\) is average power, \(V_{rms}\) is RMS voltage, and \(R\) is resistance.
  • This equation allows us to compute power when voltage fluctuates, giving a steady value we can work with.
In the exercise, we know the average power is 55 W, critical for computing the RMS and then the peak voltage.
Role of Resistance in Circuits
Resistance is a measure of how much an object opposes the flow of electric current. Measured in ohms (惟), it affects both the power consumption and voltage in the circuit.
  • The larger the resistance, the more it resists current, which often results in more heat generation.
  • Use Ohm's Law, \(V = I \times R\), to relate resistance to the voltage and current.
  • In power calculations, resistance is essential as it affects how much power is used for a given RMS voltage.
In the given exercise, the speaker resistance is 4.0 惟, which helped determine the RMS voltage from the known average power.
RMS Voltage Clarified
RMS, which stands for Root Mean Square, is a statistical measure of the magnitude of a varying quantity and is particularly useful in AC circuits. It offers a way to calculate the effective voltage or current.
  • The RMS voltage \(V_{rms}\) is essentially the square root of the average of the squares of all instantaneous voltages in a cycle.
  • It provides a meaningful average of AC voltage levels, akin to a DC voltage that would deliver the same power to a resistor.
In our calculation:\[V_{rms} = \sqrt{P_{avg} \times R},\]which gives us \(V_{rms} = \sqrt{55 \times 4.0} = 14.83\, \text{V}.\) Understanding RMS voltage is paramount for bridging between the average power and peak voltage.
Defining Peak Voltage
Peak voltage in AC circuits is the maximum voltage level achieved within a cycle. It is greater than RMS voltage and is crucial for understanding the voltage extremes.
  • The relationship between peak voltage \(V_{peak}\) and RMS voltage is expressed through the formula:\[V_{peak} = \sqrt{2} \times V_{rms}.\]
  • Peak voltage needs to be known for designing components that can withstand maximum voltage levels without damage.
For this exercise, using the RMS voltage \(14.83 \text{ V}\), the peak voltage calculates to:\[V_{peak} = \sqrt{2} \times 14.83 \approx 20.96\, \text{V}.\] Hence, it helps in understanding the maximum voltage the speaker might encounter.

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Most popular questions from this chapter

A tungsten wire has a radius of 0.075 \(\mathrm{mm}\) and is heated from 20.0 to \(1320^{\circ} \mathrm{C}\) . The temperature coefficient of resistivity is \(\alpha=4.5 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1} .\) When 120 \(\mathrm{V}\) is applied across the ends of the hot wire, a current of 1.5 \(\mathrm{A}\) is produced. How long is the wire? Neglect any effects due to thermal expansion of the wire.

Two resistances, \(R_{1}\) and \(R_{2},\) are connected in series across a \(12-\mathrm{V}\) battery. The current increases by 0.20 \(\mathrm{A}\) when \(R_{2}\) is removed, leaving \(R_{1}\) connected across the battery. However, the current increases by just 0.10 \(\mathrm{A}\) when \(R_{1}\) is removed, leaving \(R_{2}\) connected across the battery. Find \((\mathrm{a}) R_{1}\) and \((\mathrm{b}) R_{2}\)

For the three-way bulb (50 W, 100 W, 150 W) discussed in Conceptual Example 11, find the resistance of each of the two filaments. Assume that the wattage ratings are not limited by significant figures, and ignore any heating effects on the resistances.

The coil of a galvanometer has a resistance of \(20.0 \Omega,\) and its meter deflects full scale when a current of 6.20 \(\mathrm{mA}\) passes through it. To make the galvanometer into a nondigital ammeter, a 24.8 \(\mathrm{m} \Omega\) shunt resistor is added to it. What is the maximum current that this ammeter can read?

Three parallel plate capacitors are connected in series. These capacitors have identical geometries. However, they are filled with three different materials. The dielectric constants of these materials are \(3.30,5.40,\) and \(6.70 .\) It is desired to replace this series combination with a single parallel plate capacitor. Assuming that this single capacitor has the same geometry as each of the other three capacitors, determine the dielectric constant of the material with which it is filled.
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