Problem 33
Question
The average power used by a stereo speaker is \(55 \mathrm{~W}\). 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.
Step-by-Step Solution
Verified Answer
The peak voltage is approximately 20.95 V.
1Step 1: Understanding Power in AC Circuits
In AC circuits, we often use root mean square (RMS) values to simplify calculations. For a given power (P) and resistance (R), we use the formula for the average power: \( P = \frac{V_{rms}^2}{R} \). Here, \( R = 4.0 \Omega \) and \( P = 55 \mathrm{~W} \).
2Step 2: Calculating RMS Voltage
Rearrange the formula to solve for \( V_{rms} \): \( V_{rms} = \sqrt{P \times R} \). Substitute the known values: \( V_{rms} = \sqrt{55 \times 4.0} \).
3Step 3: Evaluating RMS Voltage
Calculate \( V_{rms} \): \( V_{rms} = \sqrt{220} \approx 14.83 \mathrm{~V} \).
4Step 4: Finding Peak Voltage
The peak voltage \( V_{peak} \) is related to the RMS voltage by the equation \( V_{peak} = V_{rms} \times \sqrt{2} \). Substitute \( V_{rms} = 14.83 \mathrm{~V} \): \( V_{peak} = 14.83 \times \sqrt{2} \).
5Step 5: Calculate Peak Voltage
Perform the calculation: \( V_{peak} = 14.83 \times 1.414 \approx 20.95 \mathrm{~V} \).
Key Concepts
Average Power in AC CircuitsUnderstanding RMS VoltageResistance in CircuitsPeak Voltage
Average Power in AC Circuits
In AC circuits, the concept of average power is essential to determine how much energy is used over a complete cycle. Unlike DC circuits where power is constant, AC power varies with time. Hence, we talk about the 'average power' which helps in practical analysis.
The formula most commonly used for average power in AC circuits is:
The formula most commonly used for average power in AC circuits is:
- \( P = \frac{V_{rms}^2}{R} \)
Understanding RMS Voltage
RMS (Root Mean Square) voltage is a key concept in AC circuits. It’s a kind of "average" value for voltage that allows us to use simple DC formulas to evaluate power in an AC context. The RMS value helps us grasp how an AC voltage or current affects a circuit compared to a constant DC.
Unlike peak voltage, which only quantifies the maximum at a certain instant, RMS provides a more practical measure for power calculations.
Unlike peak voltage, which only quantifies the maximum at a certain instant, RMS provides a more practical measure for power calculations.
- For instance, to find RMS voltage: \( V_{rms} = \sqrt{P \times R} \).
- It literally comes down to finding the square root of the product of power and resistance.
Resistance in Circuits
Resistance in circuits plays a pivotal role in determining how much power is dissipated and how current flows through a circuit. Measured in ohms (\( \Omega \)), resistance is that part of a circuit that resists the flow of electricity, transforming electrical energy into heat or work.
In our situation with the stereo speaker represented as a resistance:
In our situation with the stereo speaker represented as a resistance:
- We view it as a resistor with a value of \( 4.0 \Omega \).
- The relation \( P = \frac{V_{rms}^2}{R} \) also shows how critical knowing your resistance is for accurate power calculations.
Peak Voltage
Peak voltage (\( V_{peak} \)) refers to the maximum value of voltage in an AC cycle. It represents the high point of the voltage wave wherein it reaches its highest value.
To link RMS voltage to peak voltage, the following relation is used:
Knowing the peak voltage is vital for protecting electronic components, as it impacts their durability and performance. Without accounting for peak voltage, devices may improperly handle excess voltage, leading to malfunctions or damage.
To link RMS voltage to peak voltage, the following relation is used:
- \( V_{peak} = V_{rms} \times \sqrt{2} \)
Knowing the peak voltage is vital for protecting electronic components, as it impacts their durability and performance. Without accounting for peak voltage, devices may improperly handle excess voltage, leading to malfunctions or damage.
Other exercises in this chapter
Problem 31
The current in a circuit is ac and has a peak value of 2.50 A. Determine the rms current.
View solution Problem 32
The rms current in a copy machine is \(6.50 \mathrm{~A},\) and the resistance of the machine is \(18.6 \Omega\). What is (a) the average power and (b) the peak
View solution Problem 35
An electric furnace runs nine hours a day to heat a house during January ( 31 days). The heating element has a resistance of \(5.3 \Omega\) and carries a curren
View solution Problem 36
A light bulb is connected to a \(120.0-\mathrm{V}\) wall socket. The current in the bulb depends on the time \(t\) according to the relation \(I=(0.707 \mathrm{
View solution