/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 The rms current in a \(47-\Omega... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The rms current in a \(47-\Omega\) resistor is 0.50 A. What is the peak value of the voltage across this resistor?

Short Answer

Expert verified
The peak voltage across the resistor is approximately 33.2 V.

Step by step solution

01

Understand the Relationship Between RMS and Peak Values

The root mean square (RMS) value of an alternating current (AC) is related to its peak value through the formula: \( I_{rms} = \frac{I_{peak}}{\sqrt{2}} \), where \( I_{rms} \) is given as the RMS value of the current. This relationship is valid for sinusoidal waveforms.
02

Relate Current and Voltage Across the Resistor

Ohm's Law states that \( V = IR \), where \( V \) is the voltage across the resistor, \( I \) is the current through it, and \( R \) is its resistance. Thus, the RMS voltage \( V_{rms} = I_{rms} \times R \).
03

Calculate RMS Voltage

Use the values given: \( I_{rms} = 0.50 \text{ A} \) and \( R = 47 \Omega \). Substitute these into the formula \( V_{rms} = I_{rms} \times R \):\[ V_{rms} = 0.50 \, \text{A} \times 47 \, \Omega = 23.5 \, \text{V} \]
04

Determine the Peak Voltage

Given the RMS voltage calculated in the previous step, use the relationship between RMS and peak voltage for a sine wave: \( V_{rms} = \frac{V_{peak}}{\sqrt{2}} \). Solving for \( V_{peak} \), we have:\[ V_{peak} = V_{rms} \times \sqrt{2} = 23.5 \, \text{V} \times \sqrt{2} \approx 33.2 \, \text{V} \]
05

Verify the Calculation

Double-check each calculation to ensure there are no errors in arithmetic or algebraic manipulations. Confirm that each substitution and consequent result align with theoretical expectations.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in the field of electronics and electrical engineering. It defines the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit. The formula is succinctly expressed as:\[ V = I \times R \]Here, **Voltage (V)** is the force that pushes electric charges through a conductor, **current (I)** is the rate at which the charge flows, and **resistance (R)** is the opposition to the flow of current offered by the material through which it passes.
**Key Points of Ohm's Law:**
  • It applies to most conductive materials, known as ohmic materials, where the resistance remains constant.
  • In a linear, or ohmic, resistor, the current increases proportionally as the voltage across it increases, provided the temperature remains constant.
  • It's crucial for calculating parameters in both direct current (DC) and alternating current (AC) circuits.
Understanding Ohm's Law is essential as it allows us to calculate any one of the three variables—voltage, current, or resistance—if the other two are known. In AC circuits, where conditions can fluctuate, knowing how these variables interact is vital for ensuring circuit reliability and efficiency.
RMS Voltage
RMS Voltage (Root Mean Square Voltage) is an essential concept in AC circuit analysis. When dealing with AC circuits, the voltage and current fluctuate with time, typically in a sinusoidal manner. Hence, using average values can be misleading, which is why RMS values are used.**RMS Voltage Defined:**RMS voltage gives a measure of the equivalent direct current (DC) voltage that would produce the same power dissipation in a resistor as the given AC voltage.
The formula for RMS voltage in a sinusoidal signal is:\[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \]**Why RMS Voltage is Important:**
  • RMS values help standardize AC measurements, allowing us to assess AC circuits with direct comparisons.
  • It is critical for determining power calculations in AC circuits, as it aligns with how instruments read DC values.
  • RMS voltage allows engineers to design circuits and components that perform safely and efficiently across the AC and DC spectrum.
Remembering the RMS concept helps in practical applications such as electrical billing, where devices convert AC into effective electrical work, similar to its DC counterpart.
Peak Voltage
Peak Voltage is the maximum voltage a waveform reaches, either in the positive or negative direction. In the context of AC circuits, understanding peak voltage is crucial for examining how high the voltage could reach at any point during its cycle.
**How Peak Voltage Works:**- For sinusoidal AC signals, the maximum point on the wave graph represents the peak voltage.- It plays a significant role in the design and analysis of electrical systems to ensure components can withstand maximum voltage without damage.
The relationship between RMS voltage and peak voltage is described by:\[ V_{peak} = V_{rms} \times \sqrt{2} \]**Applications of Knowing Peak Voltage:**
  • Protective devices like fuses and surge protectors rely on peak voltage for operation, ensuring safety against voltage spikes.
  • Engineers and electricians use peak voltage values to specify equipment needs such as insulation levels and voltage ratings.
  • It informs the design of power electronics that convert AC power to DC, ensuring adequacy in power supply systems.
Understanding peak voltage helps in predicting and mitigating potential circuit risks, ensuring that the electrical systems or devices will operate within safe parameters.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A battery has an internal resistance of \(0.012 \Omega\) and an emf of \(9.00 \mathrm{V}\). What is the maximum current that can be drawn from the battery without the terminal voltage dropping below \(8.90 \mathrm{V} ?\)

Two resistors, 42.0 and \(64.0 \Omega,\) are connected in parallel. The current through the \(64.0-\Omega\) resistor is \(3.00 \mathrm{A}\). (a) Determine the current in the other resistor. (b) What is the total power supplied to the two resistors?

The total current delivered to a number of devices connected in parallel is the sum of the individual currents in each device. Circuit breakers are resettable automatic switches that protect against a dangerously large total current by "opening" to stop the current at a specified safe value. A \(1650-\mathrm{W}\) toaster, a \(1090-\mathrm{W}\) iron, and a \(1250-\mathrm{W}\) microwave oven are turned on in a kitchen. As the drawing shows, they are all connected through a \(20-\mathrm{A}\) circuit breaker (which has negligible resistance) to an ac voltage of \(120 \mathrm{V}\). (a) Find the equivalent resistance of the three devices. (b) Obtain the total current delivered by the source and determine whether the breaker will "open" to prevent an accident.

A \(550-\mathrm{W}\) space heater is designed for operation in Germany, where household electrical outlets supply \(230 \mathrm{V}\) ( \(\mathrm{rms}\) ) service. What is the power output of the heater when plugged into a \(120-\mathrm{V}\) (rms) electrical outlet in a house in the United States? Ignore the effects of temperature on the heater's resistance.

A \(75.0-\Omega\) and a \(45.0-\Omega\) resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is \(0.294 \mathrm{A}\). When the \(45.0-\Omega\) resistor is disconnected, the current from the battery drops to 0.116 A. Determine (a) the emf and (b) the internal resistance of the battery.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.