/*! 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 1 You have a special light bulb wi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.50 \(\mathrm{A}\) , even for an instant. What is the largest root-mean-square current you can run through this bulb?

Short Answer

Expert verified
1.06 A is the largest RMS current for the bulb.

Step by step solution

01

Understanding the Problem

The root-mean-square (rms) current is being asked, given a maximum allowable current of 1.50 A. We need to relate the rms current to this maximum instantaneous current.
02

Recall Relationship Between RMS and Maximum Current

For sinusoidal current, the relationship between the root-mean-square current \( I_{\text{rms}} \) and the peak current \( I_0 \) is given by the formula: \[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \] where \(I_0\) is the maximum current.
03

Substitute the Given Maximum Current

We know that the maximum current \( I_0 = 1.50 \) A. Substitute this into the RMS formula: \[ I_{\text{rms}} = \frac{1.50}{\sqrt{2}} \]
04

Calculate the RMS Current

Now, compute the numerical value for the root-mean-square current:\[ I_{\text{rms}} = \frac{1.50}{\sqrt{2}} \approx 1.06 \]

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.

maximum current
When dealing with electrical circuits, the term "maximum current" refers to the highest value of electric current that momentarily flows through a circuit. This represents the peak point of current that can occur during a cycle of alternating current (AC). To avoid damage or failure, it is crucial to ensure that circuits do not exceed their designated maximum current.
In practical applications, exceeding the maximum current can lead to overheating, component damage, or even a complete breakdown of the circuit, as in the case of the light bulb discussed. The bulb's filament is specially designed to handle up to 1.50 A without breaking. Understanding the maximum current helps in designing circuits that are safe and efficient.
When testing circuits, always check the maximum specified current to prevent unintended consequences that could damage the equipment or cause safety hazards.
sinusoidal current
A "sinusoidal current" is a type of alternating current (AC) that varies in magnitude and direction in a smooth, periodic manner. Its wave-like shape is known as a sine wave, and it is a fundamental concept in alternating current electricity. This type of current is prevalent because most generators produce current in this form, and it conveniently translates to effective and efficient power transmission.
In a sinusoidal current wave, there are several key points: the rising and falling parts of the wave indicate continuously changing current levels, and it completes one full cycle in the time period given.
This current form is important in understanding and implementing electrical systems that involve transformers, motors, and various types of household and industrial equipment.
peak current
The term "peak current" or peak value refers to the highest value reached by the current during one cycle of a sinusoidal wave. This is the top point in a waveform where the maximum instantaneous current occurs.
For example, in the exercise given, the peak current is specified as 1.50 A. This means that, during its cycle, the current reaches 1.50 A for a brief moment. Peak values are important because they indicate the limits at which the circuit should operate safely.
Understanding peak current helps in ensuring components like resistors and capacitors are not subjected to damaging levels of current. It also factors into calculations involving other forms of electrical analysis, such as determining the root-mean-square (RMS) current.
RMS formula
The "RMS formula" helps translate peak or maximum current values to an average-type value that is more meaningful in terms of power. RMS stands for Root-Mean-Square, which offers a measure of the "effective" value of an alternating current.
Mathematically, the RMS current (\( I_{\text{rms}}\)) for sinusoidal current is derived using the formula:
  • \( I_{\text{rms}} = \frac{I_{0}}{\sqrt{2}} \)
Here, \( I_{0} \) is the peak current. The formula provides a way to calculate how much power the circuit can effectively deliver, making it incredibly useful in both theoretical and practically applied electrical engineering.
By applying the RMS formula, engineers and technicians can estimate the safe and effective current levels for different devices, minimizing the risk of component damage due to excessive currents. In the exercise, substituting the given maximum current of 1.50 A into the RMS formula results in an RMS current of approximately 1.06 A, illustrating how these theoretical calculations apply to real-world devices.

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

An \(L-R-C\) series circuit with \(L=0.120 \mathrm{H}, R=240 \Omega\) and \(C=7.30 \mu\) F carries an rms current of 0.450 \(\mathrm{A}\) with a frequency of 400 \(\mathrm{Hz}\) (a) What are the phase angle and power factor for this circuit? (b) What is the impedance of the circuit? (c) What is the rms voltage of the source? (d) What average power is delivered by the source? (e) What is the average rate at which electrical energy is converted to thermal energy in the resistor? (f) What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor? (g) In the inductor?

An \(L-R-C\) series circuit has \(C=4.80 \mu \mathrm{F}, L=0.520 \mathrm{H}\) and source voltage amplitude \(V=56.0 \mathrm{V}\) . The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude \(80.0 \mathrm{V},\) what is the value of \(R\) for the resistor in the circuit?

In an \(L_{-} R-C\) series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance \(X_{C}\) of the capacitor is 200\(\Omega\) and the voltage amplitude across the capacitor is 600 \(\mathrm{V}\) . The circuit has \(R=300 \Omega .\) What is the voltage amplitude of the source?

A coil has a resistance of 48.0\(\Omega .\) At a frequency of 80.0 \(\mathrm{Hz}\) the voltage across the coil leads the current in it by \(52.3^{\circ} .\) Determine the inductance of the coil.

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}(\mathrm{rms})\) to a portable electronic device. The load resistance in the secondary is 5.00\(\Omega .\) (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120-\mathrm{V}\) line would draw the same power as the transformer? Show that this is equal to 5.00\(\Omega\) times the square of the ratio of primary to secondary turns.

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.