/*! 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 84 In this chapter, most examples a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In this chapter, most examples and problems involved direct current (DC). DC circuits have the current flowing in one direction, from positive to negative. When the current was changing, it was changed linearly from \(I=-I_{\max }\) to \(I=+I_{\max }\) and the voltage changed linearly from \(V=-V_{\max }\) to \(V=+V_{\max },\) where \(V_{\max }=I_{\max } R\) Suppose a voltage source is placed in series with a resistor of \(R=10 \Omega\) that supplied a current that alternated as a sine wave, for example, \(I(t)=(3.00 \mathrm{A}) \sin \left(\frac{2 \pi}{4.00 \mathrm{s}} t\right)\) What would a graph of the voltage drop across the resistor \(V(t)\) versus time look like? (b) What would a plot of \(V(t)\) versus \(I(t)\) for one period look like? (Hint: If you are not sure, try plotting \(V(t)\) versus \(I(t)\) using a spreadsheet.)

Short Answer

Expert verified
In this problem, we have a sine wave for the current, \(I(t) = (3.00 A) \sin \left(\frac{2 \pi}{4.00 s}t\right)\), and a resistor with resistance \(R = 10\Omega \). Using Ohm's law, we find the voltage function \(V(t) = (30.0V) \sin \left(\frac{2 \pi}{4.00s} t\right)\). When graphing \(V(t)\) versus time, we have an alternating current with a sinusoidal pattern, and half-cycle of AC voltage is completed in 2s. To graph \(V(t)\) versus \(I(t)\) for one period, we plot the linear equation \(V(t) = 10I(t)\), which represents a straight line with a slope equal to the resistor's value (10Ω). This graph shows that the voltage and current across the resistor are directly proportional and follow a straight line relationship.

Step by step solution

01

Understand the given information and equation

We are given a sine wave for the current, \(I(t) = (3.00 A) \sin \left(\frac{2 \pi}{4.00 s}t\right)\), and the resistance value, \(R = 10\Omega \). To find the voltage across the resistor, we'll use Ohm's law, which states that \(V(t) = I(t)R\).
02

Apply Ohm's Law

Now, we will apply Ohm's Law to find the voltage across the resistor as a function of time: \(V(t) = I(t)R\) First, we'll substitute the given current function and resistance into the equation: \(V(t) = (3.00 A) \sin \left(\frac{2 \pi}{4.00 s} t\right) \times (10\Omega)\) Then simplify the expression to find the voltage function, \(V(t)\): \(V(t) = (30.0V) \sin \left(\frac{2 \pi}{4.00s} t\right)\)
03

Graphing \(V(t)\) versus time

To graph V(t) versus time, we will plot the voltage function from step 2, \(V(t) = (30.0V) \sin \left(\frac{2 \pi}{4.00s} t\right)\), over one period of the sine wave. The period of the sine wave is \(T = 4.00 s\), so we will plot the function over the x-axis ranging from 0 to \(4.00 s\). The amplitude is 30.0 V, so the plot ranges from -30.0 V to 30.0 V along the y-axis. The graph represents an alternating current with a sinusoidal pattern, and half-cycle of AC voltage is completed in 2s.
04

Graphing of \(V(t)\) versus \(I(t)\) for one period

To create a plot of V(t) versus I(t) for one period, we'll first express V(t) in terms of I(t) using the simplified Ohm's law equation we found in step 2: \(V(t) = (30.0V) \sin \left(\frac{2 \pi}{4.00s} t\right)\) Since, \(I(t) = (3.00 A) \sin \left(\frac{2 \pi}{4.00 s}t\right)\), Now, divide both sides of V(t) expression by 10: \(V(t) = 10I(t)\) Now, plot the linear equation \(V(t) = 10I(t)\) on the \(I(t)\)-axis versus the \(V(t)\)-axis for one period of 4.00 s. The plot represents a straight line with a slope equal to the resistor's value (10Ω). The graph will show that the voltage and current across the resistor are directly proportional and follow a straight line relationship.

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.

Direct Current (DC)
Direct Current (DC) is a type of electrical flow where the current moves in only one direction. This is in contrast to Alternating Current (AC) where the current changes direction periodically. In DC circuits, electrons travel from the negative terminal to the positive terminal of a power source.
For DC, the voltage across a component is constant, making it straightforward to analyze and predict the behavior of electrical circuits. The formula used is straightforward:
  • Ohm’s Law: \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
Applications of DC include batteries, solar panels, and most portable electronic devices where a consistent and stable power supply is crucial.
Alternating Current (AC)
Alternating Current (AC) is characterized by the flow of electric charge that reverses direction periodically. This means that AC voltage and current vary with time as opposed to being constant like DC.
AC is the standard form of electricity supplied to our homes and industries because it is more efficient for power distribution over long distances.
Some key features of AC include:
  • A sinusoidal waveform, which repeats its pattern over a fixed time period;
  • The ability to easily change voltage levels using transformers, making it suitable for power distribution.
The standard formula to understand AC circuits is still rooted in Ohm's Law, but it often involves more complex mathematics due to the changing nature of the current and voltage.
Voltage and Current Relationship
The relationship between voltage and current is described by Ohm's Law, which states that the voltage across a resistor is directly proportional to the current flowing through it. This is expressed in the equation:
  • \( V = IR \)
Where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
This relationship is fundamental in both DC and AC circuits.
In AC circuits, even though the current and voltage are continually changing, they maintain a proportional relationship at any given instant, influenced by the impedance instead of just resistance. This is crucial for understanding how devices control the flow of electricity effectively and efficiently.
Sine Wave Function
A sine wave function is an essential concept in understanding AC circuits. This mathematical function describes how alternating current behaves over time. A typical sine wave is periodic, meaning it repeats itself at regular intervals.
The general form of a sine wave for AC is:
  • \( I(t) = I_{max} \sin(\omega t) \)
Where \( I_{max} \) is the maximum current, \( \omega \) is the angular frequency, and \( t \) is time.
The sine wave explains alternating current cycles through positive and negative values, which result in a smooth and continuous oscillation.
Understanding sine waves is crucial for creating, analyzing, and troubleshooting AC circuits, especially when considering power supply systems, signal processing, and radio transmission.

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

When working with high-power electric circuits, it is advised that whenever possible, you work "one-handed" or "keep one hand in your pocket" Why is this a sensible suggestion?

Common household appliances are rated at \(110 \mathrm{V}\), but power companies deliver voltage in the kilovolt range and then step the voltage down using transformers to \(110 \mathrm{V}\) to be used in homes. You will leam in later chapters that transformers consist of many turns of wire, which warm up as current flows through them, wasting some of the energy that is given off as heat. This sounds inefficient. Why do the power companies transport electric power using this method?

A wire with a resistance of \(5.0 \Omega\) is drawn out through a die so that its new length is twice times its original length. Find the resistance of the longer wire. You may assume that the resistivity and density of the material

Electrical power generators are sometimes "load tested" by passing current through a large vat of water. A similar method can be used to test the heat output of a resistor. A \(R=30 \Omega\) resistor is connected to a \(9.0-\mathrm{V}\) battery and the resistor leads are waterproofed and the resistor is placed in \(1.0 \mathrm{kg}\) of room temperature water \(\left(T=20^{\circ} \mathrm{C}\right) .\) Current nuns through the resistor for 20 minutes. Assuming all the electrical energy dissipated by the resistor is converted to heat, what is the final temperature of the water?

A heater is being designed that uses a coil of 14-gauge nichrome wire to generate \(300 \mathrm{W}\) using a voltage of \(V=110 \mathrm{V} .\) How long should the engineer make the wire?

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.