Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The formula for electric power, \(P\), is \(\mathrm{P}=1 \times V\), where \(\mathrm{I}\) is the current and \(V\) is the voltage. The formula for voltage is \(\mathrm{V}=\mathrm{I} \times R\), where \(\mathrm{I}\) is also the current and \(R\) is the resistance. How will the power of a given current be affected if the resistance is doubled and the voltage is quadrupled? (A) The power will be doubled. (B) The power will be quadrupled. (C) The power will be 8 times greater. (D) The power will be 16 times greater.

Short Answer

Expert verified
The power will be 8 times greater.

Step by step solution

01

Identify the Original Power Formula

We start with the formula for electric power given by \( P = I \times V \), where \( I \) is the current and \( V \) is the voltage.
02

Substitute Voltage Formula

The formula for voltage \( V \) is \( V = I \times R \). Substitute this into the power formula to get \( P = I \times (I \times R) = I^2 \times R \).
03

Doubling the Resistance

If the resistance \( R \) is doubled, the new resistance is \( 2R \). Substituting into the power formula gives \( P_{new1} = I^2 \times 2R = 2(I^2 \times R) \). This shows that power doubles with respect to the original resistance.
04

Quadrupling the Voltage

If the voltage \( V \) is quadrupled, the new voltage is \( 4V \). Continuing from the step where voltage is expressed as \( V = I \times R \), quadrupling gives \( V_{new} = 4(I \times R) \).
05

Calculate New Power with Adjusted Resistance and Voltage

Substitute the quadrupled voltage into the power equation: \( P_{new2} = I \times 4(I \times R) = 4(I^2 \times R) \).
06

Combine Resistance and Voltage Changes

When both the resistance is doubled and the voltage quadrupled, the increase in power is cumulative. Substitute the quadrupled power result into the doubled resistance result: \( P_{final} = 2(I^2 \times R) + 4(I^2 \times R) = 8(I^2 \times R) \). This demonstrates that power increases by a factor of 8 when both alterations are applied.

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 Vaia!

Key Concepts

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

Current and Voltage Relationship
To understand the relationship between current and voltage, we need to consider how these two elements work together within an electric circuit. Current, denoted by the symbol \(I\), is the flow of electric charge. It is measured in amperes (A). Voltage, represented as \(V\), is the electrical potential difference between two points in a circuit and is measured in volts (V). These two are inherently linked through Ohm's Law.
Ohm's Law is a fundamental principle that describes the relationship between current, voltage, and resistance in an electrical circuit. This relationship can be expressed with the formula \(V = I \times R\), where \(R\) is the resistance measured in ohms (Ω). It shows that voltage is directly proportional to both current and resistance.
Understanding this relationship helps in predicting how changes in voltage or resistance can affect current. For example, increasing voltage while keeping resistance constant will increase the current. This principle lays the groundwork for analyzing more complex scenarios in electrical power.
Electric Circuit Resistance
Resistance is a critical component of any electric circuit. It acts as a barrier to the flow of current, affecting how easily electricity can pass through a conductor. Resistance, represented by \(R\), is determined by factors such as the material of the conductor, its diameter, and length.
By manipulating the resistance of a circuit, we can control the amount of current flowing through it. The effect of resistance is so significant that it's an essential consideration when calculating power in electrical systems. For instance, when resistance doubles, as seen in many practical problems, it directly impacts the calculated power by changing the voltage relationship \(V = I \times R\).
In our case study, when the resistance is doubled, it had a direct influence on the power equation. This shows the importance of accurately accounting for resistance in any power calculations. This concept demonstrates how intricate and interlinked electrical calculations require a solid understanding of roles that resistance plays within a circuit.
Power Formula Application
Applying the power formula accurately in different scenarios is crucial for understanding how electric power behaves under changing conditions. The basic power formula \(P = I \times V\) gives a straightforward means to calculate the power consumed or produced in an electric circuit. Power, measured in watts (W), is a measure of how fast energy is being used or generated.
In our example, alterations to the circuit's components like voltage and resistance lead to a deeper analysis using an extended form of the power formula, \(P = I^2 \times R\). This derivative of the power formula helps understand how power changes when both voltage and resistance are modified.
In cases where the voltage is quadrupled and the resistance is doubled, we must consider both effects on the power equation. Each change individually affects power, but combined, they result in an eightfold increase in power usage, as shown by the calculation steps provided. Such exercises illustrate how changes in circuit components interact to affect power, entailing the necessity for precise calculations in complex electric systems.

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

Caitlin opens a checking account that earns no interest to set aside spending money for vacations. Each month she puts the same dollar amount, \(\$ 50\), into the account. Unfortunately, she does not expect to be able to take a vacation at any point in the foreseeable future. Which of the following best describes the relationship between the number of months and the total amount of money in the account? (A) A linear relationship, with the line of the relationship having a negative slope B) A linear relationship, with the line of the relationship having a positive slope (C) An exponentially increasing relationship (D) An inverse exponential relationship

Out of the following options, what is the most unexpected result of the collection of the random sets of numbers? (A) The values in both sets of data have a range of roughly between 1 and 100 . (B) The selection of 50 random numbers has a significantly higher percentage of values above 50 than does the selection of 20 random numbers. (C) The selection of 20 random numbers has a much greater range among its values than does the selection of 50 random numbers. (D) The sets of both data are portrayed as scatter plots rather than as best- fit lines.

(A) NO CHANGE (B) to go to Mars, however won't (C) to go to Mars but won't (D) to go to Mars, and, won't

What are the solution(s) to the following equation? \(5 x^{2}-15 x+10=0\) (A) 0 (B) 1,2 (C) 1,4 (D) 2,5

What is the value of \(x\) in the following equation? \(3 x+2=\frac{4}{3} x\) (A) \(-\frac{6}{5}\) (B) \(-\frac{6}{5}\) (C) \(\frac{4}{6}\) (D) \(\frac{7}{9}\)

See all solutions

Recommended explanations on English 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.

Sign-up for free