/*! 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 64 Show that \(A^{2} \times \Omega=... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 64

Show that \(A^{2} \times \Omega=W\) (amperes squared times ohms \(=\) watts).

Short Answer

Expert verified
Answer: Yes, the equation \(A^2 \times \Omega = W\) holds true by substituting Ohm's law into the power formula and simplifying, as demonstrated in the provided solution.

Step by step solution

01

Write down Ohm's law and the power formula

Ohm's law: \(V = I \times R\) Power formula: \(P = I \times V\)
02

Substitute Ohm's law into the power formula

Replace V in the power formula with \(I \times R\) from Ohm's Law: \(P = I \times (I \times R)\)
03

Simplify the equation

Multiply I by I 脳 R: \(P = I^2 \times R\)
04

Compare the equation with the given equation

We now have: \(P = I^2 \times R\) Comparing this to the given equation, \(A^2 \times \Omega = W\), we see that: \(I^2 \times R = A^2 \times \Omega\) Since I is current in amperes \(A\) and R is resistance in ohms \(\Omega\), and P is power in watts \(W\), the equation holds true. Therefore, we have proved that \(A^2 \times \Omega = W\).

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 91影视!

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 aluminum wire of diameter 2.6 mm carries a current of 12 A. How long on average does it take an electron to move \(12 \mathrm{m}\) along the wire? Assume 3.5 conduction electrons per aluminum atom. The mass density of aluminum is \(2.7 \mathrm{g} / \mathrm{cm}^{3}\) and its atomic mass is $27 \mathrm{g} / \mathrm{mol}$.
A galvanometer has a coil resistance of \(50.0 \Omega\). It is to be made into an ammeter with a full-scale deflection equal to \(10.0 \mathrm{A} .\) If the galvanometer deflects full scale for a current of \(0.250 \mathrm{mA},\) what size shunt resistor should be used?
Two copper wires, one double the diameter of the other, have the same current flowing through them. If the thinner wire has a drift speed \(v_{1},\) and the thicker wire has a drift speed \(v_{2},\) how do the drift speeds of the charge carriers compare?

A common flashlight bulb is rated at \(0.300 \mathrm{A}\) and \(2.90 \mathrm{V}\) (the values of current and voltage under operating conditions). If the resistance of the bulb's tungsten filament at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) is \(1.10 \Omega,\) estimate the temperature of the tungsten filament when the bulb is turned on.
A person can be killed if a current as small as \(50 \mathrm{mA}\) passes near the heart. An electrician is working on a humid day with hands damp from perspiration. Suppose his resistance from one hand to the other is $1 \mathrm{k} \Omega$ and he is touching two wires, one with each hand. (a) What potential difference between the two wires would cause a 50 -mA current from one hand to the other? (b) An electrician working on a "live" circuit keeps one hand behind his or her back. Why?
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Astrophysics

Read Explanation

Space Physics

Read Explanation

Magnetism

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Radiation

Read Explanation
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.