Sample Problem: Power 2

Loading...

From the course by Georgia Institute of Technology

Linear Circuits 1: DC Analysis

123 ratings

Explore Related Videos

At Coursera, you will find the best lectures in the world. Here are some of our personalized recommendations for you

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

123 ratings

This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

From the lesson

Module 1: Introduction

This module reviews background material from physics on the basic properties of electricity and electrical circuits.

Sample Problem: Ohm's Law and Power 13:39

Sample Problem: Ohm's Law and Power 22:52

Sample Problem: Power 15:34

Sample Problem: Power 23:07

Sample Problem: Power 31:40

Sample Problem: Power 41:57

Sample Problem: Voltage and Energy1:10

Sample Problem: Electric Current and Charge 11:04

Sample Problem: Electric Current and Charge 20:53

Sample Problem: Voltage Division1:24

Meet the Instructors

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer

The topic of this problem is Electric Power.

The problem is to determine the missing quantity, the power, and

the second element.

We're giving at the power in the first element is equal

to 36 watts and its absorb power.

So we have 36 watts absorbed into element one.

So if that's the case, then we know that using this well known equation for power

is equal to voltage times current, that the current in element one is going to be

Is equal to the power divided by the voltage.

And that's equal to 36 watts divided by 12 volts.

And we also know that power is positive,

so that means that our current is going to be positive flowing back in

to the complex part of the circuit, so that if it were flowing through

this circuit in this manner will be flowing through this direction and

the positive of the current would go into the positive polarity across element one.

So it's the positive current hitting the positive

polarity if the element one voltage which is 12 volts.

And so, the current in this case is going to be 3 amps.

3 amps of current flowing back into the complex circuit at the top of the circuit.

The problem is to determine what the power across the circuit element is and

whether it's absorb power or if it's supplied power.

We already know the current,

we determine that from the information were given about element one.

That is it had 12 volts across it with the polarity shown and

have 36 watts of the absorbed power.

We know that sensors absorb, it's a positive power.

And so that leads us to find the current.

Is in the direction toward the complex circuit and is a positive 3 amps.

And if that's the case, that positive 3 amps is also flowing through element two

from the bottom to the top.

And so for element two, P2 is going to be the current,

3 amps and the voltage which is minus 4 volts.

The reason we chose minus 4 volts is because, if you look at the current,

the current is flowing into the bottom side of element two or

the polarity is negative, so the negative side of the 4 voltage drop

across element two gives us the minus sign in front of the four volts.

And so we end up in this case with 12 watts,

minus 12 watts of power, and

that means that we have 12 watts of

supplied power from element two.

Explore our Catalog

Join for free and get personalized recommendations, updates and offers.

Coursera provides universal access to the world’s best education, partnering with top universities and organizations to offer courses online.

© 2019 Coursera Inc. All rights reserved.

Download on the App Store

Get it on Google Play