Sample Problem: Power 1

Loading...

En provenance du cours de Georgia Institute of Technology

Linear Circuits 1: DC Analysis

170 notes

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

170 notes

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.

À partir de la leçon

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

Rencontrer les enseignants

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.

In this problem, we want to determine the power for

each one of these elements in the circuit.

We want to determine the power itself and whether it's absorbed power or

supplied power.

So we'll start this problem with looking at the 36-volt source

on the left-hand side of the circuit.

If we look at the power in the 36-volt source, We're

going to use our well-known equation that power is equal to voltage times current.

And so in this case, we see that our voltage is 36 volts, but the current

is flowing up through this leg of the circuit, this branch of the circuit.

So that the positive 4 amps of current, I sub x,

first encounters the negative polarity of that voltage source, the 36-volt source.

So the voltage in this case is -36 volts, since we hit the negative polarity

of the voltage source first, and the current is 4 amps.

And so in this case, we end up with -144 watts

of power associated with the 36-volt source.

This is a negative power, so that tells us that we have 144 watts supplied

by this element, that is the 36-volt source.

If we look next at element 1, which is on the upper left-hand side of the circuit,

and try to determine what the power is in element 1,

P sub 1 using the same approach, is going to be the 4 amps.

And the 4 amps, a positive 4 amps, is flowing into the element 1 which

has 12 volts plus to minus, left to right, across that element.

So that 4 amps encounters the positive polarity of the 12-volt source,

or the 12 volts dropped across element 1.

So it's (12v)(4A) and that gives us 48 watts of power.

And since that is a positive 48 watts,

we have 48 watts absorbed, By element 1.

If we do the same thing for element 2, looking in the center of the circuit,

we see that there's 24 volts plus to minus, top to bottom, across element 2.

We also see that we have 2 amps positive flowing down through element 2.

The 2 amps would encounter the positive of the polarity.

The 24 volts has dropped across element 2.

So power for element 2 is going to be (24v)(2A),

and that gives us 48 watts of power once again.

And again that's a positive power, so we end up with

48 watts of absorbed power associated with element 2.

Next we can look at the current control voltage source.

So it's a dependent source at the top right-hand side of the circuit.

And so the power in that source, which is 1 times I sub x,

where I sub x is the current in the left-hand side of the circuit,

And the 1 is the amplification factor for this current control voltage source.

Then if we look at the power in that case,

we see that the voltage across it is 1I sub x, minus to plus,

so it's going to be a -1I sub x for the voltage, and the current is 2 amps.

And again, that 2-amp current is encountering the negative polarity of

the dependent source first.

So it's (-1I sub x)(2A),

and so that gives us a -2 I sub x watts.

And we know what I sub x is, I sub x is equal to 4 amps.

So we can then plug that into this and we get a -8 watts

of power associated with that current control voltage source.

And as before, if we end up with a negative power,

then we know that that is supplied power.

So we have 8 watts of supplied power.

From that dependent source.

The mixed element and the last element is element 3.

Element 3 has 28 volts dropped across it with a polarity shown,

and 2 amps of current flowing down through it.

So, using the same approach as we did for the other elements,

we see that we had 28 volts and we have 2 amps which encounters

the positive polarity of the voltage source first.

And so that again gives us 56 watts of power.

And since it has a positive power,

we know that we have 56 watts absorbed by element 3.

Explorer notre catalogue

Rejoignez-nous gratuitement et obtenez des recommendations, des mises à jour et des offres personnalisées.

Coursera propose un accès universel à la meilleure formation au monde, en partenariat avec des universités et des organisations du plus haut niveau, pour proposer des cours en ligne.

© 2019 Coursera Inc. Tous droits réservés.

Télécharger dans l'App Store

Disponible sur Google Play