Sample Problem: Series/Parallel (Independ Sources) 1

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Linear Circuits 1: DC Analysis

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Georgia Institute of Technology

Linear Circuits 1: DC Analysis

135 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 2

The module describes some basic principles used in circuit analysis.

Sample Problem: DC Op Amp 19:29

Sample Problem: DC Op Amp 25:33

Sample Problem: DC Op Amp 34:23

Sample Problem: DC Op Amp 46:22

Sample Problem: DC Op Amp 55:33

Sample Problem: Current Division 13:57

Sample Problem: Current Division 25:20

Sample Problem: Parallel and Series Resistors 12:24

Sample Problem: Parallel and Series Resistors 28:07

Sample Problem: KVL6:05

Sample Problem: KVL 25:48

Sample Problem: KVL 32:28

Sample Problem: KCL 12:12

Sample Problem: KCL 24:43

Sample Problem: KCL 32:51

Sample Problem: KCL 41:26

Sample Problem: KCL 52:10

Sample Problem: Op Amps 19:17

Sample Problem: Op Amps 25:29

Sample Problem: Op Amps 34:59

Sample Problem: Nodes, Branches, Paths, and Loops3:19

Sample Problem: Summing Op Amp6:05

Sample Problem: Differential Amp6:47

Sample Problem: Inverting Op Amp5:09

Sample Problem: Non-inverting Op Amp7:54

Sample Problem: Max Power (Depend Sources)4:06

Sample Problem: Dependent Sources 15:10

Sample Problem: Dependent Sources 24:48

Sample Problem: Series/Parallel (Independ Sources) 12:59

Sample Problem: Series/Parallel (Independ Sources) 27:22

Sample Problem: Series/Parallel (Independ Sources) 33:00

Sample Problem: Series/Parallel (Independ Sources) 44:35

Sample Problem: Op Amps 49:10

Sample Problem: Op Amps 59:29

Sample Problem: Op Amps 65:33

Sample Problem: Op Amps 74:23

Sample Problem: Op Amps 86:22

Sample Problem: Op Amps 95:33

Sample Problem: Op Amps 109:17

Sample Problem: Op Amps 115:29

Sample Problem: Op Amps 124:59

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 Series and Parallel Independent Sources, and

the problem is to determine the current in the circuit that’s drawn and

the power associated with the 5 ohm resistor.

In this problem, we have a series combination of Voltage sources and

resistors.

And we know that we can combine series voltage

sources by adding them, and so that's what we're going to do in this problem.

We first of all acknowledge that the 5V sources, all three of them,

are a series of one another.

And we combine those, and we get a 15 volt source as shown in the redrawn circuit.

And the way that we get that is by going clockwise around the circuit with

the negative polarity and the 5 volt source on the right hand side first.

That's minus 5 volts.

We could continue in the top part of the circuit.

We hit the negative polarity of the 5 volt source at the top.

So it's a minus 10 volts combining those two.

And continually around the circuit in clockwise fashion,

we hit the negative polarity of the 5V source on the right hand side.

So that's how we ended up with a 15V source on the left hand side of our

redrawn circuit.

Similarly, we also know that the resistors are in series with one another,

the 15 ohm, the 25 ohm, and the 5 ohm resistors.

So we can combine those as well, and if you add those together,

a series combination of resistors can be added, and

we have a 45 volt equivalent resistance for those three resistors.

Now we're going to try to find the current in the circuit and

we can do that with our simplified circuit very easily.

We notice that the current is going to be equal to the voltage

divided by the resistance using Ohm's law.

The voltage is going to be minus 15 volts because as we know through the passive

sign convention a positive direction of the current in this circuit is clockwise

and is hitting the negative polarity of the 15 volt source first.

That's why we have amassed 15 volts in our ohm's law equation.

We divide that by the resistance which is 45 ohms and

that gives us a current of -0.333 amps.

So, that's our current in the circuit.

Now, if we want to find the power associated with the 5 ohm resistor we can

use our well-known equation for power, where power is equal to I squared R.

Our current which we just found, is -one-third amps.

We squared that and we multiply it by the 5 ohm

resistance to get our power associated with the 5 ohm resistor.

We noticed that that power is going to be a positive power.

So we have absorbed power for this resistor remembering

using the passive sign convention that a positive power is absorbed power.

And a negative power is a supplied power.

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