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Sample Problem: KVL 3

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Linear Circuits 1: DC Analysis

Instituto de Tecnología de Georgia

4.5 (215 calificaciones) | 17K estudiantes inscritos

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.

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4.5 (215 calificaciones)

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DM

May 09, 2018

Great course, learned a lot of things that they don't teach at my school, Thanks for the opportunity.

YZ

Jan 22, 2018

Excellent course where I get to learn a lot of things that is left from my AP Physics C E&M Course.

De la lección

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

Impartido por:

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

Transcripción

The topic of this problem is Kirchhoff's Voltage Law. The problem is, to write the mesh or loop equation associated with this single loop circuit. So in order to do that, the first thing we have to do is we have to assign a current direction so that we can then assign polarities to the voltage drops across the resistors, using the passive sign convention. The passive sign convention was mentioned earlier in videos in chapter one and two, so if you need to get familiar with information related to the passive sign convention, you can look at those videos. So the first thing we do is we assign a current, and we're going to assign the current in a clockwise Direction around our single loop circuit. So if we do that, we know that through this passive sign convention, the current should enter the positive end of the voltage drops across our resistors. So we have these voltage drops assign these polarities across our resistors. Or we have a VR2 for voltage across resistor 2, VR1 across the resistor 1, and VR3 across resistor 3. Now that we have done that, we can now write our mesh, also known as loop equation, for this single loop circuit. So if we start at the lower left-hand corner of our circuit and we work our way around the circuit in a clockwise fashion, the first thing we hit is a -30 volts associated with the independent voltage source. We continue around our circuit and then we encounter the resistor. And the resistor is a positive voltage drop, + VR1. Continue past that resistor onto the next independent voltage source and it's -5 volts. Continuing around the loop to R2, we have another voltage drop, VR2. Continuing around the bottom portion of our single loop circuit, we have a -15 volts with a independent voltage source at the bottom of the circuit. And then we have one more voltage drop across VR3, and that is equal to 0. So that's our equation for this single loop circuit.

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