Sample Problem: KVL 2

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Do curso por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

97 classificações

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

97 classificações

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.

Na lição

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

Conheça os instrutores

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 Kirchhoff's voltage laws.

In this problem we have a two loop circuit, and

we're going to write the mesh or loop equations associated with this circuit.

Sometimes these equations are known as mesh equations or

equivalently loop equations.

So in order to do this problem, as we all do with all these Kirchhoff's problems,

we need to assign our current directions first so that we can use the passive sign

convention to assign the polarity drops, the voltage drops across our resistors.

So we start with our left most loop, and

we're going to assign a current direction in this direction for this loop.

So this is a current I and it's called I sub 1.

Okay, if we do that, and

we're talking about this loop on the left-hand side off the circuit.

So a voltage drop across using the passive sign convention across R1 is going to be

In this direction, V R1.

Equivalently, we have another voltage drop, V R4,

which is across the resistor R4.

And that is, again, due to the passive sign convention and

using our left-most loop and clockwise rotation of current around that loop.

On the other side of the circuit we have some additional resistors that we need to

define the voltage drop across.

And so, we're going to, again, design a current direction

in the clockwise direction around the right most loop.

And if we do that, we can then assign a voltage player at least for R2 and R3.

We're not going to go back and re-assign our voltage polarity across R4

because we've already done that, and we're going to stick with what we have.

So once you've assigned the voltage polarity, leave it that way for

the rest of the problem in order to get the correct solution.

So we assign our voltage polarities for V R2, I mean for

resistor two and for resistor three.

So we're going to start with the left most loop, and

we're going to use Kirchhoff's Voltage Law to write the loop equation for that loop.

So if we start, again, at the left lower corner of our circuit, and we work our

way around the circuit, the first thing we encounter is the dependent source.

So this is our first equation for that left loop.

And so, we end up encountering the negative polarity of

the 24 Volt independent source first, so it's a -24 Volts is

the first component of our loop equation.

Then as we continue to travel around this left most loop,

we encounter the resister R1, and we have a plus V R1 that has dropped the cross it.

We are hitting the positive polarity of that voltage drop across R1 first.

We continue around that left most loop and come down and we encounter resistor R4.

And it has a voltage drop across it, V R4,

with the positive polarity is the one we encounter first.

And continue down that center leg of our circuit, and

we encounter a -16 volts, continuing around to where we began.

And that is our equation for our loop, understanding that a loop

is defined as a path that starts and

ends at the same node without encountering any node twice.

Now we're going to move to the right loop,

the right most loop of the circuit, and we're going to write the equation for it.

We've already assigned our voltage polarities using the passive sign

convention, so we can start at the bottom left corner of that loop,

which is in the center of the circuit.

If we do that and go around that loop, we get a +16 volts,

because that's the first thing we encounter in this path or this loop.

We then go up and we encounter the voltage across R4, that's a minus V R4.

We've already assigned our polarity for the voltage drop across that resistor R4.

Continuing on up to resistor 2, we encounter a voltage drop V R2.

Continuing on to resistor R3, we encounter voltage drop V R3.

And continuing over to the independent 8 volt source, hitting

the positive polarity of that source in our path first, and we get 8 volts.

And then if we continue around the rest of the circuit back to the beginning,

that complete's our loop.

We have one more loop in this problem that we can solve for,

and that loop is the outermost loop.

It goes around the outside of our circuit.

So again, if we've start at the lower

left-hand corner of our circuit and

work our way all the way around the circuit,

we first encounter a -24 volts and

then a +V R1 on the top of the circuit + V R2 + V R3 + 8 volts = 0.

And so we now have that relationship.

So that completes our set of loop equations for this circuit.

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