Sample Problem: KCL 2

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来自 Georgia Institute of Technology 的课程

Linear Circuits 1: DC Analysis

106 个评分

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

106 个评分

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.

从本节课中

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

与讲师见面

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 Current Law and

the problem is to determine I1, I4, and I6 in the circuit that's drawn below.

In this circuit,

we have a number of different components that are involved in a complex circuit and

the cloud-like images represent a complex part of the circuit with certain

currents flowing out of that complex part of the circuit at different points.

So, at the top complex circuit, we have 60 milliamps flowing down through one leg.

We have 20 milliamps flowing down through the right-most leg.

And we have I1 which is flowing into that complex circuit from the left

to the right.

Similarly in the bottom part of the circuit there is another substance of

the circuit, which is a complex circuit which has 40 miliamps flowing out,

30 miliamps flowing in, and

I4 which we are going to determine also flowing out of that, part of the circuit.

Again, we're using Kirchhoff's Current Law to determine I1, I4 and

I6 so we treat each one of those complex circuits as a node.

And so it's kind of a node that is encompassing part of the circuit.

And flowing out of that node we have the three currents at the top,

we have I1, I or 60 milliamps and 20 milliamps.

So, we're going to start with the top and write Kirchhoff's current law for the top

and again, before we do that, let's state what Kirchhoff's current law is.

And that's, that the algebraic sum of all the currents entering any node at any

instant in time is 0, so looking at the top node,

the top node being the complex circuit element at the top, elements at the top,

we have a current I1 flowing into it, or summing currents into that node.

I1 is flowing in, it's a positive flowing in.

We have 60 flowing out so it's a minus 60 milliamps and

we also have 20 flowing out so it's a minus 20 milliamps and

that's all the currents associated with the top complex circuit node, and

we determine from that that I1 is equal to 80 milliamps.

Similarly, we can look at the bottom complex circuit node.

And the bottom complex circuit node has five currents associated with it.

If we write the equation for that, then we have I4 flowing out, so it's a minus I4.

Summing currents again, summing currents into that node.

Plus 60 milliamps flowing into it at the top, right hand side, or left hand side.

And then we also have 20 milliamps flowing in at the top right hand side.

Now looking below that complex node that was flowing out of it,

we have 40 milliamp flowing out so it's a minus 40 so

it's flowing outward, and we have 30 flowing in so it's a 30 plus 30,

That we add to our equation.

So ultimately we end up with an equation for

I4 that looks like this if we solve for I4,

then we end up with an I4 equal to 70 milliamps.

So now that we have those quantities, we can go back and we can find what I6 is.

I6 can be solved by taking the algebraic

sum of the currents flowing into the node at the left most side of the circuit.

So in this case it's going to be minus I1 since it's flowing out,

plus I4 since it's flowing in,

minus I6 because it's flowing out of that left most node.

And so if we plug in the values for

these we have minus 80 milliamps plus

70 milliamps minus I6 is equal to 0.

Solving for I6, we have an I6 equal to 80 minus 70 which

gives us 10 miliamps and that's

going to be a negative 10 miliamps because we have a negative sign in front of the I6

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