Q: excuse-me, at the lesson " step3- KCL at node B" <<-- how and where to get that i1 + is = i2 + i3; i2, i3 and iS have the same node. I still don't understand it. Please help.

The KCL equation for node b is: ( i1 + is = i2 + i3 ). This is the form of Kirchhoff's Current Law where all the currents flowing *into* the node ( i1 + is ) are set equal to all the currents flowing *out* of the node ( i2 + i3 ). The blue current arrow for i1 is drawn on the left side of the 20-ohm resistor, but that same current flows out the right end of the resistor and goes into node b. So node b has two currents flowing in, and two flowing out.

Q: Please, could any of you recommend me web site for practising such exercises

Here's a suggestion: This article and the following videos/articles on Node Voltage and Mesh Current methods all have example circuits. For each different circuit, see if you can solve them with more than one method. Another very realistic way to practice is to make up simple circuits on your own and analyze them by hand. To check your work, draw your circuit in this simulator and run a *DC* analysis. http://spinningnumbers.org/a/circuit-sandbox.html This simulator is part of a web site I've been working on since my EE Fellowship at KA completed. The articles are all updated and improved, and circuit simulation is incorporated as part of the learning process.

Q: please can i get help with how the line that follows "and crank the algebra" in the solution came about. ho did we get -540/30? please help.

The first term on the right side of the equation as a 30 in the denominator. The second term is -18. -18 is equal to -540/30. Now both terms have a 30 in the denominator.

Q: How would one label the current flow in node c? I see that it is unimportant for solving the circuit, but I can't yet wrap my head around _why_. Particularly, I don't know how I would label the current on the node between the voltage source and the 6 ohm resistor, and then the branch between the two resistors (on node c).

The thing that's causing confusion is the definition of node c. Node c is the *entire* bottom horizontal wire in the circuit. So the two segments you identified (between voltage source and 6ohm, and the segment between two resistors) are both included as part of node c. Review the definition of node and "distributed node" in this article https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/a/ee-circuit-terminology. When you account for the currents flowing into node c you will draw arrows on the vertical wire segments feeding into node c. You won't be drawing current arrows on the horizontal bits and pieces of node c. It's all one single junction.

Q: by my calculations the Vdrop over R1 is equal to: ( ( Vss / (R1 + ( ( R2 * R3 ) / ( R2 + R3) ) ) * R1 ( ( 140v / ( 20 + ( ( 6*5 ) / ( 6 / 5) ) ) ) * 20 Rtot = R1 + Rp Itot (voltage side) = Vss / Rtot Vdrop over R1 = Itot (voltage side) * R1 And this results in a Vdrop over R1 equal to 123,18 V This is ofcourse unless you count the "Current Source" as a Voltage source also which would interfere with the "voltage source given". Am i totaly wrong here?

Here's a good way to check your work. Simulate the circuit by copying this entire URL into a web browser. Then tap on **DC** to run a simulation. Compare your calculations to what the simulator thinks is happening. http://spinningnumbers.org/circuit-sandbox/index.html?value=[["v",[104,96,0],{"name":"VS","value":"dc(140)","_json_":0},["2","0"]],["i",[320,144,6],{"name":"IS","value":"dc(18)","_json_":1},["0","1"]],["r",[120,72,3],{"name":"R1","r":"20","_json_":2},["2","1"]],["r",[200,96,0],{"name":"R2","r":"6","_json_":3},["1","0"]],["r",[256,96,0],{"name":"R3","r":"5","_json_":4},["1","0"]],["w",[104,96,104,72]],["w",[104,72,120,72]],["w",[200,72,200,96]],["w",[200,72,256,72]],["w",[256,72,256,96]],["w",[256,72,320,72]],["w",[320,72,320,96]],["w",[320,144,256,144]],["w",[200,144,256,144]],["w",[200,144,104,144]],["g",[200,144,0],{"_json_":15},["0"]],["w",[168,72,200,72]],["view",7.200000000000003,-5.259999999999998,1.953125,"50","10","1G",null,"100","0.01","1000"]]

Question 1

At the point  where  KCL  is   calculated  at  node b  ,  why  is  the  voltage  same  for  currents  i2  and  i3 ?

Accepted Answer

i2 flows through the 6-ohm resistor. i3 flows through the 5-ohm resistor. The terminals of both of these resistors are connected together (both top terminals connect to node b, and both bottom terminals connect to node c). That means they have the same voltage. The voltage has been named v2. (The "2" in the voltage name is not related to the "2" in the i2 name.)

Question 2

How is it that current flows "up" through the current source in this example, from lower voltage to higher voltage?  What am I missing here?  Is it magic?

Accepted Answer

The current source in this circuit creates a current flowing up from node "c" to node "b". Like a battery, a current source is a power *generator*. Something is going on inside the current source that makes it able to force current to flow "up" against an increasing voltage. (A battery can do the same thing, it forces current to flow out its positive terminal because of a chemical reaction going on inside.)  Voltage sources (batteries) are familiar to us, since we can go to the store and buy them. Current sources are not made from a chemical reaction. They are some sort of complicated electronic circuit that you don't run across at the grocery store. They are unfamiliar, but not magic. Pretty much every analog integrated circuit has circuits that act like a current source over a limited range of voltage.

Question 3

Why isn't there a voltage over the 5 ohm resistor?

Accepted Answer

The voltage across the 5 ohm resistor is v2.
v2 is the voltage between nodes b and c, so this is the voltage appearing across both resistors and the current source.

Question 4

excuse-me, at the lesson " step3- KCL at node B" <<-- how and where to get that i1 + is = i2 + i3;
i2, i3 and iS have the same node. I still don't understand it. Please help.

Accepted Answer

The KCL equation for node b is: ( i1 + is = i2 + i3 ). This is the form of Kirchhoff's Current Law where all the currents flowing *into* the node ( i1 + is ) are set equal to all the currents flowing *out* of the node ( i2 + i3 ).  The blue current arrow for i1 is drawn on the left side of the 20-ohm resistor, but that same current flows out the right end of the resistor and goes into node b. So node b has two currents flowing in, and two flowing out.

Question 5

Please, could any of you recommend me web site for practising such exercises

Accepted Answer

Here's a suggestion: This article and the following videos/articles on Node Voltage and Mesh Current methods all have example circuits. For each different circuit, see if you can solve them with more than one method.

Another very realistic way to practice is to make up simple circuits on your own and analyze them by hand. To check your work, draw your circuit in this simulator and run a *DC* analysis.  http://spinningnumbers.org/a/circuit-sandbox.html

This simulator is part of a web site I've been working on since my EE Fellowship at KA completed. The articles are all updated and improved, and circuit simulation is incorporated as part of the learning process.

Question 6

please can i get help with how the line that follows "and crank the algebra" in the solution came about.
ho did we get -540/30? please help.

Accepted Answer

The first term on the right side of the equation as a 30 in the denominator. The second term is -18. -18 is equal to -540/30. Now both terms have a 30 in the denominator.

Question 7

How would one label the current flow in node c? I see that it is unimportant for solving the circuit, but I can't yet wrap my head around _why_.  Particularly, I don't know how I would label the current on the node between the voltage source and the 6 ohm resistor, and then the branch between the two resistors (on node c).

Accepted Answer

The thing that's causing confusion is the definition of node c. Node c is the *entire* bottom horizontal wire in the circuit. So the two segments you identified (between voltage source and 6ohm, and the segment between two resistors) are both included as part of node c. Review the definition of node and "distributed node" in this article https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/a/ee-circuit-terminology.  When you account for the currents flowing into node c you will draw arrows on the vertical wire segments feeding into node c. You won't be drawing current arrows on the horizontal bits and pieces of node c. It's all one single junction.

Question 8

by my calculations the Vdrop over R1 is equal to:

( (  Vss / (R1 + ( ( R2 * R3 ) / ( R2 + R3) ) ) * R1

(  ( 140v / ( 20 + ( ( 6*5 ) / ( 6 / 5) ) ) ) * 20

Rtot = R1 + Rp

Itot (voltage side) = Vss / Rtot

Vdrop over R1 = Itot (voltage side) * R1

And this results in a Vdrop over R1 equal to 123,18 V
This is ofcourse unless you count the "Current Source" as a Voltage source also which would interfere with the "voltage source given".
Am i totaly wrong here?

Accepted Answer

Here's a good way to check your work. Simulate the circuit by copying this entire URL into a web browser.  Then tap on **DC** to run a simulation. Compare your calculations to what the simulator thinks is happening.

http://spinningnumbers.org/circuit-sandbox/index.html?value=[["v",[104,96,0],{"name":"VS","value":"dc(140)","_json_":0},["2","0"]],["i",[320,144,6],{"name":"IS","value":"dc(18)","_json_":1},["0","1"]],["r",[120,72,3],{"name":"R1","r":"20","_json_":2},["2","1"]],["r",[200,96,0],{"name":"R2","r":"6","_json_":3},["1","0"]],["r",[256,96,0],{"name":"R3","r":"5","_json_":4},["1","0"]],["w",[104,96,104,72]],["w",[104,72,120,72]],["w",[200,72,200,96]],["w",[200,72,256,72]],["w",[256,72,256,96]],["w",[256,72,320,72]],["w",[320,72,320,96]],["w",[320,144,256,144]],["w",[200,144,256,144]],["w",[200,144,104,144]],["g",[200,144,0],{"_json_":15},["0"]],["w",[168,72,200,72]],["view",7.200000000000003,-5.259999999999998,1.953125,"50","10","1G",null,"100","0.01","1000"]]

Electrical engineering

Course: Electrical engineering > Unit 2

Application of the fundamental laws

Introduction

Step 1: Label the schematic

Step 2. Select the independent variable

Step 3. Write independent equations

KVL around the left-most mesh

KCL at node $b$ ‍

Steps 4 and 5 - Solve

Summary

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