# i need help , i'dont understand Kirchhoff's laws

Thread Starter

#### saif-aljanahi

Joined Feb 26, 2016
18
hi , firstly im saif aljanahi from iraq and i learned a lot things from this place

i have a big problem in analyze Kirchoff's current law

i dont understand any thing about it

i need someone to explained it to me from zero and by more easy way

And I appreciate your help

Thank you .

#### Jony130

Joined Feb 17, 2009
5,232
• atferrari

#### #12

Joined Nov 30, 2010
18,222
It was about 1963 when a EE said to me, "what goes into comes out of". That's the current law. Any current that moves into a component comes out of it, immediately. Current doesn't wait a while. It flows.

The voltage law in short form is, "all the voltages add up". Kirchoff's Laws are just formal ways of creating mathematical expressions of those two facts.
I'm not saying Kirchoff doesn't have a purpose. There are circuits that are much better explained in those terms, and you must practice the simple ones if you're ever going to do a complicated one.

#### WBahn

Joined Mar 31, 2012
26,398
Kirchhoff's Current Law is nothing more than a statement of the conservation of charge. Think of a bucket of water that has a bunch of hoses some of which are pumping water into the bucket and some of which are pumping it out. If the total rate of water coming into the bucket is not exactly equal to the total rate of water coming out, then the amount of water that is in the bucket will be changing by the difference of the two. But now imaging that the bucket is sealed up tight (except that the hoses are now pipes welded to the sides) and that the bucket is completely full of water (or even that is it made so small that the pipes can just barely fit to its sides). Now the amount of water in the bucket can't change, so the rate at which water comes in must equal the rate at which water goes out.

Kirchhoff's Voltage Law is nothing more than a statement of conservation of energy. Think of a building. If you carry a rock from a lower flower to a higher floor, it gains gravitational potential energy (it will hurt more if you drop it on someone's head who is standing on the ground next to the building). If you carry it higher up it will have more energy yet. If you bring it down, it will have less energy. If you carry it up, down, up, down, and so on but bring it eventually back to the same place you started, it will have the same potential energy that you started with (the person whose head you drop it on can't tell anything about the path the rock took, they only know the height from which it is eventually dropped). So the net gain of energy that the rock gets along any path is zero if path that leads back to where the path started.

#### RBR1317

Joined Nov 13, 2010
638
Once you understand the physics behind Krichhoff's Laws there is still the matter of applying them to circuit analysis. This is where node & mesh equations enter the picture. A node equation is an implicit application of the Current Law wherein the node currents are expressed in terms of the node voltages. And a mesh equation is an implicit application of the Voltage Law wherein the loop voltage drops are expressed in terms of the mesh currents. Thread Starter

#### saif-aljanahi

Joined Feb 26, 2016
18
thanks for all helping , i understand current law now and thankful of those people helping me out

but one more request i want explain of K voltage law with solving example

#### Papabravo

Joined Feb 24, 2006
16,183
Thread Starter

#### saif-aljanahi

Joined Feb 26, 2016
18
the last thing i need and please anyone help me bcz i have exam next week

i just need a steps and simple solution not complicated for this circuit in figure >>>>

Q/ find current in R2 2 ohm by thevenin and the Thevenin Equivalent i have read this circuit explanation and present here but did not understand any of it , especially the voltage a, b

#### RBR1317

Joined Nov 13, 2010
638
The simple steps are: 1)Find the Thevenin equivalent of B1, R1, & R2. 2)Use the Thevenin equivalent and voltage division to find the voltage across R3. 3)Use the voltage across R3 to get the voltage across R2. 4)Use the voltage across R2 to find the current in R2.

Thread Starter

#### saif-aljanahi

Joined Feb 26, 2016
18
The simple steps are: 1)Find the Thevenin equivalent of B1, R1, & R2. 2)Use the Thevenin equivalent and voltage division to find the voltage across R3. 3)Use the voltage across R3 to get the voltage across R2. 4)Use the voltage across R2 to find the current in R2.
can i ask you a request ?

just explain it for me again with a figure if you can do it

#### Papabravo

Joined Feb 24, 2006
16,183
I can help you get started, but I will not give you the solution.
1. There are two independent, but constant, voltage sources.
2. There are two loops with one common branch
You should be able to write two simultaneous equations, one for each loop, where the sum of the voltage drops across the resistors and the voltage rise across the source equals zero. Then you solve the system of linear equations and you have your solution.

#### WBahn

Joined Mar 31, 2012
26,398
the last thing i need and please anyone help me bcz i have exam next week

i just need a steps and simple solution not complicated for this circuit in figure >>>>

Q/ find current in R2 2 ohm by thevenin and the Thevenin Equivalent i have read this circuit explanation and present here but did not understand any of it , especially the voltage a, b
The whole idea behind Thevenin analysis is to treat one part of the circuit as a load (R2 in this case) and to replace the rest of the circuit by a simple voltage source in series with an equivalent resistance such that the load can't tell the difference.

Remove R2 from the circuit and replace it with two open terminals. Call the upper terminal A and the lower terminal B.

Now find the voltage Vab. That is your Thevenin voltage.

Now either find the resistance seen between A and B with the two voltage sources set to 0 V or place a short across A and B and find the current that would flow through it. That is the Norton current and the equivalent resistance is the ratio of the Thevenin voltage to the Norton current.

Now put the Thevenin source in series with the equivalent resistance and connect those to terminals A and B. Put the original load back between A and B and then solve for the current in the load.

Thread Starter

#### saif-aljanahi

Joined Feb 26, 2016
18
The whole idea behind Thevenin analysis is to treat one part of the circuit as a load (R2 in this case) and to replace the rest of the circuit by a simple voltage source in series with an equivalent resistance such that the load can't tell the difference.

Remove R2 from the circuit and replace it with two open terminals. Call the upper terminal A and the lower terminal B.

Now find the voltage Vab. That is your Thevenin voltage.

Now either find the resistance seen between A and B with the two voltage sources set to 0 V or place a short across A and B and find the current that would flow through it. That is the Norton current and the equivalent resistance is the ratio of the Thevenin voltage to the Norton current.

Now put the Thevenin source in series with the equivalent resistance and connect those to terminals A and B. Put the original load back between A and B and then solve for the current in the load.
thank you for your helping i really appreciate that , i understand all of this but one , how i can find the thevenin Voltage from 2 source can you teach my that by more ways like KVL or Ohm's law

Thread Starter

#### saif-aljanahi

Joined Feb 26, 2016
18
Can you not find the voltage Vth in the following circuit?

View attachment 102069

If not, then your problem has nothing to do with Thevenin equivalents and has everything to do with basic circuit analysis.
yes please all what i saying is i do not know how to find the voltages betwwen a , b can you learn me that !

#### WBahn

Joined Mar 31, 2012
26,398
Can you figure out how much current is flowing in the circuit I provided?

Thread Starter

#### saif-aljanahi

Joined Feb 26, 2016
18
Can you figure out how much current is flowing in the circuit I provided?
maybe i know it , i think by KCL analysis , is I'm right?

#### anhnha

Joined Apr 19, 2012
881
maybe i know it , i think by KCL analysis , is I'm right?
No, by KVL and Ohm's law.

#### WBahn

Joined Mar 31, 2012
26,398
maybe i know it , i think by KCL analysis , is I'm right?
No. KCL is only useful at junctions where the current can split and go more than one way. In this circuit, there are no junctions that join three or more branches, so KCL is trivially satisfied and doesn't provide any useful information.

You need to apply KVL and Ohm's Law.

It's clear that you haven't gotten the fundamentals down and that is really hurting you as you move to more complex material. That is only going to get worse. So you REALLY need to go back to about page one in an introductory circuits text and work your way through it section by section and learn these fundamentals.

Thread Starter

#### saif-aljanahi

Joined Feb 26, 2016
18
No. KCL is only useful at junctions where the current can split and go more than one way. In this circuit, there are no junctions that join three or more branches, so KCL is trivially satisfied and doesn't provide any useful information.

You need to apply KVL and Ohm's Law.

It's clear that you haven't gotten the fundamentals down and that is really hurting you as you move to more complex material. That is only going to get worse. So you REALLY need to go back to about page one in an introductory circuits text and work your way through it section by section and learn these fundamentals.
you helping me a lot , thank you but the last thing i want it , can you solve this circuit by KVL step by step ? please

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