Till now, you all must be using those conventional Kirchoff's laws to solve resistive or capacitive circuits. The method I am going to explain here is called Nodal Analysis.

Its advantages over Kirchoff's laws-

1- U wud always get one equation less than wat u get by applying Loop law. It is always easier and faster to solve 3 equations than 4 OR 4 than 5.

2- It is very simple. Equations are formed using Kirchoff's junction law means net current at a junction is zero.

3- Uses a simple concept that potential remains constant in a conductor.

4- It can be also be used to rearrange and simplify complex circuits easily.

How to apply it-

1- To find R_eff between any 2 points, put a battery(imaginary) of voltage=V across them and assign potentials 0 and V to the terminals.

2- Apply Kirchoff's junction law at any junction other than terminals.

Its difficult to post too many examples here, so I am posting an example of simplifying circuits. Its difficult to explain this concept clearly in just 1 article, so I want to u to co-operate if U want to learn this concept. Just post the circuits U find difficult to solve in the electricity forum and nudge me the link and I wud try my best to solve it using nodal analysis.

The question is shown in Figure-1. U have to find equivalent resistance between A and B. All resistances=R. Its not that tough question if U can visualise which resistances are in series and which are in parellel but this ques can be made easier by rearranging the circuit with the help of nodal analysis.

Figure-2

We put an imaginary battery across A and B of voltage V or 100(doesnt matter, I have taken 100 for simplicity. I dont like dealing with variables).

Figure-3

Now as I have mentioned earlier that it uses the fact that potential remains constant in a conductor.

1- As A,C,D are connected by a wire(conductor), these points wud have same potential as A i.e, =0.

2- G,H as connected by a wire wud have same potential lets say V1.

3- E,F as connected by a wire wud have same potential lets say V2.

Now for more simplicity I have represented the equivalent points by same numbers.

Its clear that a,b have same potential difference across them and d,e also have same potential difference across them. So we can consider a,b to be in parellel and d,e to be in parellel combination as potential remains constant in parellel.

Figure-4

Now I have redrawn the figure.

First draw terminal A. Now a,b are in parellel connection between 1 and 2. And d,e are in parellel connection between 2 and 3. And c is between 1 and 3. And f is between 3 and B.

Now you can easily solve this circuit by applying series parellel combination of resistances.

Its a very gud method and saves a lot of time and is applicable in almost every problem of circuit solving. And hence I wanted to share this topic with all of you, so I posted it here.

I know many of you must not have understood it clearly. I will try my best to teach it if U R willing to learn. I will start a thread in electricity forum and wud post some more examples there and u can also post ur circuits there only and I wud try my level best to solve them.

Hope it helps you.  Comments awaited!!!

I have started the thread here-

http://www.goiit.com/posts/list/0/electricity-nodal-analysis-67424.htm#332329

I have explained a question of circuit solving there. Check it out.