## Sunday, June 7, 2015

### Circuit Solution Methods (By Examples)

To find the equivalent resistance between A  and B.

After marking the junction

Hence equivalent resistance between A and B is $url=http://latex.codecogs.com/gif.latex?%7B%7BR%7D_%7BAB%7D%7D%3D%5Cfrac%7B3R%7D%7B5%7D&container=blogger&gadget=a&rewriteMime=image/*$

Nodal analysis:
To find the current distribution in arms

To solve this one must assume the potential of one node equal to zero and then corresponding other potentials as marked in following diagram

The solving equation for the unknown V is

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Symmetry method:

(a) Folding symmetry

(b) Input output symmetry

In such cases the current distribution follows the symmetry as shown below

Example: A battery of 10 V and negligible internal resistance is connected across the
diagonally opposite corners of a cubical network consisting of 12 resistors
each of resistance  as shown in the figure. Determine the equivalent
resistance of the network and the current along each edge of the cube.
Solution:        The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.

The paths, AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners, B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem. Next take a closed loop, say, and apply Kirchhoff’s second rule:

$-IR-(1/2)IR-IR+\varepsilon =0$
Where R is the resistance of each edge and  the emf  $\varepsilon$ of battery.
Thus,
$\varepsilon =\frac{15}{2}IR$    ${{R}_{eq}}=\frac{\varepsilon }{31}=\frac{5}{6}R$
The equivalent resistance  of the network is

For  $R=1\,\Omega ,\,\,{{R}_{eq}}=(5/6)\,\Omega$   and for  $\varepsilon =10V,$

the total current (= 3I) in the network is

$3I=10V/(5/6)\,\Omega =12A,\,i.e.,\,I=4\,A$

The current  in each edge can now be read off from the figure.