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this curretn distribution wil help u..
16 Answers
62Accross AB..
see that D and E are equipontial
while C and F are also equipotential..
so you have a paralllel between r and 3/2 r
Acrosss A and D, see that there will be no current through EF...
3sir i cant figure out equipotential points sir.help me to find out how to find equipotential points.
11see if we connect the points A and D with battery , due to symmetry the current distribution on either side of the plane will be identical and so points E and F will be at the same potential and no current will flow through them.
3at terminal E can u explain why shud the current divides by I/2 actually i am not gud at current division,i request any one to explain the theory of current division.
11
METHOD OF EQUIPOTENTIAL REDUCTION
In a given network there may be two axis of symmetry. Parallel axis of symmetry , that is, along the direction of current flow. Perpendicular axis of symmetry , that is, perpendicualr to the direction of flow of current.
For example, in the given network shown, the axis AA' is the parallel axis of symmetry , and the axis BB' is the perpendicualr axis of symmetry.
Points lying on the perpendicular axis of symmetry may have the same potential. In the given network, points 2, 0 , and 4 are at the same potential.
Points lying on the parallel axis of symmetry can never have the smae potential.
The network can be foalded about the parallel axis of symmetry. Thus as shown in the figur, the following points r at the same potential (5 and 6), (2,0,4) and (7,8).
62I have a different take on the question.. .(I hope this makes the problem far simpler for many of you.. )
62Great xyz. this part you have cracked.. [1]
what about the other part?
62Huh.. what are u saying..
the 2nd one is correct.. the first is wrong..
both are nto the same...
Think nicely..
