An electric network is formed by two hexagon and joining grid points with wires.The resistance of each side and the connecting wires is \$R\$. An ideal battery of emf \$E\$ is joined as shown in figure.The question is to find out the current passing through the battery.(Assuming that the wire connected with the cell is ideal).
I tried to find out the current distribution in the network.
Since the current cannot distinguish between the three branches hence it should divide equally. I have used the same symmetry arguments in other cases.However when i used kirchoff law in the outer loop i didnot get the correct answer.I couldnot understand where i have committed mistake
Here is my attempt after the answer below.
The simplified circuit in the answer can be redrawn as
which can also be manipulated as
The repeating unit in the above circuit is
and so the circuit simplifies to
which is similar to
But doing all this doesnot yields the correct result.Am I doing wrong somewhere.I tried to use star delta transformation but that complicated the circuit.Thanks.
Best Answer
I'm not going to do your homework for you, but maybe give you different ways of looking at this problem.
Note that this problem really comes down to is finding the apparent resistance across one of the resistors in the network.
These problems are often deliberately presented in a confusing geometry. The first thing you should do is write it down in a more accessible form:
All resistors are equal, which we simply call R. The real question comes down to what is the apparent resistance to a connection across R18. Hopefully you can see that this is a purely passive linear network, so the answer must be xR. Your job is to find x.
This is not a network that can be trivially simplified by inspection. There are various ways to attack it. One method is to solve the resistance of little pieces at a time, then combine them. Another is a numerical relaxation algorithm. It's not going to be simple. Symmetry will help somewhat, but not as much as you seem to expect.
Get crack'n.