explain current distribution in the figure.
ABCDA is a uniform circular wire of resistance 2ohm. AOC and BOD are two wires along two perpendicular diameters of the circle, each having the same resistance 1ohm. A battery of emf E and internal resistance r is connected between points A and D. Calculate the equivalent resistance of the network.
This is the question for the above figure. please explain current distribution in the figure, thus providing a detailed solution.
Here battery sends a current I then its taken I1 from A to B so current in arm AO becomes I-I1. Then from B to C current is taken to be I3 and I2 from C to D. In arm OC current becomes I2 - I3 so that after coming out from C the sum of current in arm BC and OC becomes I2. Similar is the case with arm OD. Hope this helps.