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What is resistance between point A and B
Asked by Prashant DIGHE | 22 Jul, 2019, 04:05: PM
Let us connect a battery with emf  E across points A and B as shown in figure.
Let i be the current drawn by the circuit  from battery.

At point A, main current i is divided into 5 branches as shown in figure.
By symmetry they are i1 , two divisions of current i2 and two divisions of current i3

By symmetry, at point B, currents are combining in same manner as they are getting divided at A.

At point D, combining currents i2 and i3  are getting divided into three currents i4 , i5 and i6 .

At point F currents are combining in same manner as they are getting divided at point D.

Based on this consideration, We can distribute the currents at each resistive elemnts in the circuit as shown in figure.

We have five unknown variables , i.e., i1 , i2 , i3 , i4 and i5 . Hence we need  5 equations using kirchoff's rules.
It is given that each resistive element is 1Ω .

By going through the closed path A-B-to Battery-A, we have,   2 i1 = E   or  i1 = E/2 ....................(1)

By going through the closed path, A-C-D-H-F-G-B-to Battery-A, we have , 2 i3 +2 i5 + 2 i3 = E  or  2 i3 + i5 = E ....................(2)

At node D, since sum of currents reaching the node is zero, we have, i2 + 2 i3 = i4 + i5 + i6   .........................(3)

For the closed loop A-C-D-A , we have,  2 i3 = i2 ...............................(4)

For the closed loop D-H-F-D, we have,  2 i5 = i4 ..............................(5)

By solving equations, (1) to (5), we get, i1 = E/2  , i2 = (4/11)E , i3 = (2/11)E , i4 = (3/11)E ,  i5 = (3/22)E and i6 = (3/22)E

at node A, we have,  i = i1 + 2 i2 + 2 i3 =  (E/2) + 2×(4/11)E  + 2×(2/11)E = (35/22)E ................(6)

Hence from eqn.(6), we get, effective resistance between A and B = E/i = (22/35)Ω

Answered by Thiyagarajan K | 23 Jul, 2019, 09:56: AM

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