NEET Class neet Answered

What is resistance between point A and B

Asked by Prashant DIGHE | 22 Jul, 2019, 04:05: PM

Expert Answer

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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