NEET Class neet Answered
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Asked by Prashant DIGHE | 26 Jul, 2019, 09:36: AM
Expert Answer
As shown in fig.1, net current drawn from battery is getting divided at point E as i1 , i2 and i3 as shown in fig.1.
By symmetry, at point F, currents should combine in same manner.
Hence current i1 has to flow from E to B, then B to C and finally C to F without any further division.
Similarly current i2 has to flow E to A , then A to D and finally from D to F without any further division.
Hence there will not be any current flow in the resistor connected between A and B . Similarly, no current flow between D and C.
Hence after removing the resistors connected across A - B and C - D, we get the effective circuit as shown in fig.2.
Fig.3 shows the simplified form of fig.2 and we see that the effective resistance is parallel combination of three resistors 3R, R and R.
Hence effective resistance = 
if R is 5Ω, effective resistance is (3/5)×5 = 3Ω and current drawn from battery = 6V/3Ω = 2 A
we can see that, i3 = 6 V/ 5Ω = 1.2 A , hence i2 + i3 = 2.0 - 1.2 = 0.8 A
since current i2 and i3 are getting divided symmetrically, i2 = i3 = 0.4A
Hence current in the branch AD is 0.4 A
Answered by Thiyagarajan K | 26 Jul, 2019, 10:56: AM
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