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B
R(1+√3)2
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C
7R12
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D
R12
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Solution
The correct option is AR(1+√5)2 Let the effective resistance between A & B be RE.
Since the network is infinite long, so removal of one group of resistance from the chain will not change the network.
Therefore, effective resistance between points C & D would also be RE.
The equivalent network will be as shown below.
The original infinite chain is equivalent to R in series with R & RE in parallel. So the effective resistance of infinite network will be
RE=R+RRER+RE
⇒RER+R2E=R2+2RRE
⇒R2E−RRE−R2=0
∴RE=R±√R2−4⋅1⋅(−R2)2⋅1
⇒RE=R(1±√5)2
Taking positive value, because resistance can't be negative.
∴RE=R(1+√5)2
Hence, option (a) is correct.
Key Concept:
In an infinitely long network, addition or removal of one section from the chain still leaves the chain infinite (i.e. just about the same as before).