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JEE Main 2017 · Physics · Current Electricity · hard

An infinite ladder network is made of identical repeating sections, each having a 1 Ω resistor in series along the top rail followed by a 2 Ω…

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An infinite ladder network is made of identical repeating sections, each having a 1 Ω1\ \Omega resistor in series along the top rail followed by a 2 Ω2\ \Omega resistor connected across the two rails. A battery of emf 12 V12\ \mathrm{V} and negligible internal resistance is connected to the input end. The current through the first 2 Ω2\ \Omega cross resistor is
  1. A. 3 A3\ \mathrm{A}
  2. B. 2 A2\ \mathrm{A}
  3. C. 4 A4\ \mathrm{A}
  4. D. 6 A6\ \mathrm{A}
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Answer: (A) 3 A3\ \mathrm{A}

Let the resistance of the whole infinite ladder be xx. Removing the first section leaves an identical ladder, so x=1+2x2+xx = 1 + \dfrac{2x}{2+x}, which gives x2−x−2=0x^2 - x - 2 = 0 and x=2 Ωx = 2\ \Omega. The battery current is 12/2=6 A12/2 = 6\ \mathrm{A}. The voltage across the first cross resistor is 12−6×1=6 V12 - 6\times1 = 6\ \mathrm{V}, so the current in it is 6/2=3 A6/2 = 3\ \mathrm{A}. Check: the rest of the ladder is also 2 Ω2\ \Omega at 6 V6\ \mathrm{V}, taking 3 A3\ \mathrm{A}, and 3+3=6 A3 + 3 = 6\ \mathrm{A}. The value 6 A6\ \mathrm{A} is the total current.
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