All resistances in figure are 1 Ω each. The value of current 'I' is a/5 A. The value of a is _________.

Solution

Physics sir · 2 steps

It is a symmetric tree: collapse it from the far right, one level at a time, back to the battery.

1. Spot the pattern. Every node splits into two identical 1 Ω branches, and all eight right-hand ends meet at one point. So at each level the two branches are identical, and two equal resistances in parallel give half of one.

   Level (from right)	One branch	Two branches in parallel
   Last column	1 Ω	1/2 Ω
   Third column	1 + 1/2 = 3/2 Ω	3/4 Ω
   Second column	1 + 3/4 = 7/4 Ω	7/8 Ω
   First resistor	1 + 7/8 = 15/8 Ω	(no split: this is the total)

   I = V/R_(eq) = 3/(15/8) = 24/15 = 8/5 A

2. So. I = 8/5 A, and comparing with a/5 gives a = 8.

The common slip: 1.6. Reporting the current itself gives 1.6, not 8. The question asks for a, which is 5I.

The idea to keep. For a branching tree of equal resistors, start at the far end. Each level adds one resistor in series and then halves, because two identical branches sit in parallel.

Answer from the NTA final answer key. Solution by PrepWiser.