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← JEE Main 2022 · 28 Jun · Shift 2
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JEE Main 2022 · 28 Jun · Shift 2, Physics question 55

All resistances in figure are 1 Ω\Omega each. The value of current 'I' is a5\dfrac a5 A. The value of a is ______.

A 3 V cell drives current I into a binary tree of 1 Ω resistors: one resistor, then two branches, each splitting into two, each splitting into two again (eight end resistors), all eight ends joined to one node that returns to the cell.
Answer and worked solution

Official answer: 8

Binary tree. Working from the far node: 1+12=321+\frac12=\frac32, then 1+34=741+\frac34=\frac74, then 1+78=1581+\frac78=\frac{15}8 Ω in total.

Current. 315/8=85\dfrac3{15/8}=\dfrac85 A, so a=8a=8.

Answer: 8