All resistance in figure are 1 Ω each. The value of current I is a/5 A. The value of a is . [Figure: a 3 V cell sends current I through one 1 Ω resistor to a node. From that node the circuit branches as a binary tree: 2 resistors in parallel, each of whose ends branches into 2 more (4 resistors), e…

Question

All resistance in figure are 1 Ω each. The value of current I is a/5 A. The value of a is . [Figure: a 3 V cell sends current I through one 1 Ω resistor to a node. From that node the circuit branches as a binary tree: 2 resistors in parallel, each of whose ends branches into 2 more (4 resistors), each of whose ends branches into 2 more (8 resistors). The far ends of all 8 last resistors join at a single node wired back to the cell's negative terminal. Every resistor is 1 Ω.]

Accepted Answer

8. By symmetry every resistor in one level carries the same current, so each level acts as its resistors in parallel, and the levels are in series. Level 1: 1 Ω. Level 2: two branches in parallel, 0.5 Ω. Level 3: effective 0.25 Ω. Level 4: effective 0.125 Ω. R_eq=1+0.5+0.25+0.125=(15)/8 Ω. I=3/(15/8)=8/5 A. So a=8.

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